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

Komal MiglaniUpdated on 31 Mar 2025, 04:28 AM IST

Have you ever wondered how aeroplanes change direction in the skies or how the forces act in physics? Welcome to the world of vectors, where quantities have both magnitude and direction. Vector algebra deals with mathematical operations on vectors which are having both magnitude and direction, like the addition and subtraction of two vectors, multiplication of two vectors. Also, in this chapter, you will find problems based on angles between two vectors, perpendicular distance between two vectors etc. This article covers all the important class 12 maths chapter 10 question answers based on the latest NCERT syllabus for class 12 and based on Vector Algebra from NCERT Books for class 12 Math.

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These solutions are created by our subject matter experts covering the complete solution of the 10th Chapter of the NCERT Exemplar for Class 12 Mathematics. You will find a healthy number of questions to practice here. Vectors are discussed in detail in NCERT Exemplar Class 12 Maths Solutions Chapter 10.

Class 12 Maths Chapter 10 Exemplar Solutions Exercise: 10.1
Page number: 215-219
Total questions: 45

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,
$\vec{a}=2 \hat{\imath}-\hat{\jmath}+\hat{k}$
$\overrightarrow{\mathrm{b}}=2 \hat{\jmath}+\hat{\mathrm{k}}$

Since, unit vector is needed to be found in the direction of the sum of vectors $\vec{a}$ and $\vec{b}$.
So, add vectors $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$.
Let,
$\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}$

Substituting the values of vectors $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$.
$\Rightarrow \vec{c}=(2 \hat{\imath}-\hat{\jmath}+\hat{\mathrm{k}})+(2 \hat{\jmath}+\hat{\mathrm{k}})$

$\Rightarrow \overrightarrow{\mathrm{c}}=2 \hat{\imath}-\hat{\jmath}+\hat{\mathrm{k}}+2 \hat{\jmath}+\hat{\mathrm{k}}$

$\Rightarrow \overrightarrow{\mathrm{c}}=2 \hat{\imath}$

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,

$\overrightarrow{\mathrm{b}}=2 \hat{\imath}+\hat{\jmath}-2 \hat{\mathrm{k}}$

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 $2 \vec{a}-\vec{b}$, so, we just need to substitute in the above equation.
$
\begin{aligned}
& \Rightarrow 2 \hat{a}-\hat{b}=\frac{\hat{\jmath}+6 \hat{k}}{|\hat{\jmath}+6 \hat{k}|} \\
& \text { Here },|\hat{\jmath}+6 \hat{k}|=\sqrt{1^2+6^2} \\
& \Rightarrow 2 \hat{a}-\hat{b}=\frac{\hat{\jmath}+6 \hat{k}}{\sqrt{1^2+6^2}}
\end{aligned}
$

$\begin{aligned} &\Rightarrow 2 \hat{a}-\hat{b}=\frac{\hat{\jmath}+6 \hat{\mathrm{k}}}{\sqrt{1+36}}\\ &\Rightarrow 2 \hat{a}-\hat{b}=\frac{\hat{\jmath}+6 \hat{k}}{\sqrt{37}}\\ &\text { Thus, unit vector in the direction of }\\ &2 \vec{a}-\vec{b} {i s} \frac{\hat{\jmath}+6 \hat{k}}{\sqrt{37}} \end{aligned}$

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 $=\vec{a}$

$\Rightarrow \overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{a}}$

Position vector of $B=\overrightarrow{\mathrm{b}}$
$\Rightarrow \overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{b}}$

Here, O is the origin.
We need to find the position vector of C ,
that is, $\overrightarrow{\mathrm{OC}}$
Also, we have
$\overrightarrow{\mathrm{BC}}=1.5 \overrightarrow{\mathrm{BA}} {\ldots(\mathrm{i})}$

Here, $\overrightarrow{\mathrm{BC}}=$ Positionvectorof $\mathrm{C}-$ Position vector of B
$\Rightarrow \overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}} \ldots$

And, $\overrightarrow{\mathrm{BA}}=$ Positionvectorof $A-$ Position vector of B
$\Rightarrow \overrightarrow{\mathrm{BA}}=\overrightarrow{\mathrm{OA}}-\overrightarrow{\mathrm{OB}}$

$\\ \overrightarrow{\mathrm{BC}}=1.5 \overrightarrow{\mathrm{BA}}$

$\Rightarrow \overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}}=1.5(\overrightarrow{\mathrm{OA}}-\overrightarrow{\mathrm{OB}})$

