NCERT Exemplar Class 12 Maths Solutions Chapter 10 Vector Algebra 2021

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This basic maths chapter covers the topics like vectors, properties, vector by scalar, types of vector etc.

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NCERT Exemplar Class 12 Maths Solutions Chapter 10 Vector Algebra

Edited By Ravindra Pindel | Updated on Sep 15, 2022 05:13 PM IST | #CBSE Class 12th

NCERT exemplar Class 12 Maths solutions chapter 10 Vector Algebra helps in answering questions like What is your weight? How should a footballer hit the ball to score a goal? The answers to both these questions would be very different. The answer to the former could be 50 kg, a quantity which has the presence of only one value. The second question's answer is quantities that consist of muscle strength, the direction of the foot and the ball, the air current, and the other players' distance. Such quantities are called Vectors and are discussed in detail in NCERT exemplar Class 12 Maths solutions chapter 10.

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Question:1

Find the unit vector in the direction of sum of vectors $\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}} \quad \text { and } \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$

Answer:

We have,

Since, unit vector is needed to be found in the direction of the sum of vectors and .
So, add vectors and .
Let,

Substituting the values of vectors and .
$\\ \Rightarrow \vec{c}=(2 \hat{\imath}-\hat{\jmath}+\hat{\mathrm{k}})+(2 \hat{\jmath}+\hat{\mathrm{k}}) \\ \Rightarrow \overrightarrow{\mathrm{c}}=2 \hat{\imath}-\hat{\jmath}+\hat{\mathrm{k}}+2 \hat{\jmath}+\hat{\mathrm{k}} \\ \Rightarrow \overrightarrow{\mathrm{c}}=2 \hat{\imath}-\hat{\jmath}+2 \hat{\jmath}+\hat{\mathrm{k}}+\hat{\mathrm{k}} \\ \Rightarrow \overrightarrow{\mathrm{c}}=2 \hat{\imath}+\hat{\jmath}+2 \hat{\mathrm{k}}$
We know that a unit vector in a normed vector space is a vector (often a spatial vector) of length 1.
To find a unit vector with the same direction as a given vector, we divide by the magnitude of the vector.
For finding unit vector, we have the formula:
$\\ \begin{aligned} &\hat{c}=\frac{\overrightarrow{\mathrm{c}}}{|\overrightarrow{\mathrm{c}}|}\\ &\text { Substitute the value of } \overrightarrow{\mathrm{c}} \text { . }\\ &\Rightarrow \hat{c}=\frac{2 \hat{\imath}+\hat{\jmath}+2 \hat{\mathrm{k}}}{|2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}|}\\ &\text { Here, }|2 \hat{\imath}+\hat{\jmath}+2 \hat{\mathrm{k}}|=\sqrt{2^{2}+1^{2}+2^{2}} \end{aligned}$
$\\ \Rightarrow \hat{c}=\frac{2 \hat{\imath}+\hat{\jmath}+2 \hat{\mathrm{k}}}{\sqrt{2^{2}+1^{2}+2^{2}}} \\ \Rightarrow \hat{\mathrm{c}}=\frac{2 \hat{\imath}+\hat{\jmath}+2 \hat{\mathrm{k}}}{\sqrt{4+1+4}} \\ \Rightarrow \hat{c}=\frac{2 \hat{\imath}+\hat{\jmath}+2 \hat{\mathrm{k}}}{\sqrt{9}} \\ \Rightarrow \hat{c}=\frac{2 \hat{\imath}+\hat{\jmath}+2 \hat{\mathrm{k}}}{3}$
Thus, unit vector in the direction of sum of vectors $\vec{a}{\text { and }} \vec{b}$ is $\frac{2 \hat{i}+\hat{j}+2 \hat{k}}{3}$ .

Question:2(i)

If $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}} \text { and } \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$ find the unit vector in the direction of $6 \overrightarrow{\mathrm{b}}$

Answer:

We have, $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}} \text { and } \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$
(i). We need to find the unit vector in the direction of $6 \overrightarrow{\mathrm{b}}$ .
First, let us calculate $6 \overrightarrow{\mathrm{b}}$ .
As we have,
$\\ $\overrightarrow{\mathrm{b}}=2 \hat{\imath}+\hat{\jmath}-2 \hat{\mathrm{k}}$$ $\\ \text{Multiply it by 6 on both sides.} $$\\ \Rightarrow 6 \overrightarrow{\mathrm{b}}=6(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}) $$$ We can easily multiply vector by a scalar by multiplying similar components, that is, vector’s magnitude by the scalar’s magnitude.
$\Rightarrow 6 \overrightarrow{\mathrm{b}}=12 \hat{\imath}+6 \hat{\jmath}-12 \hat{\mathrm{k}}$
We know that, a unit vector in a normed vector space is a vector (often a spatial vector) of length 1.
To find a unit vector with the same direction as a given vector, we divide by the magnitude of the vector.
For finding unit vector, we have the formula:
$6 \hat{\mathrm{b}}=\frac{6 \overrightarrow{\mathrm{b}}}{|6 \overrightarrow{\mathrm{b}}|}$
Now we know the value of $6 \overrightarrow{\mathrm{b}}$ , so just substitute the value in the above equation.
$\Rightarrow 6 \hat{\mathrm{b}}=\frac{12 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-12 \hat{\mathrm{k}}}{|12 \hat{\mathrm{l}}+6 \hat{\mathrm{j}}-12 \hat{\mathrm{k}}|} \\ \text { Here, }|12 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-12 \hat{\mathrm{k}}|=\sqrt{12^{2}+6^{2}+(-12)^{2}} \\ \Rightarrow 6 \hat{\mathrm{b}}=\frac{12 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-12 \hat{\mathrm{k}}}{\sqrt{144+36+144}} \\ \Rightarrow 6 \hat{\mathrm{b}}=\frac{12 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-12 \hat{\mathrm{k}}}{\sqrt{324}}$
$\\ \begin{aligned} &\Rightarrow 6 \hat{\mathrm{b}}=\frac{12 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-12 \hat{\mathrm{k}}}{18}\\ &\text { Let us simplify. }\\ &\Rightarrow 6 \hat{\mathrm{b}}=\frac{6(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})}{18}\\ &\Rightarrow 6 \hat{\mathrm{b}}=\frac{2 \hat{\mathrm{\imath}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}}{3} \end{aligned}$
Thus, unit vector in the direction of $6 \overrightarrow{\mathrm{b}}$ is $\frac{2 \hat{\mathrm{\imath}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}}{3}$

Question:2(ii)

If $a = \hat { i } + \hat { j } + 2 \hat { k } \text { and } b = 2 \hat { i } + \hat { j } - 2 \hat { k } \\$ find the unit vector in the direction of $2 \vec { a } - \vec { b }$

Answer:

We need to find the unit vector in the direction of $2 \vec { a } - \vec { b }$
First, let us calculate . $2 \vec { a } - \vec { b }$
As we have,
$\begin{array}{l} \overrightarrow{\mathrm{a}}=\hat{1}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}_{\ldots}(\mathrm{a}) \\ \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{\imath}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}_{\ldots(\mathrm{b})} \end{array}$
Then multiply equation (a) by 2 on both sides,
$2 \vec { a } = 2 ( \hat { \imath } + \hat { \jmath } + 2 \hat { k } )$
We can easily multiply vector by a scalar by multiplying similar components, that is, vector’s magnitude by the scalar’s magnitude.
$\Rightarrow 2 \vec { a } = 2 \hat { \imath } + 2 \hat { \jmath } + 4 \hat { k } $ \ldots $ (c)\\$
Subtract (b) from (c). We get
$\Rightarrow 2 \vec { a } - \vec { b } = 2 \hat { l } - 2 \hat { l } + 2 \hat { j } - \hat { j } + 4 \hat { k } + 2 \hat { k } \\$
$\Rightarrow 2 \vec{a}-\vec{b}=\hat{\jmath}+6 \hat{k}$
We know that, a unit vector in a normed vector space is a vector (often a spatial vector) of length 1.
To find a unit vector with the same direction as a given vector, we divide by the magnitude of the vector.
For finding unit vector, we have the formula:
$\\ 2 \hat{a}-\hat{b}=\frac{2 \vec{a}-\vec{b}}{|2 \vec{a}-\vec{b}|}$ \\Now we know the value of $2 \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}},\\$ so we just need to substitute in the above equation.\\ $\Rightarrow 2 \hat{a}-\hat{b}=\frac{\hat{\jmath}+6 \hat{k}}{|\hat{\jmath}+6 \hat{k}|}$\\ Here, $|\hat{\jmath}+6 \hat{\mathrm{k}}|=\sqrt{1^{2}+6^{2}}$ \\$\Rightarrow 2 \hat{a}-\hat{b}=\frac{\hat{\jmath}+6 \hat{k}}{\sqrt{1^{2}+6^{2}}}$$ $\begin{aligned} &\Rightarrow 2 \hat{a}-\hat{b}=\frac{\hat{\jmath}+6 \hat{\mathrm{k}}}{\sqrt{1+36}}\\ &\Rightarrow 2 \hat{a}-\hat{b}=\frac{\hat{\jmath}+6 \hat{k}}{\sqrt{37}}\\ &\text { Thus, unit vector in the direction of }\\ &2 \vec{a}-\vec{b}_{i s} \frac{\hat{\jmath}+6 \hat{k}}{\sqrt{37}} \end{aligned}$

Question:3

Find a unit vector in the direction of $\bar{PQ}$ , where P and Q have co-ordinates (5, 0, 8) and (3, 3, 2), respectively.

Answer:

We have,
Coordinates of P is (5, 0, 8).
Coordinates of Q is (3, 3, 2).
So,
Position vector of P is given by,
$\\ \begin{aligned} &\overrightarrow{\mathrm{OP}}=5 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}+8 \hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OP}}=5 \hat{\mathrm{i}}+8 \hat{\mathrm{k}}\\ &\text { Position vector of } Q \text { is given by, }\\ &\overrightarrow{\mathrm{OQ}}=3 \hat{\imath}+3 \hat{\jmath}+2 \hat{\mathrm{k}} \end{aligned}$
To find unit vector in the direction of PQ, we need to find position vector of PQ.
Position vector of PQ is given by,
$\\ \overrightarrow{\mathrm{PQ}}=\text { Position vector of } \mathrm{Q}-\text { Position vector of } \mathrm{P} \\ \Rightarrow \overrightarrow{\mathrm{PQ}}=\overrightarrow{\mathrm{OQ}}-\overrightarrow{\mathrm{OP}} \\ \Rightarrow \overrightarrow{\mathrm{PQ}}=(3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})-(5 \hat{\mathrm{l}}+8 \hat{\mathrm{k}}) \\ \Rightarrow \overrightarrow{\mathrm{PQ}}=3 \hat{\mathrm{i}}-5 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}-8 \hat{\mathrm{k}}$
$\Rightarrow \overrightarrow{\mathrm{PQ}}=-2 \hat{\imath}+3 \hat{\jmath}-6 \hat{\mathrm{k}}$
We know that a unit vector in a normed vector space is a vector (often a spatial vector) of length 1.
To find a unit vector with the same direction as a given vector, we divide by the magnitude of the vector.
For finding unit vector, we have the formula:

$\\ \widehat{\mathrm{PQ}}=\frac{\overrightarrow{\mathrm{PQ}}}{|\overrightarrow{\mathrm{PQ}}|} \\ \Rightarrow \widehat{\mathrm{PQ}}=\frac{-2 \hat{\imath}+3 \hat{\jmath}-6 \hat{\mathrm{k}}}{|-2 \hat{\imath}+3 \hat{\jmath}-6 \hat{\mathrm{k}}|} \\ \text { Here, }|-2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}|=\sqrt{(-2)^{2}+3^{2}+(-6)^{2}} \\ \Rightarrow \widehat{\mathrm{PQ}}=\frac{-2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}}{\sqrt{(-2)^{2}+3^{2}+(-6)^{2}}}$
$\\ \Rightarrow \widehat{\mathrm{PQ}}=\frac{-2 \hat{\imath}+3 \hat{\jmath}-6 \hat{\mathrm{k}}}{\sqrt{4+9+36}} \\ \Rightarrow \widehat{\mathrm{PQ}}=\frac{-2 \hat{\mathrm{l}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}}{\sqrt{49}} \\ \Rightarrow \widehat{\mathrm{PQ}}=\frac{-2 \hat{\mathrm{l}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}}{7}$
Thus, unit vector in the direction of PQ is $\frac{-2 \hat{\mathrm{l}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}}{7}$

Question:4

If $\vec{a}$ and $\vec{b}$ are the position vectors of A and B, respectively, find the position vector of a point C in BA produced such that BC = 1.5 BA.

Answer:

We have been given that,
Position vector of A $=\vec{a}$
$\begin{aligned} &\Rightarrow \overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{a}}\\ &\text { Position vector of } B=\overrightarrow{\mathrm{b}} \end{aligned}$ $\\ $$ \Rightarrow \overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{b}} $$\\$Here, O is the origin.\\ \text{We need to find the position vector of } $\mathrm{C},\\$ \text{that is}, $\overrightarrow{\mathrm{OC}}$\\ \text{Also, we have}$ $\\$\overrightarrow{\mathrm{BC}}=1.5 \overrightarrow{\mathrm{BA}}_{\ldots .(\mathrm{i})}$\\ Here, $\overrightarrow{\mathrm{BC}}=$ $Position vector of $\mathrm{C}$ -$ $Position vector of $\mathrm{B}$\\ $\Rightarrow \overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}}$...(2)$ $\\ And, $\overrightarrow{\mathrm{BA}}=$ $ Position vector of A-Position vector of $\mathrm{B}$\\ $\Rightarrow \overrightarrow{\mathrm{BA}}=\overrightarrow{\mathrm{OA}}-\overrightarrow{\mathrm{OB}}_{\ldots(\mathrm{iii})}$$ $\\ \overrightarrow{\mathrm{BC}}=1.5 \overrightarrow{\mathrm{BA}} \\ \Rightarrow \overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}}=1.5(\overrightarrow{\mathrm{OA}}-\overrightarrow{\mathrm{OB}}) \\ \Rightarrow \overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}}=1.5 \overrightarrow{\mathrm{OA}}-1.5 \overrightarrow{\mathrm{OB}} \\ \Rightarrow \overrightarrow{\mathrm{OC}}=1.5 \overrightarrow{\mathrm{OA}}-1.5 \overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{OB}}$
$\\ \begin{aligned} &\Rightarrow \overrightarrow{\mathrm{OC}}=1.5 \overrightarrow{\mathrm{OA}}-0.5 \overrightarrow{\mathrm{OB}}\\ &\Rightarrow \overrightarrow{\mathrm{OC}}=1.5 \overrightarrow{\mathrm{a}}-0.5 \overrightarrow{\mathrm{b}}\\ &[\because \text { it is given that } \overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{a}} \text { and } \overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{b}}]\\ &\Rightarrow \overrightarrow{\mathrm{OC}}=\frac{15 \overrightarrow{\mathrm{a}}}{10}-\frac{5 \overrightarrow{\mathrm{b}}}{10} \end{aligned}$
$\\ \Rightarrow \overrightarrow{\mathrm{OC}}=\frac{3 \overrightarrow{\mathrm{a}}}{2}-\frac{\overrightarrow{\mathrm{b}}}{2} \\ \Rightarrow \overrightarrow{\mathrm{OC}}=\frac{3 \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}}{2}$
Thus, position vector of point C is $=\frac{3 \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}}{2}$

Question:5

Using vectors, find the value of k such that the points (k, – 10, 3), (1, –1, 3) and (3, 5, 3) are collinear.

Answer:

Let the points be A (k, -10, 3), B (1, -1, 3) and C (3, 5, 3).
Let us find the position vectors of these points.
Assume that O is the origin.
Position vector of A is given by,
$\\ \overrightarrow{\mathrm{OA}}=\mathrm{ki}-10 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\ $Position vector of $B$ is given by, \\ \overrightarrow{\mathrm{OB}}=\hat{\imath}-\hat{\jmath}+3 \hat{\mathrm{k}}\\ \text{Position vector of } $\mathrm{C}$ is given by,$ $\vec { OC } = 3 \hat { l } + 5 \hat { j } + 3 \hat { k }$
Know that, two vectors are said to be collinear, if they lie on the same line or parallel lines.
Since, A (k, -10, 3), B (1, -1, 3) and C (3, 5, 3) are collinear, we can say that:
Sum of modulus of any two vectors will be equal to the modulus of third vector.
This means, we need to find $| \vec { AB } |$
To find : $| \vec { AB } |$
Position vector of B-Position vector of A
$\Rightarrow \vec { AB } = \vec { OB } - \vec { OA }$
$\\ \begin{aligned} &\Rightarrow \overrightarrow{\mathrm{AB}}=(\hat{\imath}-\hat{\jmath}+3 \hat{\mathrm{k}})-(\mathrm{k} \hat{\imath}-10 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})\\ &\Rightarrow \overrightarrow{\mathrm{AB}}=\hat{\imath}-k \hat{\imath}-\hat{\jmath}+10 \hat{\jmath}+3 \hat{\mathrm{k}}-3 \hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{AB}}=(1-\mathrm{k}) \hat{\imath}+9 \hat{\mathrm{j}}\\ &\text { Now, }\\ &|\overrightarrow{\mathrm{AB}}|=\sqrt{(1-\mathrm{k})^{2}+9^{2}} \end{aligned}$
$\\ \begin{aligned} &\Rightarrow|\overrightarrow{\mathrm{AB}}|=\sqrt{(1-\mathrm{k})^{2}+81}_{\ldots(\mathrm{i})}\\ &\text { To find }|\overrightarrow{\mathrm{BC}}|_{:}\\ &\overrightarrow{\mathrm{BC}}=\text { position vector of C-Position vector of } \mathrm{B}\\ &\Rightarrow \overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}} \end{aligned}$
$\\ \begin{aligned} &\Rightarrow \overrightarrow{\mathrm{BC}}=(3 \hat{\imath}+5 \hat{\jmath}+3 \hat{\mathrm{k}})-(\hat{\imath}-\hat{\jmath}+3 \hat{\mathrm{k}})\\ &\Rightarrow \overrightarrow{\mathrm{BC}}=3 \hat{\imath}-\hat{\imath}+5 \hat{\jmath}+\hat{\jmath}+3 \hat{\mathrm{k}}-3 \hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{BC}}=2 \hat{\imath}+6 \hat{j}\\ &\text { Now, }\\ &|\overrightarrow{\mathrm{BC}}|=\sqrt{2^{2}+6^{2}} \end{aligned}$
$\\ \begin{aligned} &\Rightarrow|\overrightarrow{\mathrm{BC}}|=\sqrt{4+36}\\ &\Rightarrow|\overrightarrow{\mathrm{BC}}|=\sqrt{40}\\ &\text { To find }\\ &|\overrightarrow{\mathrm{AC}}| \end{aligned}$
$\overrightarrow{\mathrm{AC}}$ = Position vector of C-Position vector of A
$\\ \Rightarrow \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}} \\ \Rightarrow \overrightarrow{\mathrm{AC}}=(3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})-(\mathrm{ki}-10 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \\ \Rightarrow \overrightarrow{\mathrm{AC}}=3 \hat{\mathrm{i}}-\mathrm{k} \hat{\mathrm{I}}+5 \hat{\mathrm{j}}+10 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}-3 \hat{\mathrm{k}} \\ \Rightarrow \overrightarrow{\mathrm{AC}}=(3-\mathrm{k}) \hat{\mathrm{i}}+15 \hat{\mathrm{j}}$
$\\ \begin{aligned} &\text { Now, }\\ &|\overrightarrow{\mathrm{AC}}|=\sqrt{(3-\mathrm{k})^{2}+15^{2}}\\ &\Rightarrow|\overrightarrow{\mathrm{AC}}|=\sqrt{(3-\mathrm{k})^{2}+225}...(iii) \end{aligned}$
$\\ Take,\\ $$ |\overrightarrow{\mathrm{AB}}|+|\overrightarrow{\mathrm{BC}}|=|\overrightarrow{\mathrm{AC}}|\\ $$$ Substitute values of $|\overrightarrow{\mathrm{AB}}|,|\overrightarrow{\mathrm{BC}}| \text { and }|\overrightarrow{\mathrm{AC}}|$ from (i), (ii) and (iii) respectively. We get, $$$ \sqrt{(1-\mathrm{k})^{2}+81}+\sqrt{40}=\sqrt{(3-\mathrm{k})^{2}+225} $$\\$ $\\ Or\\ $\Rightarrow \sqrt{(3-\mathrm{k})^{2}+225}-\sqrt{40}=\sqrt{(1-\mathrm{k})^{2}+81}$\\ $\Rightarrow \sqrt{\left(9+\mathrm{k}^{2}-6 \mathrm{k}\right)+225}-\sqrt{40}=\sqrt{\left(1+\mathrm{k}^{2}-2 \mathrm{k}\right)+81}$\\ $\left[\because\right.$ \text{by algebraic identity,} $\left.(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$\\$ $\\$\Rightarrow \sqrt{\mathrm{k}^{2}-6 \mathrm{k}+225+9}-\sqrt{40}=\sqrt{\mathrm{k}^{2}-2 \mathrm{k}+81+1}$ $$\\ \Rightarrow \sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234}-\sqrt{40}=\sqrt{\mathrm{k}^{2}-2 \mathrm{k}+82} $$\\ $Squaring on both sides, $$ \\\Rightarrow\left[\sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234}-\sqrt{40}\right]^{2}=\left[\sqrt{\mathrm{k}^{2}-2 \mathrm{k}+82}\right]^{2} $$$ $\\ \begin{aligned} &\Rightarrow\left[\sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234}\right]^{2}+[\sqrt{40}]^{2}-2\left[\sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234}\right][\sqrt{40}]=\mathrm{k}^{2}-2 \mathrm{k}+82\\ &\left[\because \text { by algebraic identity, }(a-b)^{2}=a^{2}+b^{2}-2 a b\right]\\ &\Rightarrow \mathrm{k}^{2}-6 \mathrm{k}+234+40-2\left[\sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234}\right][\sqrt{40}]=\mathrm{k}^{2}-2 \mathrm{k}+82\\ &\Rightarrow \mathrm{k}^{2}-6 \mathrm{k}+234+40-\mathrm{k}^{2}+2 \mathrm{k}-82=2\left[\sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234}\right][\sqrt{40}] \end{aligned}$
$\\ \Rightarrow \mathrm{k}^{2}-\mathrm{k}^{2}-6 \mathrm{k}+2 \mathrm{k}+234+40-82=2\left[\sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234}\right][\sqrt{40}] \\ \Rightarrow-4 \mathrm{k}+192=2\left[\sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234}\right][\sqrt{40}] \\ \Rightarrow-4 \mathrm{k}+192=2\left[\sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234}\right] .2[\sqrt{10}] \\ \Rightarrow 4(-\mathrm{k}+48)=4\left[\sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234}\right][\sqrt{10}]$
$\\ \begin{aligned} &\Rightarrow-\mathrm{k}+48=\sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234} \cdot \sqrt{10}\\ &\text { Again, squaring on both sides, we get }\\ &[48-\mathrm{k}]^{2}=\left[\sqrt{\mathrm{k}^{2}-6 \mathrm{k}+234} \cdot \sqrt{10}\right]^{2}\\ &\Rightarrow(48)^{2}+k^{2}-2(48)(k)=\left(k^{2}-6 k+234\right)(10)\left[\because \text { by algebraic identity },(a-b)^{2}=a^{2}+b^{2}-2 a b\right] \end{aligned}$
$\\ \begin{aligned} &\Rightarrow 2304+\mathrm{k}^{2}-96 \mathrm{k}=10 \mathrm{k}^{2}-60 \mathrm{k}+2340\\ &\Rightarrow 10 k^{2}-k^{2}-60 k+96 k+2340-2304=0\\ &\Rightarrow 9 \mathrm{k}^{2}+36 \mathrm{k}+36=0\\ &\Rightarrow 9\left(\mathrm{k}^{2}+4 \mathrm{k}+4\right)=0\\ &\Rightarrow \mathrm{k}^{2}+4 \mathrm{k}+4=0 \end{aligned}$
$\\ \Rightarrow \mathrm{k}^{2}+2 \mathrm{k}+2 \mathrm{k}+4=0 \\ \Rightarrow \mathrm{k}(\mathrm{k}+2)+2(\mathrm{k}+2)=0 \\ \Rightarrow(\mathrm{k}+2)(\mathrm{k}+2)=0 \\ \Rightarrow \mathrm{k}=-2 \text { or } \mathrm{k}=-2$
Thus, value of k is -2.

Question:6

A vector $\overrightarrow{\mathrm{r}}$ is inclined at equal angles to the three axes. If the magnitude of $\overrightarrow{\mathrm{r}}$ is $2\sqrt3$ units, find $\overrightarrow{\mathrm{r}}$ .

Answer:

Given that,
Magnitude of $\overrightarrow{\mathrm{r}}$ = $2\sqrt3$
$\Rightarrow|\vec{r}|=2 \sqrt{3}$
Also, given that
Vector $\overrightarrow{\mathrm{r}}$ is equally inclined to the three axes.
This means, direction cosines of the unit vector $\overrightarrow{\mathrm{r}}$ will be same. The direction cosines are (l, m, n).
$\mathrm{l}=\mathrm{m}=\mathrm{n}$
The direction cosines of a vector are simply the cosines of the angles between the vector and the three coordinate axes.
We know the relationship between direction cosines is,
$\\ l^{2}+\mathrm{m}^{2}+\mathrm{n}^{2}=1 \\ \Rightarrow l^{2}+l^{2}+l^{2}=1[\because \mathrm{l}=\mathrm{m}=\mathrm{n}] \\ \Rightarrow 3.l^{2}=1 \\ \Rightarrow l=\pm \frac{1}{\sqrt{3}}$
Also, we know that $\overrightarrow{\mathrm{r}}$ is represented in terms of direction cosines as,
$\\ \hat{\mathrm{r}}=l \hat{\mathrm{u}}+\mathrm{m} \hat{\mathrm{j}}+\mathrm{n} \hat{\mathrm{k}}\\ \Rightarrow \hat{\mathrm{r}}=\pm \frac{1}{\sqrt{3}} \hat{\imath} \pm \frac{1}{\sqrt{3}} \hat{\jmath} \pm \frac{1}{\sqrt{3}} \hat{\mathrm{k}}$
$\\ \text{We are familiar with the formula,} $$ \hat{\mathrm{r}}=\frac{\overrightarrow{\mathrm{r}}}{|\overrightarrow{\mathrm{r}}|} $$$ $\\ \hat{\mathrm{r}}=\frac{\overrightarrow{\mathrm{r}}}{|\overrightarrow{\mathrm{r}}|} \\\text{To find } \overrightarrow{\mathrm{r}} \\ \Rightarrow \overrightarrow{\mathrm{r}}=\hat{\mathrm{r}}|\overrightarrow{\mathrm{r}}| \\ $Substituting values of$ |\overrightarrow{\mathrm{r}}| and \hat{\mathrm{r}}$
$\\ \begin{aligned} &\overrightarrow{\mathrm{r}}=\left(\pm \frac{1}{\sqrt{3}} \hat{\imath} \pm \frac{1}{\sqrt{3}} \hat{\jmath} \pm \frac{1}{\sqrt{3}} \hat{\mathrm{k}}\right)(2 \sqrt{3})\\ &\Rightarrow \overrightarrow{\mathrm{r}}=\pm \frac{1}{\sqrt{3}}(\hat{\imath}+\hat{\jmath}+\hat{\mathrm{k}})(2 \sqrt{3})\\ &\Rightarrow \overrightarrow{\mathrm{r}}=\pm 2(\hat{\imath}+\hat{\jmath}+\hat{\mathrm{k}})\\ &\text { Thus, the value of } \overrightarrow{\mathrm{r}}_{\text {is }} \pm 2(\hat{\imath}+\hat{\jmath}+\hat{\mathrm{k}}) \end{aligned}$

Question:7

A vector $\overrightarrow{\mathrm{r}}$ has magnitude 14 and direction ratios 2, 3, –6. Find the direction cosines and components of $\overrightarrow{\mathrm{r}}$ , given that $\overrightarrow{\mathrm{r}}$ makes an acute angle with x-axis.

Answer:

Given that,
Magnitude of vector $\overrightarrow{\mathrm{r}}$ = 14
$\Rightarrow|\overrightarrow{\mathrm{r}}|=14$
Also, direction ratios = 2 : 3 : -6
$\\ \begin{array}{l} \overrightarrow{\mathrm{a}}=2 \mathrm{k} \\ \overrightarrow{\mathrm{b}}=3 \mathrm{k} \end{array} \\ \overrightarrow{\mathrm{c}}=-6 \mathrm{k}\\$
Also $\overrightarrow{\mathrm{r}}$ can be defined as, $\overrightarrow{\mathrm{r}}=a \hat{\imath}+\mathrm{b} \hat{\jmath}+c \hat{\mathrm{k}}$
Know that, the direction cosines of a vector are the cosines of the angles between the vector and the three coordinate axes.