$\Rightarrow \overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}}=1.5 \overrightarrow{\mathrm{OA}}-1.5 \overrightarrow{\mathrm{OB}}$

$\Rightarrow \overrightarrow{\mathrm{OC}}=1.5 \overrightarrow{\mathrm{OA}}-1.5 \overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{OB}}$
$\\ \begin{aligned} &\Rightarrow \overrightarrow{\mathrm{OC}}=1.5 \overrightarrow{\mathrm{OA}}-0.5 \overrightarrow{\mathrm{OB}}\\ &\Rightarrow \overrightarrow{\mathrm{OC}}=1.5 \overrightarrow{\mathrm{a}}-0.5 \overrightarrow{\mathrm{b}}\\ &[\because \text { it is given that } \overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{a}} \text { and } \overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{b}}]\\ &\Rightarrow \overrightarrow{\mathrm{OC}}=\frac{15 \overrightarrow{\mathrm{a}}}{10}-\frac{5 \overrightarrow{\mathrm{b}}}{10} \end{aligned}$
$\Rightarrow \overrightarrow{\mathrm{OC}}=\frac{3 \overrightarrow{\mathrm{a}}}{2}-\frac{\overrightarrow{\mathrm{b}}}{2}$

$\Rightarrow \overrightarrow{\mathrm{OC}}=\frac{3 \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}}{2}$
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,
$\overrightarrow{\mathrm{OB}}=\hat{\imath}-\hat{\jmath}+3 \hat{\mathrm{k}}$

Position vector of C is given by,

$\vec { OC } = 3 \hat { i } + 5 \hat { j } + 3 \hat { k }$
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 $|\overrightarrow{\mathrm{AB}}|,|\overrightarrow{\mathrm{BC}}|$ 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
$\begin{aligned}& \Rightarrow \sqrt{(3-\mathrm{k})^2+225}\sqrt{40}=\sqrt{(1-\mathrm{k})^2+81} \\& \Rightarrow \sqrt{\left(9+\mathrm{k}^2-6 \mathrm{k}\right)+225}-\sqrt{40}=\sqrt{\left(1+\mathrm{k}^2-2 \mathrm{k}\right)+81}\end{aligned}$
$\left[\because\right.$ by algebraic identity, $\left.(a-b)^2=a^2+b^2-2 a b\right]$
$\begin{aligned}& \Rightarrow \sqrt{\mathrm{k}^2-6 \mathrm{k}+225+9}-\sqrt{40}=\sqrt{\mathrm{k}^2-2 \mathrm{k}+81+1} \\& \Rightarrow \sqrt{\mathrm{k}^2-6 \mathrm{k}+234}-\sqrt{40}=\sqrt{\mathrm{k}^2-2 \mathrm{k}+82}\end{aligned}$

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,
$\\ \hat{\mathrm{r}}=l \hat{\mathrm{i}}+\mathrm{m} \hat{\mathrm{j}}+\mathrm{n} \hat{\mathrm{k}}\\ \Rightarrow \hat{\mathrm{r}}=\pm \frac{1}{\sqrt{3}} \hat{\imath} \pm \frac{1}{\sqrt{3}} \hat{\jmath} \pm \frac{1}{\sqrt{3}} \hat{\mathrm{k}}$

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 vector 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 r is given by, vector $=6 \times \frac{\vec{r}}{\mid \vec{r}}$ $\Rightarrow$ vector $=6 \times \frac{-\hat{\imath}+2 \hat{\jmath}+2 \hat{k}}{|-\hat{l}+2 \hat{j}+2 \hat{k}|}$
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, parallelograms on the same base and between the same parallels are equal in area, so we have

$|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}|=|\overrightarrow{\mathrm{b}} \times \vec{\rightarrow} \overrightarrow{\mathrm{c}}| \rightarrow$

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
$\\ \begin{aligned} &=\hat{\imath}+\hat{\jmath}-\hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OA}}=\hat{\imath}+\hat{\jmath}-\hat{\mathrm{k}}\\ &\text { Position vector of } \mathrm{B}=2 \hat{\imath}-\hat{\jmath}+3 \hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OB}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}} \end{aligned}$
$\\ \begin{aligned} &\text { Position vector of } c=2 \hat{\imath}-3 \hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OC}}=2 \hat{\mathrm{\imath}}-3 \hat{\mathrm{k}}\\ &\text { Position vector of } D=3 \hat{\imath}-2 \hat{\jmath}+\hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OD}}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \end{aligned}$