∴, the direction cosines l, m and n are
$\\ \mathrm{l}=\frac{\overrightarrow{\mathrm{a}}}{|\overrightarrow{\mathrm{r}}|} \\ \Rightarrow l=\frac{2 \mathrm{k}}{14}[\because]=2 \mathrm{k} \text { and }|\overrightarrow{\mathrm{r}}|=14_{\mathrm{J}} \\ \Rightarrow l=\frac{\mathrm{k}}{7} \\ \mathrm{~m}=\frac{\overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{r}}|}$
$\\ \Rightarrow \mathrm{m}=\frac{3 \mathrm{k}}{14}[\because \overrightarrow{\mathrm{b}}=3 \mathrm{k} \text { and }|\overrightarrow{\mathrm{r}}|=14 \mathrm{~g} \\ \mathrm{n}=\frac{\overrightarrow{\mathrm{c}}}{|\overrightarrow{\mathrm{r}}|} \\ \Rightarrow \mathrm{n}=-\frac{6 \mathrm{k}}{14}[\because]{\mathrm{c}}=-6 \mathrm{k} \text { and }|\overrightarrow{\mathrm{r}}|=14]$
$\\ \begin{aligned} &\Rightarrow \mathrm{n}=-\frac{3 \mathrm{k}}{7}\\ &\text { And we know that, }\\ &l^{2}+m^{2}+n^{2}=1\\ &\Rightarrow\left(\frac{\mathrm{k}}{7}\right)^{2}+\left(\frac{3 \mathrm{k}}{14}\right)^{2}+\left(-\frac{3 \mathrm{k}}{7}\right)^{2}=1 \end{aligned}$
$\\ \Rightarrow \frac{\mathrm{k}^{2}}{49}+\frac{9 \mathrm{k}^{2}}{196}+\frac{9 \mathrm{k}^{2}}{49}=1 \\ \Rightarrow \frac{4 \mathrm{k}^{2}+9 \mathrm{k}^{2}+36 \mathrm{k}^{2}}{196}=1 \\ \Rightarrow \frac{49 \mathrm{k}^{2}}{196}=1 \\ \Rightarrow 49 \mathrm{k}^{2}=196$
$\\ \Rightarrow \mathrm{k}^{2}=\frac{196}{49} \\ \Rightarrow \mathrm{k}^{2}=4 \\ \Rightarrow \mathrm{k}=\pm \sqrt{4} \\ \Rightarrow \mathrm{k}=\pm 2$
Since, $\overrightarrow{\mathrm{r}}$ makes an acute angle with x-axis, then k will be positive.
$\\ \begin{aligned} &\Rightarrow \mathrm{k}=2\\ &\text { The direction cosines are }\\ &l=\frac{\mathrm{k}}{7}=\frac{2}{7} \end{aligned}$
$\\ \begin{aligned} &\mathrm{m}=\frac{3 \mathrm{k}}{14}=\frac{3 \times 2}{14}=\frac{3}{7}\\ &\mathrm{n}=-\frac{3 \mathrm{k}}{7}=-\frac{3 \times 2}{7}=-\frac{6}{7}\\ &\text { The components of } \overrightarrow{\mathrm{r}} \text { can be found out by, }\\ &\overrightarrow{\mathrm{r}}=\hat{\mathrm{r}} \cdot|\overrightarrow{\mathrm{r}}| \end{aligned}$
$\\ \Rightarrow \overrightarrow{\mathrm{r}}=(l \hat{\imath}+\mathrm{m} \hat{\mathrm{j}}+\mathrm{n} \hat{\mathrm{k}})(14) \\ \Rightarrow \overrightarrow{\mathrm{r}}=\left(\frac{2}{7} \hat{\mathrm{r}}+\frac{3}{7} \hat{\mathrm{j}}-\frac{6}{7} \hat{\mathrm{k}}\right)(14) \\ \Rightarrow \overrightarrow{\mathrm{r}}=14 \times \frac{2}{7} \hat{\mathrm{i}}+14 \times \frac{3}{7} \hat{\mathrm{j}}-14 \times \frac{6}{7} \hat{\mathrm{k}} \\ \Rightarrow \overrightarrow{\mathrm{r}}=4 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-12 \hat{\mathrm{k}}$
Thus, the direction cosines (l, m, n) are $\left(\frac{2}{7}, \frac{3}{7},-\frac{6}{7}\right)$ ; and the components of $\overrightarrow{\mathrm{r}}$ are (4,6,-12)

Question:8

Find a vector of magnitude 6, which is perpendicular to both the vectors $2 \hat{i}-\hat{j}+2 \hat{k}$ and $4\hat{i}-\hat{j}+3 \hat{k}$

Answer:

Let the vectors be $\\ \vec{a}$ and $\\ \vec{b}$ , such that
$\\ \vec{a}=2 \hat{\imath}-\hat{\jmath}+2 \hat{k} \\ \vec{b}=4 \hat{\imath}-\hat{\jmath}+3 \hat{k}$
We need to find a vector perpendicular to both the vectors $\\ \vec{a}$ and $\\ \vec{b}$

Any vector perpendicular to both $\\ \vec{a}$ and $\\ \vec{b}$ can be given as,
$\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & -1 & 2 \\ 4 & -1 & 3 \end{array}\right| \\ \begin{aligned} \Rightarrow \vec{a} \times \vec{b}=\hat{i}((-1)(3)-(2)(-1))-\hat{j}((2)(3)-(2)(4)) \\ +\hat{k}((2)(-1)-(-1)(4)) \end{aligned} \\ \Rightarrow \vec{a} \times \overrightarrow{\mathrm{b}}=\hat{\imath}(-3+2)-\hat{\jmath}(6-8)+\hat{\mathrm{k}}(-2+4) \\ \Rightarrow \vec{a} \times \overrightarrow{\mathrm{b}}=-\hat{\imath}+2 \hat{\jmath}+2 \hat{\mathrm{k}}$
Let
$\\ \overrightarrow{\mathrm{r}}=-\hat{\mathrm{l}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$

As we know, a unit vector in a normed vector space is a vector (often a spatial vector) of length 1.
So,
$\\ $A vector of magnitude 6 in the direction of$ \mathrm{r}$ is given by, vector $=6 \times \frac{\overrightarrow{\mathrm{r}}}{|\overrightarrow{\mathrm{r}}|}$ \\$\Rightarrow$ vector $=6 \times \frac{-\hat{\imath}+2 \hat{\jmath}+2 \hat{\mathrm{k}}}{|-\hat{\mathrm{l}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}|}$ \\Here, $|-\hat{\imath}+2 \hat{\jmath}+2 \hat{k}|=\sqrt{(-1)^{2}+(2)^{2}+(2)^{2}}$$ $\\ \Rightarrow \text { vector }=6 \times \frac{-\hat{1}+2 \hat{\jmath}+2 \hat{\mathrm{k}}}{\sqrt{(-1)^{2}+(2)^{2}+(2)^{2}}} \\ \Rightarrow \text { vector }=6 \times \frac{(-\hat{\imath}+2 \hat{\jmath}+2 \hat{\mathrm{k}})}{\sqrt{1+4+4}} \\ \Rightarrow \text { vector }=6 \times \frac{-\hat{\mathrm{i}}+2 \hat{\jmath}+2 \hat{\mathrm{k}}}{\sqrt{9}}$
$\\ \begin{aligned} &\Rightarrow \text { vector }=6 \times \frac{-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}}{3}\\ &\Rightarrow \text { vector }=2 \times(-\hat{1}+2 \hat{\jmath}+2 \hat{\mathrm{k}})\\ &\Rightarrow \text { vector }=-2 \hat{\imath}+4 \hat{\jmath}+4 \hat{k}\\ &\text { Thus, required vector is }-2 \hat{\imath}+4 \hat{\jmath}+4 \hat{k} \end{aligned}$

Question:9

Find the angle between the vectors $2 \hat{i}-\hat{j}+\hat{k}$ and $3 \hat{i}+4 \hat{j}-\hat{k}$ .

Answer:

Given:
$\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{b} =3 \hat{i}+4 \hat{j}-\hat{k}$
Assume $\theta$ is angle between $\vec{a}$ and $\vec{b}$ .
$\\ \cos \theta =\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \\ \\ =\frac{(2 \hat{i}-\hat{j}+\hat{k}) \cdot(3 \hat{i}+4 \hat{j}-\hat{k})}{\sqrt{4+1+1} \sqrt{9+16+1}} \\\\ =\frac{6-4-1}{\sqrt{6} \sqrt{26}}=\frac{1}{2 \sqrt{39}} \\\\ \theta =\cos ^{-1} \frac{1}{2 \sqrt{39}}$

Question:10

If $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=0, \text { show that } \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}$ . Interpret the result geometrically?

Answer:

Given that,
$\begin{aligned} &\vec{a}+\vec{b}+\vec{c}=0\\ &\text { Find the value of } \overrightarrow{\mathrm{b}}\\ &\Rightarrow \overrightarrow{\mathrm{b}}=-\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{c}} \end{aligned}$
$\\ \text { Take } \vec{a} \times \vec{b} \\ \vec{a} \times \vec{b}=\vec{a} \times(-\vec{a}-\vec{c}) \\ \Rightarrow \vec{a} \times \vec{b}=(-\vec{a} \times \vec{a})-\vec{a} \times \vec{c} \\ \Rightarrow \vec{a} \times \vec{b}=0-\vec{a} \times \vec{c}$
$\\ {[\because \vec{a} \times \vec{a}=0]} \\ \Rightarrow \vec{a} \times \vec{b}=-\vec{a} \times \vec{c} \\ \Rightarrow \vec{a} \times \vec{b}=\vec{c} \times \vec{a}_{\ldots(i)}$
$\\ \text { [\% by anti-commutative law, }-\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}_{1} \\ \text { Now, take } \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}} \text { . } \\ \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=(-\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{c}}) \times \overrightarrow{\mathrm{c}} \\ \Rightarrow \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=-\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{c}}$
$\\ \qquad\left[\because \overrightarrow{\mathrm{C}} \times \overrightarrow{\mathrm{c}}=0\right]. \\ \Rightarrow \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=-\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}-0 \\ \Rightarrow \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=-\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}} \\ \Rightarrow \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}...(ii)$
$\\ \begin{aligned} &\left[\because\right. \text { by anti-commutative law, }-\vec{a} \times \vec{c}=\vec{c} \times \vec{a}_{]}\\ &\text { From equations (i) and (ii), we have }\\ &\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}} \text { and } \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}\\ &\Rightarrow \vec{a} \times \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}} \end{aligned}$
Now, let us interpret the result graphically.
Let there be a parallelogram ABCD.

Here, and .
And AB and AD sides are making angle θ between them.
Area of parallelogram is given by,
Area of parallelogram = Base × Height
So from the diagram, area of parallelogram ABCD can be written as,
$\\ \begin{aligned} &\text { Area of parallelogram }=|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \sin \theta\\ &\text { Or, }\\ &\text { Area of parallelogram }=|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}| \end{aligned}$
Since, parallelogram on the same base and between the same parallels are equal in area, so we have
$\\ $$ |\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}|=|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}| $$\\ $This also implies that, $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}$\\ $ Thus, $ it is represented graphically.$

Question:11

Find the sine of the angle between the vectors $\overrightarrow{\mathrm{a}}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}_{\text {and }} \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$

Answer:

We have
$\overrightarrow{\mathrm{a}}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}_{\text {and }} \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$
Let these vectors be represented as,
$\begin{array}{l} \overrightarrow{\mathrm{a}}=\mathrm{a}_{1} \hat{\imath}+\mathrm{a}_{2} \hat{\jmath}+\mathrm{a}_{3} \hat{\mathrm{k}} \\ \overrightarrow{\mathrm{b}}=\mathrm{b}_{1} \hat{\imath}+\mathrm{b}_{2} \hat{\jmath}+\mathrm{b}_{3} \hat{\mathrm{k}} \end{array}$
Comparing the vectors, we get
$\begin{array}{l} a_{1}=3, a_{2}=1 \text { and } a_{3}=2 \\ b_{1}=2, b_{2}=-2 \text { and } b_{3}=4 \end{array}$
To find sine of the angle between the vectors $\vec{a} \text { and } \vec{b}$ , we can first find out cosine of the angle between them.
Cosine of the angle between $\vec{a} \text { and } \vec{b}$ is given by,
$\\\cos \theta=\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{a}}| \overrightarrow{\mathrm{b}} \mid} \\ =\frac{(3 \hat{\imath}+\hat{\jmath}+2 \hat{\mathrm{k}})(2 \hat{\imath}-2 \hat{\jmath}+4 \hat{\mathrm{k}})}{|3 \hat{\imath}+\hat{\jmath}+2 \hat{\mathrm{k}}||2 \hat{\imath}-2 \hat{\jmath}+4 \hat{\mathrm{k}}|} \\ =\frac{(3 \hat{\imath})(2 \hat{\imath})+(\hat{j})(-2 \hat{\jmath})+(2 \hat{\mathrm{k}})(4 \hat{\mathrm{k}})}{\sqrt{3^{2}+1^{2}+2^{2}} \sqrt{2^{2}+(-2)^{2}+4^{2}}}$
$\\ \begin{aligned} &\because\text {we know that, } \hat{i} \times \hat{i}=\hat{\jmath} \times \hat{\jmath}=\hat{\mathrm{k}} \times \hat{\mathrm{k}}=1_{\text {and }} \hat{i} \times \hat{\mathrm{j}}=\hat{i} \times \hat{\mathrm{k}}=\hat{\mathrm{j}} \times \hat{\mathrm{k}}=0\\ &\text { So, }\left(a_{1} \hat{\imath}+a_{2} \hat{\jmath}+a_{3} \hat{k}\right)\left(b_{1} \hat{\imath}+b_{2} \hat{\jmath}+b_{3} \hat{k}\right)=a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}\\ &\text { Also, we know that, } \left.|a \hat{\imath}+b \hat{\jmath}+c \hat{k}|=\sqrt{a^{2}+b^{2}+c^{2}}\right] \end{aligned}$
$\\ =\frac{6-2+8}{\sqrt{9+1+4} \sqrt{4+4+16}} \\ =\frac{12}{\sqrt{14} \sqrt{24}} \\ =\frac{12}{2 \sqrt{14} \sqrt{6}} \\ =\frac{6}{\sqrt{14 \times 6}}$
$\\ =\frac{6}{\sqrt{84}} \\ =\frac{6}{2 \sqrt{21}} \\ =\frac{3}{\sqrt{21}}$
By algebraic identity, we have
$\\ \sin \theta=\sqrt{1-\cos ^{2} \theta} \\ =\sqrt{1-\left(\frac{3}{\sqrt{21}}\right)^{2}} \\ =\sqrt{1-\frac{9}{21}} \\ =\sqrt{\frac{21-9}{21}}$
$\\ =\sqrt{\frac{12}{21}} \\ =\sqrt{\frac{4}{7}} \\ =\frac{2}{\sqrt{7}}$
Thus, sine of the angle between the vectors is $\frac{2}{\sqrt{7}}$ .

Question:12

If A, B, C, D are the points with position vectors

$\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}, 2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}, 2 \hat{\mathrm{i}}-3 \hat{\mathrm{k}}, 3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ respectively, find the projection of $\overrightarrow{\mathrm{AB}}$ along $\overrightarrow{\mathrm{CD}}$ .

Answer:

Given are points, A, B, C and D.
Let O be the origin.
We have,
Position vector of A
$\\ \begin{aligned} &=\hat{\imath}+\hat{\jmath}-\hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OA}}=\hat{\imath}+\hat{\jmath}-\hat{\mathrm{k}}\\ &\text { Position vector of } \mathrm{B}=2 \hat{\imath}-\hat{\jmath}+3 \hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OB}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}} \end{aligned}$
$\\ \begin{aligned} &\text { Position vector of } c=2 \hat{\imath}-3 \hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OC}}=2 \hat{\mathrm{\imath}}-3 \hat{\mathrm{k}}\\ &\text { Position vector of } D=3 \hat{\imath}-2 \hat{\jmath}+\hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OD}}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \end{aligned}$
$\\ $Now, let us find out $\overrightarrow{\mathrm{AB}}$ and $\overrightarrow{\mathrm{CD}}$\\ $\overrightarrow{\mathrm{AB}}=$ position vector of B-Position vector of $\mathrm{A}$$ $\\ \Rightarrow \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}} \\ \Rightarrow \overrightarrow{\mathrm{AB}}=(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}})-(\hat{\imath}+\hat{\mathrm{j}}-\hat{\mathrm{k}}) \\ \Rightarrow \overrightarrow{\mathrm{AB}}=2 \hat{\mathrm{i}}-\hat{1}-\hat{\mathrm{j}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}+\hat{\mathrm{k}} \\ \Rightarrow \overrightarrow{\mathrm{AB}}=\hat{\imath}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$ $\\ $And, $\overrightarrow{\mathrm{CD}}=$ position vector of $D$ -Position vector of $\mathrm{C}$\\ $\Rightarrow \overrightarrow{\mathrm{CD}}=\overrightarrow{\mathrm{OD}}-\overrightarrow{\mathrm{OC}}$\\ $\Rightarrow \overrightarrow{\mathrm{CD}}=(3 \hat{\imath}-2 \hat{\jmath}+\hat{\mathrm{k}})-(2 \hat{\imath}-3 \hat{\mathrm{k}})$ \\$\Rightarrow \overrightarrow{\mathrm{CD}}=3 \hat{\imath}-2 \hat{\imath}-2 \hat{\jmath}+\hat{\mathrm{k}}+3 \hat{\mathrm{k}}$$ $\\ \begin{aligned} &\Rightarrow \overrightarrow{\mathrm{CD}}=\hat{\imath}-2 \hat{\jmath}+4 \hat{\mathrm{k}}\\ &\text { The projection of } \overrightarrow{\mathrm{AB}} \text { along } \overrightarrow{\mathrm{CD}} \text { is given by, }\\ &\text { Projection }=\frac{\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{CD}}}{|\overrightarrow{\mathrm{CD}}|} \end{aligned}$
$\\ \begin{aligned} &\Rightarrow \text { Projection }=\frac{(\hat{\imath}-2 \hat{\jmath}+4 \hat{\mathrm{k}})(\hat{\imath}-2 \hat{\jmath}+4 \hat{\mathrm{k}})}{|\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}|}\\ &\Rightarrow \text { Projection }=\frac{(\hat{\imath})(\hat{i})+(-2 \hat{\jmath})(-2 \hat{\jmath})+(4 \hat{\mathrm{k}})(4 \hat{\mathrm{k}})}{\sqrt{1^{2}+(-2)^{2}+4^{2}}}\\ &\text { "we know that, } \hat{i} \times \hat{i}=\hat{\jmath} \times \hat{\jmath}=\hat{\mathrm{k}} \times \hat{\mathrm{k}}=1 \text { and } \hat{i} \times \hat{\mathrm{j}}=\hat{i} \times \hat{\mathrm{k}}=\hat{\mathrm{j}} \times \hat{\mathrm{k}}=0 \end{aligned}$
$\\ \begin{aligned} &\text { So }\\ &\left(a_{1} \hat{\imath}+a_{2} \hat{\jmath}+a_{3} \hat{k}\right)\left(b_{1} \hat{\imath}+b_{2} \hat{\jmath}+b_{3} \hat{k}\right)=a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}\\ &\text { Also, we know that, } \left.|a \hat{\imath}+b \hat{\jmath}+c \hat{k}|=\sqrt{a^{2}+b^{2}+c^{2}}\right]\\ &\Rightarrow \text { Projection }=\frac{1+4+16}{\sqrt{1+4+16}}\\ &\Rightarrow \text { Projection }=\frac{21}{\sqrt{21}} \end{aligned}$
Multiply numerator and denominator by √21.
$\\ \begin{aligned} &\Rightarrow \text { Projection }=\frac{21}{\sqrt{21}} \times \frac{\sqrt{21}}{\sqrt{21}}\\ &\Rightarrow \text { Projection }=\frac{21 \times \sqrt{21}}{21}\\ &\Rightarrow \text { Projection }=\sqrt{21}\\ &\text { Thus, projection of } \overrightarrow{\mathrm{AB}} \text { along } \overrightarrow{\mathrm{CD}} \text { is } \sqrt {21} \text { units. } \end{aligned}$

Question:13

Using vectors, find the area of the triangle ABC with vertices A(1, 2, 3), B(2, – 1, 4) and C(4, 5, – 1).

Answer:

We have,

The coordinates of points A, B and C are (1, 2, 3), (2, -1, 4) and (4, 5, -1) respectively.
We need to find the area of this triangle ABC.
We have the formula given as,
$\\ \begin{aligned} &\text { Area of } \Delta \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|_{\ldots \text { (i) }}\\ &\text { Let us find out } \overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{AC}} \text { first. } \end{aligned}$
$\\ \begin{aligned} &\text { We can say, }\\ &\text { Position vector of } \mathrm{A}=\hat{i}+2 \hat{\jmath}+3 \hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OA}}=\hat{\imath}+2 \hat{\jmath}+3 \hat{\mathrm{k}}\\ &\text { Position vector of } \mathrm{B}=2 \hat{\mathrm{l}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OB}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}} \end{aligned}$
$\\ \begin{aligned} &\text { Position vector of } \mathrm{c}=4 \hat{\imath}+5 \hat{\jmath}-\hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OC}}=4 \hat{\imath}+5 \hat{\jmath}-\hat{\mathrm{k}}\\ &\overrightarrow{\mathrm{AB}}\\ &\overrightarrow{\mathrm{AB}}=\text { position vector of B-Position vector of } \mathrm{A} \end{aligned}$
$\\ \Rightarrow \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}} \\ \Rightarrow \overrightarrow{\mathrm{AB}}=(2 \hat{\imath}-\hat{\jmath}+4 \hat{\mathrm{k}})-(\hat{\imath}+2 \hat{\jmath}+3 \hat{\mathrm{k}}) \\ \Rightarrow \overrightarrow{\mathrm{AB}}=2 \hat{\imath}-\hat{\imath}-\hat{\jmath}-2 \hat{\jmath}+4 \hat{\mathrm{k}}-3 \hat{\mathrm{k}} \\ \Rightarrow \overrightarrow{\mathrm{AB}}=\hat{i}-3 \hat{\jmath}+\hat{\mathrm{k}} \\ \text { For } \overrightarrow{\mathrm{AC}}$
$\\ \begin{aligned} &\overrightarrow{\mathrm{AC}}=\text { position vector of } \mathrm{C} \text { -Position vector of } \mathrm{A}\\ &\Rightarrow \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}}\\ &\Rightarrow \overrightarrow{\mathrm{AC}}=(4 \hat{\imath}+5 \hat{\jmath}-\hat{\mathrm{k}})-(\hat{\imath}+2 \hat{\jmath}+3 \hat{\mathrm{k}})\\ &\Rightarrow \overrightarrow{\mathrm{AC}}=4 \hat{\imath}-\hat{\imath}+5 \hat{\jmath}-2 \hat{\jmath}-\hat{\mathrm{k}}-3 \hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{AC}}=3 \hat{\imath}+3 \hat{\jmath}-4 \hat{\mathrm{k}} \end{aligned}$
$\\ \begin{aligned} &\Rightarrow \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=\hat{\imath}(12-3)-\hat{\jmath}(-4-3)+\hat{\mathrm{k}}(3+9)\\ &\Rightarrow \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=9 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}\\ &\text { And, }\\ &|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=|9 \hat{\imath}+7 \hat{\jmath}+12 \hat{\mathrm{k}}|\\ &\Rightarrow|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=\sqrt{9^{2}+7^{2}+12^{2}} \end{aligned}$
$\\ \Rightarrow|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=\sqrt{81+49+144} \Rightarrow|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=\sqrt{274}$
$From equation (i), we get $\\ Area of \triangle \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|$ \\$\Rightarrow$ Area of $\Delta \mathrm{ABC}=\frac{1}{2} \times \sqrt{274}$ Thus, area of yriangle ABC is $\frac{\sqrt{274}}{2}{} $ sq units$ .

Question:14

Using vectors, prove that the parallelogram on the same base and between the same parallels are equal in area.

Answer:

We have,

Given:
There are more than 1 parallelogram, and their bases can be taken as common and they are between same parallels.
To Prove:
These parallelograms whose bases are same and are between the same parallel sides have equal area.
Proof:
Let ABCD and ABFE be two parallelograms on the same base AB and between same parallel lines AB and DF.
Here,
AB || DC and AE || BF
We can represent area of parallelogram ABCD as,
$\text { Area of parallelogram } \mathrm{ABCD}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}_{\ldots \text { .. } \mathrm{i} \text { ) }}$
Now, area of parallelogram ABFE can be represented as,
Area of parallelogram ABFE
$\\ \begin{aligned} &=\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AE}}\\ &=\overrightarrow{\mathrm{AB}} \times(\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{DE}})\\ &[\because \text { in right-angled } \triangle A D E, \overrightarrow{A E}=\overrightarrow{A D}+\overrightarrow{D E}]\\ &\Rightarrow \text { Area of parallelogram } \mathrm{ABFE}=\overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{b}}+\mathrm{ka})\\ &[\because \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{AD}}=\overrightarrow{\mathrm{b}} \text { and } \overrightarrow{\mathrm{DE}}=\mathrm{ka}, \text { where } \mathrm{k} \text { is scalar; } \overrightarrow{\mathrm{DE}} \text { is parallel }\\ &\overrightarrow{\mathrm{AB}}_{\text {and }}\\ &\text { hence } \overrightarrow{\mathrm{DE}}=\mathrm{ka}_{1} \end{aligned}$
$\\ \begin{aligned} &=\vec{a} \times \vec{b}+\vec{a} \times k \vec{a}\\ &=\vec{a} \times \vec{b}+k(\vec{a} \times \vec{a})\\ &[\because \text { a scalar term can be taken out of a vector product] }\\ &=\vec{a} \times \vec{b}+k \times 0\\ &\left[\because \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{a}}=0\right]. \end{aligned}$
⇒Area of parallelogram ABFE $=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$ …(ii)
From equation (i) and (ii), we can conclude that
Area of parallelogram ABCD = Area of parallelogram ABFE
Thus, parallelogram on same base and between same parallels are equal in area.
Hence, proved.

Question:15

Prove that in any triangle ABC, $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$ where a, b, c are the magnitudes of the sides opposite to the vertices A, B, C, respectively.

Answer:

Given:
a, b, c are magnitudes of the sides opposite to the vertices A, B, C respectively.
⇒ AB = c, BC = a and CA = b
To Prove:
In triangle ABC,
$\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$
Construction: We have constructed a triangle ABC and named the vertices according to the question.
Note the height of the triangle, BD.
If ∠BAD = A
Then, BD = c sin A
$\\ \qquad \sin \mathrm{A}=\frac{\text { perpendicular }}{\text { hypotenuse }} \text { in } \Delta \mathrm{BAD} \\ \qquad \because \sin \mathrm{A}=\frac{\mathrm{BD}}{\mathrm{c}} \\ \Rightarrow \mathrm{BD}=\mathrm{c} \sin \mathrm{A}$
$\\ \text { And, } A D=c \cos A \\ \qquad \cos A=\frac{\text { base }}{\text { hypotenuse }} \text { in } \Delta B A D \\ \because \\ \Rightarrow \cos A=\frac{A D}{C} \\ \Rightarrow A D=c \cos A$

Proof:
Here, components of c which are:
c sin A
c cos A
are drawn on the diagram.
Using Pythagoras theorem which says that,
(hypotenuse)² =(perpendicular)² + (base)²
Take triangle BDC, which is a right-angled triangle.
Here,
Hypotenuse = BC
Base = CD
Perpendicular = BD
We get, $\\ (BC)\textsuperscript{2} = (BD)\textsuperscript{2} + (CD)\textsuperscript{2}\\ \\ $ \Rightarrow $ a\textsuperscript{2} = (c sin A)\textsuperscript{2} + (CD)\textsuperscript{2} [\because $ from the diagram, BD = c sin A]\\ \\ $ \Rightarrow $ a\textsuperscript{2} = c\textsuperscript{2} sin\textsuperscript{2} A + (b - c cos A)\textsuperscript{2}\\ $\because$ from the diagram, AC = CD + AD\\$ $\\ \\$ \Rightarrow $ CD = AC - AD\\ $ \Rightarrow $ CD = b - c cos A]\\ \\ $ \Rightarrow $ a\textsuperscript{2} = c\textsuperscript{2} sin\textsuperscript{2} A + (b\textsuperscript{2} + (-c cos A)\textsuperscript{2} - 2bc cos A) [\because$ from algebraic identity, (a -b)\textsuperscript{2} = a\textsuperscript{2} + b\textsuperscript{2} - 2ab]\\ \\ $ \Rightarrow $ a\textsuperscript{2} = c\textsuperscript{2} sin\textsuperscript{2} A + b\textsuperscript{2} + c\textsuperscript{2} cos\textsuperscript{2} A - 2bc cos A\\ \\ $ \Rightarrow $ a\textsuperscript{2} = c\textsuperscript{2} sin\textsuperscript{2} A + c\textsuperscript{2} cos\textsuperscript{2} A + b\textsuperscript{2} - 2bc cos A\\$ $\\ \\ $ \Rightarrow $ a\textsuperscript{2} = c\textsuperscript{2} (sin\textsuperscript{2} A + cos\textsuperscript{2} A) + b\textsuperscript{2} - 2bc cos A\\ \\ $ \Rightarrow $ a\textsuperscript{2} = c\textsuperscript{2} + b\textsuperscript{2} - 2bc cos A [\because$ from trigonometric identity, sin\textsuperscript{2} $ \theta $ + cos\textsuperscript{2} $ \theta $ = 1]\\ \\ $ \Rightarrow $ 2bc cos A = c\textsuperscript{2} + b\textsuperscript{2} - a\textsuperscript{2}\\ \\ $ \Rightarrow $ 2bc cos A = b\textsuperscript{2} + c\textsuperscript{2} - a\textsuperscript{2}\\$ . $\\\Rightarrow \cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$
Hence proved

Question:16

Answer:

Let $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}$ are vertices of a triangle ABC.
Also, we get
Position vector of A $=\overrightarrow{\mathrm{a}}$
Position vector of B $=\overrightarrow{\mathrm{b}}$
Position vector of C $=\overrightarrow{\mathrm{c}}$
We need to show that,
$\frac{1}{2}[\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}]$ gives the vector are of the triangle.
We know that,
Vector area of triangle ABC is given as,
$\\ Area of \Delta \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|$\\ Here, $\overrightarrow{\mathrm{AB}}=$ Position vector of $\mathrm{B}-$ Position vector of $\mathrm{A}$\\ $\Rightarrow \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}$\\ $\overrightarrow{\mathrm{AC}}=$ Position vector of $\mathrm{C}-$ Position vector of $\mathrm{A}$\\$ $\\ \Rightarrow \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}} \\ \therefore \text { Area of } \Delta \mathrm{ABC}=\frac{1}{2}|(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}) \times(\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}})| \\ \Rightarrow \text { Area of } \Delta \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{a}}| \\ {[\because-\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}},-\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}} \text { and } \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{a}}=0]}$
$\Rightarrow \text { Area of } \Delta \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|_{\ldots(\mathrm{j})}$
Thus, shown.
We know that, two vectors are collinear if they lie on the same line or parallel lines.
For $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}$ to be collinear, area of the ?ABC should be equal to 0.
⇒ Area of ?ABC = 0
$\\ \Rightarrow \frac{1}{2}|\vec{b} \times \vec{c}+\vec{c} \times \vec{a}+\vec{a} \times \vec{b}|=0 \\ \Rightarrow \vec{b} \times \vec{c}+\vec{c} \times \vec{a}+\vec{a} \times \vec{b}=0$
Thus, this is the required condition for $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}$ to be collinear.
Now, we need to find the unit vector normal to the plane of the triangle.
Let $\vec{\pi}$ be the unit vector normal to the plane of the triangle.
$\\ \overrightarrow{\mathrm{n}}=\frac{\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}}{|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|}$ \\Note that, $\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}_{\text {from equation }(\mathrm{i})}$ \\And, $|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|_{\text {from equation }(\mathrm{i})}$$ $So, \overrightarrow{\mathrm{n}}=\frac{\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|}$$ Thus, unit vector normal to the plane of the triangle is $\frac{\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|}$$