$\begin{aligned} & \text { Now, let us find out } \overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{CD}} \\ & \Rightarrow \overrightarrow{\mathrm{AB}}=\text { position vector of } B-\text { Position vector of } \mathrm{A} \\ & \Rightarrow \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}} \\ & \Rightarrow \overrightarrow{\mathrm{AB}}=(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}})-(\hat{\imath}+\hat{\mathrm{j}}-\hat{\mathrm{k}}) \\ & \Rightarrow \overrightarrow{\mathrm{AB}}=2 \hat{\mathrm{i}}-\hat{1}-\hat{\mathrm{j}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}+\hat{\mathrm{k}} \\ & \Rightarrow \overrightarrow{\mathrm{AB}}=\hat{\imath}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \\ & \text { And, } \overrightarrow{\mathrm{CD}}=\text { position vector of } D-\text { Position vector of } \mathrm{C} \\ & \Rightarrow \overrightarrow{\mathrm{CD}}=\overrightarrow{\mathrm{OD}}-\overrightarrow{\mathrm{OC}} \\ & \Rightarrow \overrightarrow{\mathrm{CD}}=(3 \hat{\imath}-2 \hat{\jmath}+\hat{\mathrm{k}})-(2 \hat{\imath}-3 \hat{\mathrm{k}}) \\ & \Rightarrow \overrightarrow{\mathrm{CD}}=3 \hat{\imath}-2 \hat{\imath}-2 \hat{\jmath}+\hat{\mathrm{k}}+3 \hat{\mathrm{k}}\end{aligned}$

$\\ \begin{aligned} &\Rightarrow \overrightarrow{\mathrm{CD}}=\hat{\imath}-2 \hat{\jmath}+4 \hat{\mathrm{k}}\\ &\text { The projection of } \overrightarrow{\mathrm{AB}} \text { along } \overrightarrow{\mathrm{CD}} \text { is given by, }\\ &\text { Projection }=\frac{\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{CD}}}{|\overrightarrow{\mathrm{CD}}|} \end{aligned}$
$\\ \begin{aligned} &\Rightarrow \text { Projection }=\frac{(\hat{\imath}-2 \hat{\jmath}+4 \hat{\mathrm{k}})(\hat{\imath}-2 \hat{\jmath}+4 \hat{\mathrm{k}})}{|\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}|}\\ &\Rightarrow \text { Projection }=\frac{(\hat{\imath})(\hat{i})+(-2 \hat{\jmath})(-2 \hat{\jmath})+(4 \hat{\mathrm{k}})(4 \hat{\mathrm{k}})}{\sqrt{1^{2}+(-2)^{2}+4^{2}}}\\ &\text { "we know that, } \hat{i} \times \hat{i}=\hat{\jmath} \times \hat{\jmath}=\hat{\mathrm{k}} \times \hat{\mathrm{k}}=1 \text { and } \hat{i} \times \hat{\mathrm{j}}=\hat{i} \times \hat{\mathrm{k}}=\hat{\mathrm{j}} \times \hat{\mathrm{k}}=0 \end{aligned}$
$\\ \begin{aligned} &\text { So }\\ &\left(a_{1} \hat{\imath}+a_{2} \hat{\jmath}+a_{3} \hat{k}\right)\left(b_{1} \hat{\imath}+b_{2} \hat{\jmath}+b_{3} \hat{k}\right)=a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}\\ &\text { Also, we know that, } \left.|a \hat{\imath}+b \hat{\jmath}+c \hat{k}|=\sqrt{a^{2}+b^{2}+c^{2}}\right]\\ &\Rightarrow \text { Projection }=\frac{1+4+16}{\sqrt{1+4+16}}\\ &\Rightarrow \text { Projection }=\frac{21}{\sqrt{21}} \end{aligned}$
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,
$\\ \begin{aligned} &\text { Area of } \Delta \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|_{\ldots \text { (i) }}\\ &\text { Let us find out } \overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{AC}} \text { first. } \end{aligned}$
$\\ \begin{aligned} &\text { We can say, }\\ &\text { Position vector of } \mathrm{A}=\hat{i}+2 \hat{\jmath}+3 \hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OA}}=\hat{\imath}+2 \hat{\jmath}+3 \hat{\mathrm{k}}\\ &\text { Position vector of } \mathrm{B}=2 \hat{\mathrm{l}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OB}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}} \end{aligned}$
$\\ \begin{aligned} &\text { Position vector of } \mathrm{c}=4 \hat{\imath}+5 \hat{\jmath}-\hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{OC}}=4 \hat{\imath}+5 \hat{\jmath}-\hat{\mathrm{k}}\\ &\overrightarrow{\mathrm{AB}}\\ &\overrightarrow{\mathrm{AB}}=\text { position vector of B-Position vector of } \mathrm{A} \end{aligned}$
$\\ \Rightarrow \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}} \\ \Rightarrow \overrightarrow{\mathrm{AB}}=(2 \hat{\imath}-\hat{\jmath}+4 \hat{\mathrm{k}})-(\hat{\imath}+2 \hat{\jmath}+3 \hat{\mathrm{k}}) \\ \Rightarrow \overrightarrow{\mathrm{AB}}=2 \hat{\imath}-\hat{\imath}-\hat{\jmath}-2 \hat{\jmath}+4 \hat{\mathrm{k}}-3 \hat{\mathrm{k}} \\ \Rightarrow \overrightarrow{\mathrm{AB}}=\hat{i}-3 \hat{\jmath}+\hat{\mathrm{k}} \\ \text { For } \overrightarrow{\mathrm{AC}}$
$\\ \begin{aligned} &\overrightarrow{\mathrm{AC}}=\text { position vector of } \mathrm{C} \text { -Position vector of } \mathrm{A}\\ &\Rightarrow \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}}\\ &\Rightarrow \overrightarrow{\mathrm{AC}}=(4 \hat{\imath}+5 \hat{\jmath}-\hat{\mathrm{k}})-(\hat{\imath}+2 \hat{\jmath}+3 \hat{\mathrm{k}})\\ &\Rightarrow \overrightarrow{\mathrm{AC}}=4 \hat{\imath}-\hat{\imath}+5 \hat{\jmath}-2 \hat{\jmath}-\hat{\mathrm{k}}-3 \hat{\mathrm{k}}\\ &\Rightarrow \overrightarrow{\mathrm{AC}}=3 \hat{\imath}+3 \hat{\jmath}-4 \hat{\mathrm{k}} \end{aligned}$
$\\ \begin{aligned} &\Rightarrow \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=\hat{\imath}(12-3)-\hat{\jmath}(-4-3)+\hat{\mathrm{k}}(3+9)\\ &\Rightarrow \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=9 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}\\ &\text { And, }\\ &|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=|9 \hat{\imath}+7 \hat{\jmath}+12 \hat{\mathrm{k}}|\\ &\Rightarrow|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=\sqrt{9^{2}+7^{2}+12^{2}} \end{aligned}$
$\\ \Rightarrow|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=\sqrt{81+49+144} \Rightarrow|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=\sqrt{274}$