Question:17

Show that area of the parallelogram whose diagonals are given by $\overrightarrow{\mathrm{a}} \text { and } \overrightarrow{\mathrm{b}} \text { is } \frac{|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|}{2}$ Also find the area of the parallelogram whose diagonals are $2 \hat{i}-\hat{j}+\hat{k} \text { and } \hat{i}+3 \hat{j}-\hat{k}$

Answer:

We have,

Let ABCD be a parallelogram.
In ABCD,
$\\ \mathrm{AB}=\overrightarrow{\mathrm{p}} \\ \mathrm{AD}=\overrightarrow{\mathrm{q}} $$$ And since, AD || BC $\mathrm{So}, \mathrm{BC}=\overrightarrow{\mathrm{q}}$ We need to show that, $\\ $ Area of parallelogram $ \mathrm{ABCD}=\frac{|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|}{2}$ $\\$Where, \overrightarrow{\mathrm{a}} and \overrightarrow{\mathrm{b}} \text{are diagonals of the parallelogram} \mathrm{ABCD}\\ \text{Now, by triangle law of addition, we get} \\\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}$

$\\ \Rightarrow \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}} \\ \Rightarrow \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{a}}(\mathrm{say})_{\ldots(\mathrm{i})} \\ \Rightarrow \overrightarrow{\mathrm{BD}}=\overrightarrow{\mathrm{BA}}+\overrightarrow{\mathrm{AD}} \\ \text { Similarly, } \\ \Rightarrow \overrightarrow{\mathrm{BD}}=-\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}$
$\\ \begin{aligned} &\Rightarrow \overrightarrow{\mathrm{BD}}=\overrightarrow{\mathrm{b}}(\mathrm{say}) \ldots(\mathrm{ii})\\ &\text { Adding equations (i) and (ii), we get }\\ &\vec{a}+\vec{b}=(\vec{p}+\vec{q})+(-\vec{p}+\vec{q}) \end{aligned}$
$\\ \begin{aligned} &\Rightarrow \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{q}}\\ &\Rightarrow \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=2 \overrightarrow{\mathrm{q}}\\ &\Rightarrow \overrightarrow{\mathrm{q}}=\frac{1}{2}(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})\\ &\text { And, }\\ &\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}=(\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}})-(-\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}) \end{aligned}$
$\\ \begin{aligned} &\Rightarrow \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}-\overrightarrow{\mathrm{q}}\\ &\Rightarrow \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}=2 \overrightarrow{\mathrm{p}}\\ &\Rightarrow \overrightarrow{\mathrm{p}}=\frac{1}{2}(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}})\\ &\text { Now, } \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}} \text { can be written as, } \end{aligned}$
$\\ \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\left(\frac{1}{2}(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}})\right) \times\left(\frac{1}{2}(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})\right) \\ \Rightarrow \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\frac{1}{4}(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}) \times(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \\ \Rightarrow \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\frac{1}{4}(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{b}})$
$\\ \\ \quad\left[\because \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{b}}=0_{\text {and }}-\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}\right]$
$\begin{array}{l} \Rightarrow \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\frac{2}{4}(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \\ \Rightarrow \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\frac{1}{2}(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \end{array}$
We know that,Vector area of parallelogram ABCD is given by,

Area of parallelogram ABCD $\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}$
$\\ =\frac{1}{2}|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}| \\ =\frac{|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|}{2}$
Hence, shown.
Now, we need to find the area of parallelogram whose diagonals are $2 \hat{\imath}-\hat{\jmath}+\hat{k}_{\text {and }} \hat{\imath}+3 \hat{\jmath}-\hat{k}$
We have already derived the relationship between area of parallelogram and diagonals of parallelogram, which is
$\\ \begin{aligned} &\text { Area of parallelogram }=\frac{|\vec{a} \times \vec{b}|}{2}\\ &\text { Here, } \vec{a}=2 \hat{\imath}-\hat{\jmath}+\hat{k}\\ &\text { And, } \overrightarrow{\mathrm{b}}=\hat{\imath}+3 \hat{\jmath}-\hat{\mathrm{k}}\\ &\Rightarrow \text { Area of parallelogram }=\frac{|(2 \hat{i}-\hat{j}+\hat{k}) \times(\hat{i}+3 \hat{\jmath}-\hat{k})|}{2} \end{aligned}$
$\begin{array}{l} =\frac{1}{2}|| \begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{\mathrm{k}} \\ 2 & -1 & 1 \\ 1 & 3 & -1 \end{array}|| \\ =\frac{1}{2}|[\hat{\imath}((-1)(-1)-(1)(3))-\hat{\jmath}((2)(-1)-(1)(1))+\hat{\mathrm{k}}((2)(3)-(-1)(1))]| \\ =\frac{1}{2}|[\hat{\imath}(1-3)-\hat{\jmath}(-2-1)+\hat{\mathrm{k}}(6+1)]| \\ =\frac{1}{2}|-2 \hat{\imath}+3 \hat{\jmath}+7 \hat{\mathrm{k}}| \end{array}$
$\\ \begin{aligned} &=\frac{1}{2} \sqrt{(-2)^{2}+3^{2}+7^{2}}\\ &=\frac{1}{2} \sqrt{4+9+49}\\ &=\frac{1}{2} \sqrt{62}\\ &\text { Thus, area of required parallelogram is } \frac{1}{2} \sqrt{62} \text { sq units } \end{aligned}$

Question:18

If $\\ \overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}\\ \overrightarrow{b}=\hat{j}-\hat{k}$ find a vector $\begin{aligned} \\ &\overrightarrow{\mathrm{c}} \text { such that } &\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}}_{\text {and }} \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=3\end{aligned}$ .

Answer:

Given that,
$\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k} \text { and } \overrightarrow{b}=\hat{j}-\hat{k}$ $\\ \begin{aligned} &\text { We need to find vector } \overrightarrow{\mathrm{C}} \text { . }\\ &\text { Let } \overrightarrow{\mathrm{c}}=\mathrm{x} \hat{1}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}, \text { where } \mathrm{x}, \mathrm{y}, \mathrm{z} \text { be any scalars. }\\ &\text { Now, for } \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}}\\ &\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=(\hat{\imath}+\hat{\jmath}+\hat{\mathrm{k}}) \times(\mathrm{x} \hat{\mathrm{l}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}) \end{aligned}$
$\\ \Rightarrow(\hat{\imath}+\hat{\jmath}+\hat{\mathrm{k}}) \times(\mathrm{xi}+\mathrm{y} \hat{\jmath}+\mathrm{z} \hat{\mathrm{k}})=\hat{\mathrm{j}}-\hat{\mathrm{k}} \\ \Rightarrow\left|\begin{array}{lll} \hat{\imath} & \hat{\jmath} & \hat{\mathrm{k}} \\ 1 & 1 & 1 \\ \mathrm{x} & \mathrm{y} & \mathrm{z} \end{array}\right|=\hat{\mathrm{j}}-\hat{\mathrm{k}} \\ \Rightarrow \hat{1}((1)(\mathrm{z})-(1)(\mathrm{y}))-\hat{\jmath}((1)(\mathrm{z})-(1)(\mathrm{x}))+\hat{\mathrm{k}}(1)(\mathrm{y})-(1)(\mathrm{x}))=\hat{\mathrm{j}}-\hat{\mathrm{k}} \\ \Rightarrow \hat{\mathrm{i}}(\mathrm{z}-\mathrm{y})-\hat{\mathrm{j}}(\mathrm{z}-\mathrm{x})+\hat{\mathrm{k}}(\mathrm{y}-\mathrm{x})=\hat{\mathrm{j}}-\hat{\mathrm{k}}$
Comparing Left Hand Side and Right Hand Side, we get
From coefficient of $\hat{i}$ ⇒ z-y = 0 …(i)
From coefficient of $\hat{j}$ ⇒ -(z-x) = 1
⇒ x-z = 1 …(ii)
From coefficient of $\hat{k}$ ⇒ y-x = -1
⇒ x-y = 1 …(iii)
$\\ \begin{aligned} &\text { Also, for } \overrightarrow{a} \overrightarrow{c}=3\\ &\overrightarrow{a} \overrightarrow{c}=(\hat{i}+\hat{j}+\hat{k}) \cdot(\mathrm{x} \hat{\imath}+\mathrm{y} \hat{\jmath}+\mathrm{z} \hat{\mathrm{k}})\\ &\Rightarrow(\hat{\imath}+\hat{\jmath}+\hat{\mathrm{k}}) \cdot(\mathrm{x} \hat{\imath}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}})=3\\ &\text { since }_{2}(\hat{\imath}+\hat{\jmath}+\hat{\mathrm{k}}) \cdot(\mathrm{x} \hat{\imath}+\mathrm{y} \hat{\jmath}+\mathrm{z} \hat{\mathrm{k}})=\mathrm{x}+\mathrm{y}+\mathrm{z}, \text { as } \hat{1} . \hat{\imath}=\hat{\mathrm{j}} \cdot \hat{\mathrm{j}}=\hat{\mathrm{k}} . \hat{\mathrm{k}}=1 \text { and other dot }\\ &\text { multiplication is zero. We get, } \end{aligned}$ $\\x + y + z = 3 $ \ldots $ (iv)\\ \\$ Now, add equations (ii) and (iii), we get $\\ (x - z) + (x - y) = 1 + 1\\ \\ $ \Rightarrow $ x + x - y - z = 2\\ \\ $ \Rightarrow $ 2x - y - z = 2 $ \ldots $ (v)\\ \\$ Add equations (iv) and (v), we get $\\ (x + y + z) + (2x - y - z) = 3 + 2\\ \\ $ \Rightarrow $ x + 2x + y - y + z - z = 5\\ \\ $ \Rightarrow $ 3x = 5\\ \\$ $\\ \Rightarrow x=\frac{5}{3}\\ $Put value of x in equation (iii), we get Equation (iii)$\\ \Rightarrow x-y=1\ \Rightarrow \frac{5}{3}-\mathrm{y}=1 $$ \Rightarrow \mathrm{y}=\frac{5}{3}-1 $$$ $\\\begin{aligned} &\Rightarrow \mathrm{y}=\frac{5-3}{3}\\ &\Rightarrow \mathrm{y}=\frac{2}{3}\\ &\text { Put this value of } y \text { in equation (i), we get } \end{aligned}$ $\\ $Equation $ (\mathrm{i}) \Rightarrow \mathrm{z}-\mathrm{y}=0$ $\Rightarrow \mathrm{z}-\frac{2}{3}=0$ $\Rightarrow \mathrm{z}=\frac{2}{3}$\\ since, $\overrightarrow{\mathrm{c}}=\mathrm{x} \hat{1}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}\\$ By putting the values of $x, y$ and $z,$ we get$ $\\ \begin{aligned} &\overrightarrow{\mathrm{c}}=\frac{5}{3} \hat{\imath}+\frac{2}{3} \hat{\jmath}+\frac{2}{3} \hat{\mathrm{k}}\\ &\text { Thus, we have found the vector } \overrightarrow{\mathrm{c}} \text { . } \end{aligned}$

Question:19

The vector in the direction of the vector $\hat{i}-2 \hat{j}+2 \hat{k}$ that has magnitude 9 is
$\\ A. \hat{i}-2 \hat{j}+2 \hat{k}\\ B.\frac{\hat{i}-2 \hat{j}+2 \hat{k}}{3}\\ C.3(\hat{i}-2 \hat{j}+2 \hat{k})\\ D. 9(\hat{i}-2 \hat{j}+2 \hat{k})\\$

Answer:

C)
Given is the vector $\hat{i}-2 \hat{j}+2 \hat{k}$
Let this vector be $\vec{a}$ , such that
$\vec{a}=\hat{i}-2 \hat{j}+2 \hat{k}$
Let us first find the unit vector in the direction of this vector $\vec{a}$ .
We know that, a unit vector in a normed vector space is a vector (often a spatial vector) of length 1.
To find a unit vector with the same direction as a given vector, we divide by the magnitude of the vector.
Unit vector in the direction of the vector $\vec{a}$ is given as,
$\\ \begin{aligned} &\hat{a}=\frac{\overrightarrow{\mathrm{a}}}{|\overrightarrow{\mathrm{a}}|}\\ &\text { As, we have } \overrightarrow{\mathrm{a}}=\hat{\imath}-2 \hat{\jmath}+2 \hat{\mathrm{k}}\\ &\text { Then, }\\ &|\overrightarrow{\mathrm{a}}|=|\hat{\imath}-2 \hat{\jmath}+2 \hat{\mathrm{k}}|\\ &\Rightarrow|\vec{a}|=\sqrt{1^{2}+(-2)^{2}+2^{2}}\\ &[\because \text { if }|\vec{p}|=|x \hat{\imath}+y \hat{\jmath}+z \hat{k}| \end{aligned}$
$\\ \Rightarrow|\vec{p}|=\sqrt{x^{2}+y^{2}+z^{2}} \\ \Rightarrow|\vec{a}|=\sqrt{1+4+4} \\ \Rightarrow|\vec{a}|=\sqrt{9} \\ \Rightarrow|\vec{a}|=3$
Therefore,
$\\ \hat{\mathrm{a}}=\frac{\hat{1}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}}{3} \\ {\left[\because \overrightarrow{\mathrm{a}}=\hat{1}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}_{\text {and }}|\overrightarrow{\mathrm{a}}|=3 \right].}$
We have found unit vector in the direction of the vector $\hat{i}-2 \hat{j}+2 \hat{k}$ , but we need to find the unit vector in the direction of $\hat{i}-2 \hat{j}+2 \hat{k}$ but also with the magnitude 9.
We have the formula:
Vector in the direction of $\vec{a}$ with a magnitude of 9 $=9 \times \frac{\overrightarrow{\mathrm{a}}}{|\overrightarrow{\mathrm{a}}|}$
$\\ \begin{aligned} &=9 \times \widehat{a}\\ &\text { And } \hat{\mathrm{a}}=\frac{\hat{1}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}}{3} \text { as just found. }\\ &\text { So, }\\ &\Rightarrow \text { Vector in the direction of } \overrightarrow{\mathrm{a}} \text { with a magnitude of } 9=9 \times \frac{\hat{1}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}}{3} \end{aligned}$
$\\ \begin{aligned} &=3(\hat{\imath}-2 \hat{\jmath}+2 \hat{\mathrm{k}})\\ &\text { Thus, vector in the direction of vector }\\ &\hat{\imath}-2 \hat{\jmath}+2 \hat{k}_{\text {and }} \text { has magnitude } 9 \text { is } 3(\hat{\imath}-2 \hat{\jmath}+2 \hat{\mathrm{k}}) \end{aligned}$

Question:20

The position vector of the point which divides the join of points $2 \vec{a}-3 \vec{b}_{\text {and }} \vec{a}+\vec{b}$ in the ratio 3 : 1 is
$\\A.\frac{3 \vec{a}-2 \vec{b}}{2}$\\ B. $\frac{7 \overrightarrow{\mathrm{a}}-8 \overrightarrow{\mathrm{b}}}{4}$ \\C.$\frac{3 \overrightarrow{\mathrm{a}}}{4}$ \\D. $\frac{5 \vec{a}}{4}$$

Answer:

D)
We are given points $2 \vec{a}-3 \vec{b}_{\text {and }} \vec{a}+\vec{b}$
Let these points be
$A(2 \vec{a}-3 \vec{b})_{a n d} B(\vec{a}+\vec{b})$ .
Also, given in the question that,
A point divides AB in the ratio of 3 : 1.
Let this point be C.
⇒ C divides AB in the ratio = 3 : 1
We need to find the position vector of C.
We know the position vector of a point C dividing the line segment joining the points P and Q, whose position vectors are p and q in the ratio m : n internally is given by,
$\text { Position vector }=\frac{\mathrm{m} \overrightarrow{\mathrm{q}}+\mathrm{n} \overrightarrow{\mathrm{p}}}{\mathrm{m}+\mathrm{n}}$
According to the question, here
m : n = 3 : 1
⇒ m = 3 and n = 1
$\\ Also, $\overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}$\\ And $\overrightarrow{\mathrm{p}}=2 \overrightarrow{\mathrm{a}}-3 \overrightarrow{\mathrm{b}}$ $\\ $Substituting these values in the formula above, we get\\ Position vector of $C=\frac{3(\vec{a}+\vec{b})+1(2 \vec{a}-3 \vec{b})}{3+1}$ $\\=\frac{3 \vec{a}+3 \vec{b}+2 \vec{a}-3 \vec{b}}{4}$$ . $\begin{aligned} &=\frac{3 \overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}}-3 \overrightarrow{\mathrm{b}}}{4}\\ &=\frac{5 \vec{a}}{4}\\ &\text { Thus, position vector of the point is } \frac{5 \vec{a}}{4} \text { . } \end{aligned}$

Question:21

The vector having initial and terminal points as (2, 5, 0) and (–3, 7, 4), respectively is
$\\A. -\hat{i}+12 \hat{j}+4 \hat{k}\\ B. 5 \hat{i}+2 \hat{j}-4 \hat{k}\\ C.-5 \hat{i}+2 \hat{j}+4 \hat{k}\\ D.\hat{i}+\hat{j}+\hat{k}$

Answer:

C)
Let initial point be A(2,5,0) and terminal point be B(-3,7,4).So, the required vector joining A and B is the vector $\vec{AB}$ .
$\\ \Rightarrow \vec{\mathrm{AB}}=(-3-2) \hat{1}+(7-5) \hat{\mathrm{j}}+(4-0) \hat{\mathrm{k}} \\ =-5 \hat{\imath}+2 \hat{\jmath}+4 \hat{\mathrm{k}}$

Question:22

The angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitudes $\sqrt3$ and 4, respectively, and $\vec{a}.\vec{b}=2\sqrt3$ is

$\\A. \frac{\pi}{6}\\\\ B.\frac{\pi}{3}\\\\ C. \frac{\pi}{2}$ $\\\\ D. \frac{5\pi}{2}$

Answer:

Answer :(B)
Given that, $|\vec{\mathrm{a}}|=\sqrt{3},|\vec{\mathrm{b}}|=4 \text { and } \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}=2 \sqrt{3}$
Let θ be the angle between vector a and b.
$\\ \text { Then, } \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}=|\vec{\mathrm{a}}||\vec{\mathrm{b}}| \cos \theta \\ \Rightarrow 2 \sqrt{3}=\sqrt{3} .4 \cos \theta \\ \Rightarrow \quad \cos \theta=\frac{2 \sqrt{3}}{\sqrt{3} .4}=\frac{1}{2} \\ \Rightarrow \theta=\frac{\pi}{3}$

Question:23

Find the value of λ such that the vectors $\vec{\mathrm{a}}=2 \hat{\mathrm{i}}+\lambda \hat{\mathrm{j}}+\hat{\mathrm{k}}_{\text {and }} \vec{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ are orthogonal

A. 0
B. 1
C. $\frac{3}{2}$
D. $-\frac{5}{2}$

Answer:

D)
Given that, $\vec{\mathrm{a}}{\text { and }} \vec{\mathrm{b}}$ are orthogonal.
$\\ \Rightarrow \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}=0 \\ \Rightarrow \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}=(2 \hat{\imath}+\lambda \hat{\jmath}+\hat{\mathrm{k}}) \cdot(\hat{\imath}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=0 \\ \Rightarrow 2+2 \lambda+3=0(\because \hat{\mathrm{i}} . \hat{\mathrm{j}}=0, \hat{\mathrm{j}} . \hat{\mathrm{k}}=0, \hat{\mathrm{k}} \cdot \hat{\mathrm{i}}=0) \\ \Rightarrow 2 \lambda=-5 \\ \Rightarrow \lambda=-\frac{5}{2}$

Question:24

The value of λ for which the vectors $3 \hat{i}-6 \hat{j}+\hat{k} \text{ and } 2 \hat{i}-4 \hat{j}+\lambda \hat{k}$ are parallel is
A. $\frac{2}{3}$

B. $\frac{3}{2}$
C. $\frac{5}{2}$
D. $\frac{2}{5}$

Answer:

Answer:(A)
Given that, $3 \hat{i}-6 \hat{j}+\hat{k} \text{ and } 2 \hat{i}-4 \hat{j}+\lambda \hat{k}$ are parallel
$\\ \Rightarrow \frac{3}{2}=\frac{-6}{-4}=\frac{1}{\lambda} \\ \Rightarrow \lambda=\frac{2}{3}$

Question:25

The vectors from origin to the points A and B are $\vec{\mathrm{a}}=2 \hat{\mathrm{i}}-3\hat{\mathrm{j}}+2\hat{\mathrm{k}} \text { and }\vec{\mathrm{b}}=2\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ respectively, then the area of triangle OAB is
A. 340
B. $\sqrt{25}$
C. $\sqrt{229}$
D. $\frac{1}{2}\sqrt{229}$

Answer:

Answer :(D)
Given that, vector from origin to the point A, $\vec{OA}=2 \hat{\mathrm{i}}-3\hat{\mathrm{j}}+2\hat{\mathrm{k}}$ and vector from origin to the point B, $\vec{OB}=2\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}$
$\\ \text { Area of } \Delta \mathrm{OAB}=\frac{1}{2}|\overrightarrow{\mathrm{OA}} \times \overrightarrow{\mathrm{OB}}| \\ =\frac{1}{2}|(2 \hat{\mathrm{l}}-3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \times(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}})|$
$\\ =\frac{1}{2}\left|\begin{array}{ccc} \hat{1} & \hat{\jmath} & \hat{\mathrm{k}} \\ 2 & -3 & 2 \\ 2 & 3 & 1 \end{array}\right| \\ =\frac{1}{2}|\hat{\imath}(-3-6)-\hat{\jmath}(2-4)+\hat{\mathrm{k}}(6+6)| \\ =\frac{1}{2}|(-9 \hat{\imath}+2 \hat{\jmath}+12 \hat{\mathrm{k}})| \\ =\frac{1}{2} \sqrt{(-9)^{2}+(2)^{2}+(12)^{2}}$
$\\ =\frac{1}{2} \sqrt{81+4+144} \\ =\frac{1}{2} \sqrt{229}$

Question:26

For any vector $\vec{a}$ , the value of $(\vec{a} \times \hat{i})^{2}+(\vec{a} \times \hat{j})^{2}+(\vec{a} \times \hat{k})^{2}$ is equal to
A. $\vec{a} ^2$
B. $3\vec{a} ^2$
C. $4\vec{a} ^2$
D. $2\vec{a} ^2$

Answer:

Answer :(D)
$\text { Let } \overrightarrow{a}=\mathrm{a}_{1} \hat{i}+\mathrm{a}_{2} \hat{j}+\mathrm{a}_{3} \hat{k} \text { then }\\ \overrightarrow{\mathrm{a}} \times \hat{\mathrm{i}}=\left(\mathrm{a}_{1} \hat{\imath}+\mathrm{a}_{2} \hat{\jmath}+\mathrm{a}_{3} \hat{\mathrm{k}}\right) \times \hat{\mathrm{i}}=\mathrm{a}_{1} \hat{\imath} \times \hat{\mathrm{i}}+\mathrm{a}_{2} \hat{\jmath} \times \hat{\mathrm{i}}+\mathrm{a}_{3} \hat{\mathrm{k}} \times \hat{\mathrm{i}} \\ \Rightarrow \overrightarrow{a} \times \hat{\mathrm{i}}=0-\mathrm{a}_{2} \hat{\mathrm{k}}+\mathrm{a}_{3} \hat{\mathrm{j}}(\because \hat{\mathrm{i}} \times \hat{\mathrm{i}}=0, \hat{\mathrm{j}} \times \hat{\mathrm{i}}=-\hat{\mathrm{k}}, \hat{\mathrm{k}} \times \hat{\mathrm{i}}=\hat{\mathrm{j}})\\ \Rightarrow \overrightarrow{\mathrm{a}} \times \hat{i}=-\mathrm{a}_{2} \hat{k}+\mathrm{a}_{3} \hat{j}$ $\\ \begin{aligned} &\Rightarrow|\vec{a} \times \hat{i}|^{2}=a_{2}^{2}+a_{3}^{2}\\ &\text { Similarly, we get }\\ &\Rightarrow|\vec{a} \times \hat{\jmath}|^{2}=a_{1}^{2}+a_{3}^{2}\\ &\Rightarrow|\overrightarrow{\mathrm{a}} \times \hat{\mathrm{k}}|^{2}=\mathrm{a}_{1}^{2}+\mathrm{a}_{2}^{2}\\ &\therefore|\vec{a} \times \hat{\imath}|^{2}+|\vec{a} \times \hat{\jmath}|^{2}+|\vec{a} \times \hat{k}|^{2}=a_{2}^{2}+a_{3}^{2}+a_{1}^{2}+a_{3}^{2}+a_{1}^{2}+a_{2}^{2} \end{aligned}$
$=2\left(\mathrm{a}_{1}^{2}+\mathrm{a}_{2}^{2}+\mathrm{a}_{3}^{2}\right)=2|\overrightarrow{\mathrm{a}}|^{2}\left(\because|\overrightarrow{\mathrm{a}}|=\sqrt{\mathrm{a}_{1}^{2}+\mathrm{a}_{2}^{2}+\mathrm{a}_{3}^{2}}\right)$

Question:27

$\text { If }|\vec{\mathrm{a}}|=10,|\vec{\mathrm{b}}|=2 \text { and } \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}=12, \text { then value of }|\vec{\mathrm{a}} \times \vec{\mathrm{b}}|{\text { is }}$
A. 5
B. 10
C. 14
D. 16

Answer:

Answer :(D)
Given that, $|\vec{\mathrm{a}}|=10,|\vec{\mathrm{b}}|=2 \text { and } \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}=12$
Let θ be the angle between vector a and b.
$\\ \begin{aligned} &\text { Then, }\\ &\vec{\mathrm{a}} \cdot \vec{\mathrm{b}}=|\vec{\mathrm{a}}||\vec{\mathrm{b}}| \cos \theta\\ &\Rightarrow 12=10 \times 2 \cos \theta\\ &\Rightarrow \cos \theta=\frac{12}{20}=\frac{3}{5}\\ &\sin \theta=\sqrt{1-\cos ^{2} \theta}=\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\sqrt{1-\frac{9}{25}} \end{aligned}$
$\\ \Rightarrow \sin \theta=\pm \frac{4}{5} \\ \text { Now, }|\vec{\mathrm{a}} \times \vec{\mathrm{b}}|=|\vec{\mathrm{a}}||\vec{\mathrm{b}}| \sin \theta \\ \Rightarrow|\vec{\mathrm{a}} \times \vec{\mathrm{b}}|=10 \times 2 \times \frac{4}{5}=16$

Question:28

The vectors $\lambda \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}, \hat{\mathrm{i}}+\lambda \hat{\mathrm{j}}-\hat{\mathrm{k}}{\text { and }} 2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\lambda \hat{\mathrm{k}}$ coplanar if

$\\A. \lambda = -2\\ B. \lambda = 0\\ C. \lambda = 1\\ D. \lambda = -1\\$

Answer:

Answer :(A)
Given that, $\lambda \hat{\imath}+\hat{\jmath}+2 \hat{\mathrm{k}}, \hat{\imath}+\lambda \hat{\jmath}-\hat{\mathrm{k}} \text { and } 2 \hat{\imath}-\hat{\jmath}+\lambda \hat{\mathrm{k}} \text { are coplanar. }$
$\\ \begin{aligned} &\text { Let } \vec{a}=\lambda \hat{\imath}+\hat{\jmath}+2 \hat{k}, \vec{b}=\hat{\imath}+\lambda \hat{\jmath}-\hat{k} \text { and } \vec{c}=2 \hat{\imath}-\hat{\jmath}+\lambda \hat{k}\\ &\text { Now, }\\ &\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}} \text { and } \overrightarrow{\mathrm{c}} \text { are coplanar } \end{aligned}$
If $\begin{aligned} &\left|\begin{array}{ccc} \lambda & 1 & 2 \\ 1 & \lambda & -1 \\ 2 & -1 & \lambda \end{array}\right|=0\\ \end{aligned}$
$\\ $ \Rightarrow $ $ \lambda $ ($ \lambda $ \textsuperscript{2}-1) -1($ \lambda $ +2) +2(-1-2 $ \lambda $ )=0\\ \\ $ \Rightarrow $ $ \lambda $ \textsuperscript{3}- $ \lambda $ -$ \lambda $ -2-2-4 $ \lambda $ =0\\ \\ $ \Rightarrow $ $ \lambda $ \textsuperscript{3}- 6$ \lambda $ -4 =0\\ \\ $ \Rightarrow $ ($ \lambda $ +2)( $ \lambda $ \textsuperscript{2}-2 $ \lambda $ -2)=0\\ \\$ $\\ \Rightarrow \lambda=-2 \text { and } \lambda=\frac{2 \pm \sqrt{(-2)^{2}-4 \times 1 \times-2}}{2}=\frac{2 \pm \sqrt{12}}{2} \\ \Rightarrow \lambda=-2 \text { and } \lambda=1 \pm \sqrt{3}$

Question:29

If $\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}$ are unit vectors such that $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=\overrightarrow{0}$ , then the value of $\vec{\mathrm{a}} \cdot \vec{\mathrm{b}}+\vec{\mathrm{b}} \cdot \vec{\mathrm{c}}+\vec{\mathrm{c}} \cdot \vec{\mathrm{a}}$ is
A. 1
B. 3
C. $-\frac{3}{2}$
D. None of these

Answer:

Answer :(C)
$\begin{aligned} &\text { Given that, }\\ &\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}} \text { and } \overrightarrow{\mathrm{c}} \text { are unit vectors } \Rightarrow \overrightarrow{\mid \mathrm{a}}|=| \overrightarrow{\mathrm{b}}|=\overrightarrow{\mid \mathrm{c}}|=1{\text { and }}\\ &\vec{a}+\vec{b}+\vec{c}=0 \end{aligned}$
$\\ \begin{aligned} &\Rightarrow(\vec{a}+\vec{b}+\vec{c}) \cdot(\vec{a}+\vec{b}+\vec{c})=0\\ &\Rightarrow \vec{\mathrm{a}}^{2}+\vec{\mathrm{a}} \cdot \vec{\mathrm{b}}+\vec{\mathrm{a}} \cdot \vec{\mathrm{c}}+\vec{\mathrm{b}} \cdot \vec{\mathrm{a}}+\vec{\mathrm{b}}^{2}+\vec{\mathrm{b}} . \vec{\mathrm{c}}+\vec{\mathrm{c}} \cdot \vec{\mathrm{a}}+\vec{\mathrm{c}} \cdot \vec{\mathrm{b}}+\vec{\mathrm{c}}^{2}=0\\ &\Rightarrow \vec{\mathrm{a}}^{2}+\vec{\mathrm{b}}^{2}+\vec{\mathrm{c}}^{2}+2(\vec{\mathrm{a}} \cdot \vec{\mathrm{b}}+\vec{\mathrm{b}} \cdot \vec{\mathrm{c}}+\vec{c} \cdot \vec{\mathrm{a}})=0\\ \end{aligned}$
$\\ \begin{aligned} &\Rightarrow 1+1+1+2(\vec{\mathrm{a}} \cdot \vec{\mathrm{b}}+\vec{\mathrm{b}} . \vec{\mathrm{c}}+\vec{\mathrm{c}} \cdot \vec{\mathrm{a}})=0(\because \vec{\mathrm{a}}, \vec{\mathrm{b}} \text { and } \vec{\mathrm{c}} \text { are unit vectors }) \end{aligned}$
$\Rightarrow \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}+\vec{\mathrm{b}} \cdot \vec{\mathrm{c}}+\vec{\mathrm{c}} \cdot \vec{\mathrm{a}}=-\frac{3}{2}$

Question:30

Projection vector of $\vec{a}$ on $\vec{b}$ is

$\\A. \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^{2}}\right) \vec{b}$\\ B. $\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{b}}|}$\\C. $\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{a}}|}$\\ D. $\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^{2}}\right) \hat{b}$$

Answer:

Answer :(A)

Let θ be the angle between $\vec {a} $ and $\vec{b}$
From figure we can see that, $\text{length }\mathrm{OL}$ is the projection of $\\\vec{a} on \overrightarrow{\mathrm{b}} and \overrightarrow{\mathrm{O}} \text{is the projection vector of} \overrightarrow{\mathrm{a}} on \overrightarrow{\mathrm{b}} \\ In \Delta OLA, \text{we have}\\ \cos \theta=\frac{O L}{O A}\\ \Rightarrow \mathrm{OL}=\mathrm{OA} \cos \theta$
$\\ \Rightarrow \mathrm{OL}=|\overrightarrow{\mathrm{a}}| \cos \theta \\ \qquad \mathrm{OL}=|\overrightarrow{\mathrm{a}}|\left\{\frac{(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}})}{|\overrightarrow{\mathrm{a}}||\mathrm{b}|}\right\}\left(\because \cos \theta=\frac{(\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}})}{|\overrightarrow{\mathrm{a}}||\mathrm{b}|}\right) \\ \Rightarrow \mathrm{OL}=\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{b}}|}$
$\\ \begin{aligned} &\text { Now, }\\ &\overrightarrow{\mathrm{OL}}=(\mathrm{OL}) \hat{\mathrm{b}}\\ &\Rightarrow \overrightarrow{\mathrm{OL}}=\left\{\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{b}}|}\right\} \hat{\mathrm{b}}\\ &\overrightarrow{\mathrm{OL}}=\left\{\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{b}}|}\right\} \frac{\overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{b}}|}=\left\{\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{b}}|^{2}}\right\} \overrightarrow{\mathrm{b}} \end{aligned}$

Question:31

If $\vec{a},\vec{b},\vec{c}$ are three vectors such that $\vec{a}+\vec{b}+\vec{c}=0$ and $|\overrightarrow{\mathrm{a}}|=2,|\overrightarrow{\mathrm{b}}|=3,|\overrightarrow{\mathrm{c}}|=5$ then value of $\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}=0$ is
A. 0
B. 1
C. –19
D. 38

Answer:

Answer :(C)
Given that, $\vec{a}+\vec{b}+\vec{c}=0$ and $|\overrightarrow{\mathrm{a}}|=2,|\overrightarrow{\mathrm{b}}|=3,|\overrightarrow{\mathrm{c}}|=5$
$\\ \Rightarrow(\vec{a}+\vec{b}+\vec{c}) \cdot(\vec{a}+\vec{b}+\vec{c})=0 \\ \Rightarrow \vec{a}^{2}+\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c}+\vec{b} \cdot \vec{a}+\vec{b}^{2}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}+\vec{c} \cdot \vec{b}+\vec{c}^{2}=0 \\ \Rightarrow \vec{a}^{2}+\vec{b}^{2}+\vec{c}^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0$
$\\ \Rightarrow 4+9+25+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0(\because \vec{a}|=2,| \vec{b}|=3, \overrightarrow{\mid c}|=5) \\ \Rightarrow \vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=-\frac{38}{2}=-19$

Question:32

If $\left | \vec{a} \right |=4$ and $-3\leq\lambda\leq2$ , then the range of $\left | \lambda a \right |$ is
A. [0, 8]
B. [–12, 8]
C. [0, 12]
D. [8, 12]

Answer:

Answer :(C)
Given that, $\left | \vec{a} \right |=4$ and $-3\leq\lambda\leq2$ ,
$\\ \begin{aligned} &\text { We know that, }|\lambda \vec{\mathrm{a}}|=|\lambda||\vec{\mathrm{a}}|\\ &\Rightarrow|\lambda \vec{\mathrm{a}}|=|-3||\vec{\mathrm{a}}|=3.4=12 \text { at } \lambda=-3\\ &\Rightarrow|\lambda \vec{\mathrm{a}}|=|0||\vec{\mathrm{a}}|=0.4=0 \text { at } \lambda=0\\ &\Rightarrow|\lambda \vec{\mathrm{a}}|=|2||\vec{\mathrm{a}}|=2.4=8 \text { at } \lambda=2\\ &\text { Hence, the range of }|\lambda \vec{\mathrm{a}}| \text { is }[0,12] \end{aligned}$

Question:33

The number of vectors of unit length perpendicular to the vectors $\vec{\mathrm{a}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}\text { and } \vec{\mathrm{b}}=\hat{\mathrm{j}}+\hat{\mathrm{k}}$ is
A. one
B. two
C. three
D. infinite

Answer:

Answer :(B)
Given that , $\vec{\mathrm{a}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}\text { and } \vec{\mathrm{b}}=\hat{\mathrm{j}}+\hat{\mathrm{k}}$
Now, a vector which is perpendicular to both $\vec{a} \text{ and } \vec{b}$ is given by
$\\ \vec{a} \times \vec{b}=\left|\begin{array}{lll}\hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 1 & 2 \\ 0 & 1 & 1\end{array}\right|=\hat{\imath}(1-2)-\hat{\jmath}(2-0)+\hat{\mathrm{k}}(2-0)=-\hat{\imath}-2 \hat{\jmath}+2 \hat{\mathrm{k}}$\\ Now, $|\vec{a} \times \vec{b}|=\sqrt{(-1)^{2}+(-2)^{2}+(2)^{2}}=\sqrt{1+4+4}=\sqrt{9}=3$ \\$\therefore$ the required unit vector$ $\\ =\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}=\frac{-\hat{i}-2 \hat{\jmath}+2 \hat{k}}{3}=\frac{-1}{3} \hat{\imath}-\frac{2}{3} \hat{\jmath}+\frac{2}{3} k$\\ There are two perpendicular directions to any plane. Thus, another unit vector perpendicular to $\vec{a}$ and $\vec{b}$ is given $b y-\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}=\frac{\vec{b} \times \vec{a}}{|\vec{b} \times \vec{a}|}$\\ $\Rightarrow \frac{\vec{b} \times \vec{a}}{|\vec{b} \times \vec{a}|}=\frac{1}{3} \hat{\imath}+\frac{2}{3} \hat{\jmath}-\frac{2}{3} k$$ Hence, there are two unit length perpendicular to the $\vec{a} \text{ and } \vec{b}$ .

Question:34

Fill in the blanks
The vector $\vec{a} + \vec{b}$ bisects the angle between the non-collinear vectors $\vec{a}$ and $\vec{b}$ if ________

Answer:

Let $\vec{a}$ and $\vec{b}$ are two non-collinear vectors.

Let $\vec{a} + \vec{b}$ bisects the angle between $\vec{a}$ and $\vec{b}$ .
$\\ \Rightarrow \theta_{1}=\theta_{2} \\ \qquad \cos \theta_{1}=\frac{\vec{a} \cdot(\vec{a}+\vec{b})}{|\vec{a}||\vec{a}+\bar{b}|} \text { and } \cos \theta_{2}=\frac{\vec{b} \cdot(\vec{a}+\vec{b})}{|\vec{b}||\vec{a}+\vec{b}|} \\ \text { since, } \theta_{1}=\theta_{2} \Rightarrow \cos \theta_{1}=\cos \theta_{2} \\ \therefore \quad \frac{\vec{a} \cdot(\vec{a}+\vec{b})}{|\vec{a}||\vec{a}+\vec{b}|}=\frac{\vec{b} \cdot(\vec{a}+\vec{b})}{|\vec{b} \| \vec{a}+\vec{b}|} \\ \Rightarrow |\vec{a}|=|\vec{b}|$
Thus, the vector $\vec{a} + \vec{b}$ bisects the angle between the non-collinear vectors $\vec{a}$ and $\vec{b}$ if they are equal.

Question:35

Fill in the blanks
If $\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{a}}=0, \overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{b}}=0,$ and $\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{c}}=0$ for some non-zero vector $\overrightarrow{\mathrm{r}}$ , then the value of $\overrightarrow{\mathrm{a}} \cdot(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})$ is _________

Answer:

Given that, $\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{a}}=0, \overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{b}}=0,$ and $\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{c}}=0$ for some non-zero vector $\overrightarrow{\mathrm{r}}$
$\\ \Rightarrow \vec{r} $ is perpendicular to $\vec{a}, \vec{b}$ and $\vec{c}$\\ $\Rightarrow \vec{a}, \vec{b}$ and $\vec{c}$ are coplanar.\\ $\Rightarrow \vec{a} \cdot(\vec{b} \times \vec{c})=0$\\$

Question:36

Fill in the blanks
The vectors $\overrightarrow{\mathrm{a}}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}} \text { and } \overrightarrow{\mathrm{b}}=-\hat{\mathrm{i}}-2 \hat{\mathrm{k}}$ are the adjacent sides of a parallelogram. The acute angel between its diagonals is ____________.

Answer:

Given that , $\overrightarrow{\mathrm{a}}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}} \text { and } \overrightarrow{\mathrm{b}}=-\hat{\mathrm{i}}-2 \hat{\mathrm{k}}$
$\\ Let \overrightarrow{d_{1}}$ and $\overrightarrow{d_{2}}$ are two diagonals of parallelogram.\\ $$ \Rightarrow \overrightarrow{d_{1}}=\vec{a}+\vec{b}=(3 \hat{\imath}-2 \hat{\jmath}+2 \hat{k})+(-\hat{\imath}-2 \hat{k})=(3-1) \hat{\imath}+(-2+0) \hat{\jmath}+(2-2) \hat{k}\\ $$ $\Rightarrow \overrightarrow{d_{1}}=2 \hat{\imath}-2 \hat{j}$$ $\\ \Rightarrow \overrightarrow{d_{2}}=\vec{a}-\vec{b}=(3 \hat{\imath}-2 \hat{\jmath}+2 \hat{k})-(-\hat{\imath}-2 \hat{k})=(3+1) \hat{\imath}+(-2-0) \hat{\jmath}+(2+2) \hat{k}$ $\Rightarrow \overrightarrow{d_{2}}=4 \hat{\imath}-2 \hat{\jmath}+4 \hat{k}$ \\Let $\theta$ be the angle between diagonals $\overrightarrow{d_{1}}$ and $\overrightarrow{d_{2}}$. \\Then, $\overrightarrow{d_{1}} \cdot \overrightarrow{d_{2}}=\left|\overrightarrow{d_{1}} \| \overrightarrow{d_{2}}\right| \cos \theta$ \\$\Rightarrow \cos \theta=\frac{\overrightarrow{d_{1}} \cdot \overrightarrow{d_{2}}}{\left|\overline{d_{1}}\right|\left|\overline{d_{2}}\right|}$$ $\\ \cos \theta=\frac{(2 \hat{\imath}-2 \hat{\jmath}) \cdot(4 \hat{i}-2 \hat{\jmath}+4 \hat{k})}{\sqrt{2^{2}+2^{2}} \sqrt{4^{2}+(-2)^{2}+4^{2}}}=\frac{8+4}{\sqrt{8} \sqrt{16+4+16}}(\because \hat{\imath} . \hat{\jmath}=0, \hat{\jmath} \cdot \hat{k}=0, \hat{k} \cdot \hat{\imath}=0) \\ \Rightarrow \cos \theta=\frac{12}{2 \sqrt{2} .6}=\frac{1}{\sqrt{2}} \\ \Rightarrow \theta=\frac{\pi}{4}\left(\because \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}\right)$

Question:37

Fill in the blanks
The values of k for which $|k \vec{a}|<|\vec{a}| \text { and } k \vec{a}+\frac{1}{2} \vec{a}$ is parallel to $\vec{a}$ holds true are _______.

Answer:

Given that, $|k \vec{a}|<|\vec{a}|$
$\\ \begin{aligned} &\Rightarrow|k||\vec{a}|<|\vec{a}|\\ &\Rightarrow|k|<1\\ &\Rightarrow-1<k<1\\ &\text { Also, }\\ &k \vec{a}+\frac{1}{2} \vec{a} \text { is parallel to } \vec{a} \end{aligned}$
⇒ k cannot be equal to $-\frac{1}{2}$ , otherwise it will become null vector and then it will not be parallel to $\vec{a}$ .
Since, k is along the direction of $\vec{a}$ and not in its opposite direction.
$\therefore \mathrm{k} \in(-1,1)-\left\{-\frac{1}{2}\right\}$

Question:38

Fill in the blanks
The value of the expression $|\vec{a} \times \vec{b}|^{2}+(\vec{a} \cdot \vec{b})^{2}$ is ______.