$\begin{gathered}\text {From equation }(i) \text {, weget } \\ \text { Area of } \triangle \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}| \\ \Rightarrow \text { Areaof } \triangle \mathrm{ABC}=\frac{1}{2} \times \sqrt{274}\end{gathered}$

Thus, area of triangle 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
$\\ \sin \mathrm{A}=\frac{\text { perpendicular }}{\text { hypotenuse }} \text { in } \Delta \mathrm{BAD} \\ \qquad \because \sin \mathrm{A}=\frac{\mathrm{BD}}{\mathrm{c}} \\ \Rightarrow \mathrm{BD}=\mathrm{c} \sin \mathrm{A}$
$\\ \text { And, } A D=c \cos A \\ \qquad \cos A=\frac{\text { base }}{\text { hypotenuse }} \text { in } \Delta B A D \\ \because \\ \Rightarrow \cos A=\frac{A D}{C} \\ \Rightarrow A D=c \cos A$

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

$
\begin{aligned}
& (B C)^2=(B D)^2+(C D)^2 \\
& \Rightarrow a^2=(c \sin A)^2+(C D)^2[\because \text { from the diagram, } B D=\text { csin } A] \\
& \Rightarrow a^2=c^2 \sin ^2 A+(b-\cos A)^2 \\
& \because \text { fromthediagram, } A C=C D+A D \\
& \Rightarrow C D=A C-A D \\
& \Rightarrow C D=b-\cos A] \\
& \Rightarrow a^2=c^2 \sin ^2 A+\left(b^2+(-\cos A)^2-2 b \cos A\right)[\because \text { from algebraic identity, }(a- \\
& \left.b)^2=a^2+b^2-2 a b\right] \\
& \Rightarrow a^2=c^2 \sin ^2 A+b^2+c^2 \cos ^2 A-2 b c \cos A \\
& \Rightarrow a^2=c^2 \sin ^2 A+c^2 \cos ^2 A+b^2-2 b c \cos A \\
& \Rightarrow a^2=c^2\left(\sin ^2 A+\cos ^2 A\right)+b^2-2 b c \cos A \\
& \Rightarrow a^2=c^2+b^2-2 b c \cos ^2\left[\because \text { from trigonometric identity, } \sin ^2 \theta+\cos ^2 \theta=1\right] \\
& \Rightarrow 2 b c \cos A=c^2+b^2-a^2 \\
& \Rightarrow 2 b c \cos A=b^2+c^2-a^2 \\
& \Rightarrow \cos A=\frac{b^2+c^2-a^2}{2 b c}
\end{aligned}
$

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,

$\begin{aligned} & \text { Areaof } \triangle \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}| \\ & \text { Here } \overrightarrow{\mathrm{AB}}=\text { Positionvectorof } \mathrm{B}-\text { Positionvectorof } \mathrm{A} \\ & \Rightarrow \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}\end{aligned}$

$\\ \Rightarrow \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}} \\ \therefore \text { Area of } \Delta \mathrm{ABC}=\frac{1}{2}|(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}) \times(\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}})| \\ \Rightarrow \text { Area of } \Delta \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{a}}| \\ {[\because-\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}},-\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}} \text { and } \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{a}}=0]}$
$\Rightarrow \text { Area of } \Delta \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|_{\ldots(\mathrm{j})}$
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 $\triangle \mathrm{ABC}$ should be equal to 0.
⇒ Area of $\triangle \mathrm{ABC} = 0$
$\\ \Rightarrow \frac{1}{2}|\vec{b} \times \vec{c}+\vec{c} \times \vec{a}+\vec{a} \times \vec{b}|=0 \\ \Rightarrow \vec{b} \times \vec{c}+\vec{c} \times \vec{a}+\vec{a} \times \vec{b}=0$
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, $\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$ from equation (i) And, $|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|{\text {from equation (i) }}$
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{\vec{b} \times \vec{c}+\vec{c} \times \vec{a}+\vec{a} \times \vec{b}}{|\vec{b} \times \vec{c}+\vec{c} \times \vec{a}+\vec{a} \times \vec{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, $\mathrm{AD} \| \mathrm{BC}$

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 $\overrightarrow{\mathrm{b}}$ are diagonals of the parallelogramABCD 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 $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{j}-\hat{k}$, find a vector $\vec{c}$ such that $\vec{a} \times \vec{c}=\vec{b}$ and $\vec{a} . \vec{c}=3$.

Answer:

Given that,$\begin{aligned} \vec{a} & =\hat{i}+\hat{j}+\hat{k} and \vec{b} & =\hat{j}-\hat{k}\end{aligned}$

$\\ \begin{aligned} &\text { We need to find vector } \overrightarrow{\mathrm{C}} \text { . }\\ &\text { Let } \overrightarrow{\mathrm{c}}=\mathrm{x} \hat{1}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}, \text { where } \mathrm{x}, \mathrm{y}, \mathrm{z} \text { be any scalars. }\\ &\text { Now, for } \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}}\\ &\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=(\hat{\imath}+\hat{\jmath}+\hat{\mathrm{k}}) \times(\mathrm{x} \hat{\mathrm{l}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}) \end{aligned}$
$\\ \Rightarrow(\hat{\imath}+\hat{\jmath}+\hat{\mathrm{k}}) \times(\mathrm{xi}+\mathrm{y} \hat{\jmath}+\mathrm{z} \hat{\mathrm{k}})=\hat{\mathrm{j}}-\hat{\mathrm{k}} \\ \Rightarrow\left|\begin{array}{lll} \hat{\imath} & \hat{\jmath} & \hat{\mathrm{k}} \\ 1 & 1 & 1 \\ \mathrm{x} & \mathrm{y} & \mathrm{z} \end{array}\right|=\hat{\mathrm{j}}-\hat{\mathrm{k}} \\ \Rightarrow \hat{1}((1)(\mathrm{z})-(1)(\mathrm{y}))-\hat{\jmath}((1)(\mathrm{z})-(1)(\mathrm{x}))+\hat{\mathrm{k}}(1)(\mathrm{y})-(1)(\mathrm{x}))=\hat{\mathrm{j}}-\hat{\mathrm{k}} \\ \Rightarrow \hat{\mathrm{i}}(\mathrm{z}-\mathrm{y})-\hat{\mathrm{j}}(\mathrm{z}-\mathrm{x})+\hat{\mathrm{k}}(\mathrm{y}-\mathrm{x})=\hat{\mathrm{j}}-\hat{\mathrm{k}}$
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)