Answer:

$\\ \text { We have, }|\vec{a} \times \vec{b}|^{2}+(\vec{a} \cdot \vec{b})^{2}=|\vec{a}|^{2}|\vec{b}|^{2} \sin ^{2} \theta+(\vec{a} \cdot \vec{b})^{2} \\ =|\vec{a}|^{2}|\vec{b}|^{2} \sin ^{2} \theta+(\vec{a} \cdot \vec{b})^{2}=|\vec{a}|^{2}|\vec{b}|^{2}\left(1-\cos ^{2} \theta\right)+(\vec{a} \cdot \vec{b})^{2} \\ =|\vec{a}|^{2}|\vec{b}|^{2}-|\vec{a}|^{2}|\vec{b}|^{2} \cos ^{2} \theta+(\vec{a} \cdot \vec{b})^{2}$
$\\ =|\vec{a}|^{2}|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})^{2}+(\vec{a} \cdot \vec{b})^{2} \\ =|\vec{a}|^{2}|\vec{b}|^{2} \\ \text { Thus, }|\vec{a} \times \vec{b}|^{2}+(\vec{a} \cdot \vec{b})^{2}=|\vec{a}|^{2}|\vec{b}|^{2}$

Question:39

if $\left | \vec{a} \times \vec{b} \right |^{2}+\left | (\vec{a} \cdot \vec{b}) \right |^{2}=144$ and $\left |\vec{a} \right |=4$ then $\left |\vec{b} \right |$ is equal to ________.

Answer:

Given that, $\left | \vec{a} \times \vec{b} \right |^{2}+\left | (\vec{a} \cdot \vec{b}) \right |^{2}=144$ and $\left |\vec{a} \right |=4$
$\\|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=144 \\ \Rightarrow|\vec{a}|^{2}|\vec{b}|^{2} \sin ^{2} \theta+|\vec{a}|^{2}|\vec{b}|^{2} \cos ^{2} \theta=144 \\ \Rightarrow|\vec{a}|^{2}|\vec{b}|^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=144$
$\\ \Rightarrow|\vec{a}|^{2}|\vec{b}|^{2}(1)=144 \\ \Rightarrow|\vec{a}|^{2}|\vec{b}|^{2}=144 \\ \Rightarrow|\vec{a}||\vec{b}|=12 \\ \Rightarrow \quad 4 \cdot|\vec{b}|=12 \\ \Rightarrow|\vec{b}|=3$

Question:40

Fill in the blanks
If $\vec{\mathrm{a}}$ is any non-zero vector, then $(\vec{\mathrm{a}} \cdot \hat{\mathrm{i}}) \hat{\mathrm{i}}+(\vec{\mathrm{a}} \cdot \hat{\mathrm{j}}) \hat{\mathrm{j}}+(\vec{\mathrm{a}} \cdot \hat{\mathrm{k}}) \hat{\mathrm{k}}$ equals ______.

Answer:

$\\ \begin{aligned} &\text { Let } \vec{a}=a_{1} \hat{\imath}+a_{2} \hat{\jmath}+a_{3} \hat{k}\\ &\text { Now, taking dot product of } \vec{a} \text { with } \hat{\imath}, \text { we get }\\ &\vec{a} . \hat{\imath}=\left(a_{1} \hat{\imath}+a_{2} \hat{\jmath}+a_{3} \hat{k}\right) . \hat{\imath}=a_{1} \hat{\imath} . \hat{\imath}+a_{2} \hat{\jmath} . \hat{\imath}+a_{3} \hat{k} . \hat{\imath}\\ &\Rightarrow \vec{a} . \hat{\imath}=a_{1} \hat{\imath} . \hat{\imath}+a_{2} \cdot 0+a_{3} \cdot 0(\because \hat{\jmath} \cdot \hat{\imath}=\hat{k} \cdot \hat{\imath}=0) \end{aligned}$
$\\ \begin{aligned} &\Rightarrow \vec{a} . \hat{\imath}=a_{1} \hat{\imath} . \hat{\imath}+a_{2} .0+a_{3} .0(\because \hat{\jmath} . \hat{\imath}=\hat{k} . \hat{\imath}=0)\\ &\Rightarrow \vec{a} \cdot \hat{\imath}=a_{1}\\ &\text { Similarly, taking dot product of } \vec{a} \text { with } \hat{\jmath} \text { and } \hat{k} \text { , we get }\\ &\vec{a} . \hat{\jmath}=a_{2} \text { and } \vec{a} . \hat{k}=a_{3} \\&\Rightarrow(\vec{a} \cdot \hat{\imath}) \hat{\imath}+(\vec{a} . \hat{\jmath}) \hat{\jmath}+(\vec{a} . \hat{k}) \hat{k}=a_{1} \hat{\imath}+a_{2} \hat{\jmath}+a_{3} \hat{k}=\vec{a} \end{aligned}$

Question:41

True and False
If $|\vec{a}|=|\vec{b}|$ , then necessarily at implies $\vec{a}=\pm\vec{b}$ .

Answer:

False
Explanation:
$\\ \begin{aligned} &\text { Let } \vec{a}=\hat{\imath}-2 \hat{\jmath}-3 \hat{k} \text { and } \vec{b}=\hat{\imath}+2 \hat{\jmath}+3 \hat{k}\\ &\Rightarrow|\vec{a}|=\sqrt{(1)^{2}+(-2)^{2}+(-3)^{2}}=\sqrt{1+4+9}=\sqrt{14}\\ &\text { and }|\vec{b}|=\sqrt{(1)^{2}+(2)^{2}+(3)^{2}}=\sqrt{1+4+9}=\sqrt{14}\\ &\text { Now, we observe that }\\ &|\vec{a}|=|\vec{b}| \text { but } \vec{a} \neq \vec{b} \end{aligned}$

Question:42

True and False
Position vector of a point P is a vector whose initial point is origin.

Answer:

True
Explanation:
Consider a point P in space, having coordinates (x, y, z) with respect to the origin O(0, 0, 0). Then, the vector $\vec{OP}$ having O and P as its initial and terminal points, respectively, is called the position vector of the point P with respect to O.

Question:43

True and False
If $|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|$ , then the vectors $\vec{a}$ and $\vec{b}$ are orthogonal.

Answer:

True
Explanation:
Given that, $|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|$
On squaring both the sides, we get
$\\ \Rightarrow|\vec{a}+\vec{b}|^{2}=|\vec{a}-\vec{b}|^{2} \\ \Rightarrow \vec{a}^{2}+2 \vec{a} \cdot \vec{b}+\vec{b}^{2}=\vec{a}^{2}-2 \vec{a} \cdot \vec{b}+\vec{b}^{2} \\ \Rightarrow 2 \vec{a} \cdot \vec{b}=-2 \vec{a} \cdot \vec{b} \\ \Rightarrow 2 \vec{a} \cdot \vec{b}+2 \vec{a} \cdot \vec{b}=0 \\ \Rightarrow 4 \vec{a} \cdot \vec{b}=0\\ \Rightarrow \vec{a} \cdot \vec{b}=0$
Hence, $\vec{a}$ and $\vec{b}$ are orthogonal.

Question:44

True and False
The formula $(\vec{a}+\vec{b})^{2}=\vec{a}^{2}+\vec{b}^{2}+2 \vec{a} \times \vec{b}$ is valid for non-zero vectors $\vec{a} \text{ and } \vec{b}$

Answer:

False
Explanation:
$\\ (\vec{a}+\vec{\mathrm{b}})^{2}=(\vec{\mathrm{a}}+\vec{\mathrm{b}}) \cdot(\vec{\mathrm{a}}+\vec{\mathrm{b}}) \\ \Rightarrow \vec{\mathrm{a}}^{2}+\vec{\mathrm{a}} \cdot \vec{\mathrm{b}}+\vec{\mathrm{b}} \cdot \vec{\mathrm{a}}+\vec{\mathrm{b}}^{2} \\ \Rightarrow \vec{\mathrm{a}}^{2}+\vec{\mathrm{a}} \cdot \vec{\mathrm{b}}+\vec{\mathrm{a}} \cdot \vec{\mathrm{b}}+\vec{\mathrm{b}}^{2}(\because \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}=\vec{\mathrm{b}} . \vec{\mathrm{a}}) \\ \Rightarrow \vec{\mathrm{a}}^{2}+2 \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}+\vec{\mathrm{b}}^{2}$

Question:45

True and False
If $\vec{a}.$ and $\vec{b}$ are adjacent sides of a rhombus, then $\vec{a}.\vec{b}=0$ .

Answer:

False
Explanation:
Given that, $\vec{\mathrm{a}} \cdot \vec{\mathrm{b}}=0 \Rightarrow \vec{\mathrm{a}} \text { and } \vec{\mathrm{b}}$ are perpendicular to each other.
But, adjacent sides of rhombus are not perpendicular.

More About NCERT Exemplar Class 12 Maths Solutions Chapter 10

NCERT exemplar Class 12 Maths chapter 10 solutions would help achieve academic excellence, be it in the 12 board exams or additional exams, get rid of doubts in fundamental concepts, and promote intellectual curiosity. The students can also use NCERT exemplar Class 12 Maths solutions chapter 10 PDF download, to access the solutions offline.

Sub-topics Covered in NCERT exemplar solutions for Class 12 Maths chapter 10 Vector Algebra

The sub-topics that are covered under the Class 12 Maths NCERT exemplar solutions chapter 10 Vector Algebra are:

Basic concepts
Directed line
Terminal point
Magnitude
Position Vectors
Direction cosines
Types of Vectors
Zero Vectors
Unit Vectors
Co-initial Vectors
Collinear Vectors
Equal Vectors
Negative of a Vector
Addition of Vectors
Properties of Vector Addition
Multiplication of a Vector by a Scalar
Components of a Vector
Vector joining two points
Section formula
Product of two Vectors
Scalar(or dot) product
Projection of a Vector on a line
Vector(or cross) product

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What can you learn in NCERT Exemplar Class 12 Maths Chapter 10 Solutions?

Class 12 Maths NCERT exemplar solutions chapter 10 explore the real-world concepts and applications of Vectors. Vectors have several real-world uses in Aerospace, Fluid concepts, Complex calculations, sports, etc.
They have applications in electromagnetism, hydrodynamics, blood flow, calculation of the distance between aircraft in space and the angle between their paths.
Their broad range of reach also includes setting up Solar panels considering the roof's location and the direction of the sunbeams.
Vectors are used to launch satellites, calculate the distance between their panels, find their prospective trajectory, develop new targeting techniques in warfare, count on-location services like GPS, calculate the ball's trajectory on a field, and a thousand other areas. Knowledge of all the aspects of Vector algebra can help one do well in academics and find out how the world works.

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NCERT Exemplar Class 12 Maths Solutions

Chapter 1 Relations and Functions

Chapter 2 Inverse Trigonometric Functions

Chapter 3 Matrices

Chapter 4 Determinants

Chapter 5 Continuity and Differentiability

Chapter 6 Applications of Derivatives

Chapter 7 Integrals

Chapter 8 Applications of Integrals

Chapter 9 Differential Equations

Chapter 11 Three dimensional Geometry

Chapter 12 Linear Programming

Chapter 13 Probability

Benefits of NCERT Exemplar Class 12 Maths Chapter 10 Solutions

NCERT exemplar solutions for Class 12 Maths chapter 10 define Vector algebra as a type of algebra that involves vector form elements (having magnitude and direction), and their algebraic expression follows vector laws.
If quantities revolving around velocity, displacement, acceleration, force, weight, spark immediate curiosity in you, Class 12 Maths NCERT exemplar solutions chapter 10 will be fun
In this chapter, we will explore some concepts of vectors, the basic algebraic operations of vector and scalar addition and multiplication, their properties related to vector algebra, the geometric properties, the properties to be followed for the proper visualisation of vectors, and their applications and uses in the modern world.

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NCERT Exemplar Class 12 Solutions

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Also, check NCERT Solutions for questions given in the book:

Chapter 1	Relations and Functions
Chapter 2	Inverse Trigonometric Functions
Chapter 3	Matrices
Chapter 4	Determinants
Chapter 5	Continuity and Differentiability
Chapter 6	Application of Derivatives
Chapter 7	Integrals
Chapter 8	Application of Integrals
Chapter 9	Differential Equations
Chapter 10	Vector Algebra
Chapter 11	Three Dimensional Geometry
Chapter 12	Linear Programming
Chapter 13	Probability

Must Read NCERT Solution subject wise

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As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

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Frequently Asked Question (FAQs)

1. What are the topics covered in the chapter?

This basic maths chapter covers the topics like vectors, properties, vector by scalar, types of vector etc.

2. Who can benefit from these solutions?

Students appearing for the 12 board exams and those who are preparing from entrance exams for engineering can make use of NCERT exemplar Class 12 Maths solutions chapter 10.

3. How to download these solutions?

One can use NCERT exemplar Class 12 Maths solutions chapter 10 pdf download by an online webpage to pdf converter

4. How many questions are solved in these solutions?

Our experts have covered all the questions and have given NCERT exemplar solutions for Class 12 Maths chapter 10 for main exercise of the chapter.

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NCERT Exemplar Class 12 Physics Solutions Chapter 8 Electromagnetic Waves

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Read Complete Answer

Ashvadha 12 July,2024

Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Read Complete Answer

kumardurgesh1802 10 July,2024

I have completed my 12th with arts cbse. I want to repeat my 12th with pcmb. How to become a private candidate and what are deadline to be able to appear in 2025 exam?

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
You have to appear for the 2025 12th board exams.
Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
Aim to register before late October to avoid extra fees.
Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

Good luck with your studies!

Read Complete Answer

kumardurgesh1802 10 July,2024

I passed my 12th cbse boards in 2023 , and first time appear for jee mains and advance in 2024 so , am i eligible for 2025 jee mains and advance ?

Hello Aspirant , Hope your doing great . As per your query , your eligible for JEE mains in the year of 2025 , Every candidate can appear for the JEE Main exam 6 times over three consecutive years . The JEE Main exam is held two times every year, in January and April.

Read Complete Answer

satyamercedes0626 23 July,2024

I've passed my class 12th board exam from cbse in PCB in 2024. Now I'll give mathematics exam from cbse as a additional subject in 2025. However, they'll provide separate marksheet for maths. Will i be eligible for jee?? And how I'll fill the form??

Hi there,

Hope you are doing fine

Yes you are certainly eligible for giving the jee exam in the year 2025. You must pass the maths exam with at least 75% criteria as required by jee and provide the marksheet and the passing certificate while registering for the exam.

Pursuing maths as an additional subject while taking biology as your main subject does not offer any hindrance in you appearing for the jee examination. It is indeed an privilege to pursue both maths and biology as the subjects and prepare for the same.

There will be no issue in filling the form while registering for the exam as it will only require your basic details and marksheet which you can provide by attaching the marksheet of maths also. Also, a detailed roadmap is also available on the official websites on how to fill the registration form. So you can fill the form easily.

Hope this resolves your query.

Read Complete Answer

Abhishek 12 July,2024

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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NCERT Exemplar Class 12 Maths Solutions Chapter 10 Vector Algebra

More About NCERT Exemplar Class 12 Maths Solutions Chapter 10

Applications for Admissions are open.

What can you learn in NCERT Exemplar Class 12 Maths Chapter 10 Solutions?

Tallentex 2025 - ALLEN's Talent Encouragement Exam

Aakash iACST Scholarship Test 2024

NCERT Exemplar Class 12 Maths Solutions

Benefits of NCERT Exemplar Class 12 Maths Chapter 10 Solutions

NCERT Exemplar Class 12 Solutions

Also, check NCERT Solutions for questions given in the book:

Must Read NCERT Solution subject wise

JEE Main Important Mathematics Formulas

Read more NCERT Notes subject wise

Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Question (FAQs)

Also Read

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Upcoming School Exams