Also, for $\vec{a} \vec{c}=3$
$\begin{aligned}
& \vec{a} \vec{c}=(\hat{i}+\hat{j}+\hat{k}) \cdot(x \hat{\imath}+\mathrm{y} \hat{\jmath}+\mathrm{z} \hat{\mathrm{k}}) \\
& \Rightarrow(\hat{\imath}+\hat{\jmath}+\hat{\mathrm{k}}) \cdot(\mathrm{x} \hat{\imath}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}})=3
\end{aligned}$
$\text { since }2(\hat{\imath}+\hat{\jmath}+\hat{\mathrm{k}}) \cdot(\mathrm{x} \hat{\imath}+\mathrm{y} \hat{\jmath}+\mathrm{z} \hat{\mathrm{k}})=\mathrm{x}+\mathrm{y}+\mathrm{z}, \text { as } \hat{1} \cdot \hat{\imath}=\hat{\mathrm{j}} \cdot \hat{\mathrm{j}}=\hat{\mathrm{k}} \cdot \hat{\mathrm{k}}=1 \text { and other dot }$

multiplication is zero. We get,
$\begin{array}{lllllll}
x+y+z=3 \ldots(i v) \text { Now, } & \text { add } & \text { equations } & \text { (ii) } \quad \text { and } & \text { (iii), we } & \text { wet } \\
(x-z)+(x-y)=1+1 & & & & & &
\end{array}$
$\begin{aligned}
& \Rightarrow x+x-y-z=2 \quad \text { Add equations } \\
& \Rightarrow 2 x-y-z=2 \ldots(v) \\
& (x+y+z)+(2 x-y-z)=3+2 \\
& \Rightarrow x+2 x+y-y+z-z=5 \\
& \Rightarrow 3 x=5 \\
& \Rightarrow x=\frac{5}{3}
\end{aligned}$

Put value of x in equation (iii), we get Equation (iii)
$\begin{aligned}
& \Rightarrow x-y=1 \Rightarrow \frac{5}{3}-\mathrm{y}=1 \Rightarrow \mathrm{y}=\frac{5}{3}-1 \\
& \Rightarrow \mathrm{y}=\frac{5-3}{3} \\
& \Rightarrow \mathrm{y}=\frac{2}{3}
\end{aligned}$
(iv) and
(v), we get

Put this value of $y$ in equation (i), we get
Equation(i) $\Rightarrow \mathrm{z}-\mathrm{y}=0 \Rightarrow \mathrm{z}-\frac{2}{3}=0 \Rightarrow \mathrm{z}=\frac{2}{3}$
since, $\vec{c}=x \hat{1}+y \hat{j}+z \hat{k}$
By putting the values of x, y and z, we get
$\overrightarrow{\mathrm{c}}=\frac{5}{3} \hat{\imath}+\frac{2}{3} \hat{\jmath}+\frac{2}{3} \hat{\mathrm{k}}$

Thus, we have found the vector $\vec{c}$.

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
$A(2 \vec{a}-3 \vec{b}){\text { and }} B(\vec{a}+\vec{b})$.
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)

Let $\vec{a}=\mathrm{a}_1 \hat{i}+\mathrm{a}_2 \hat{j}+\mathrm{a}_3 \hat{k}$ then
$\begin{aligned}
& \overrightarrow{\mathrm{a}} \times \hat{\mathrm{i}}=\left(\mathrm{a}_1 \hat{\imath}+\mathrm{a}_2 \hat{\jmath}+\mathrm{a}_3 \hat{\mathrm{k}}\right) \times \hat{\mathrm{i}}=\mathrm{a}_1 \hat{\imath} \times \hat{\mathrm{i}}+\mathrm{a}_2 \hat{\jmath} \times \hat{\mathrm{i}}+\mathrm{a}_3 \hat{\mathrm{k}} \times \hat{\mathrm{i}} \\
& \Rightarrow \vec{a} \times \hat{\mathrm{i}}=0-\mathrm{a}_2 \hat{\mathrm{k}}+\mathrm{a}_3 \hat{\mathrm{j}}(\because \hat{\mathrm{i}} \times \hat{\mathrm{i}}=0, \hat{\mathrm{j}} \times \hat{\mathrm{i}}=-\hat{\mathrm{k}}, \hat{\mathrm{k}} \times \hat{\mathrm{i}}=\hat{\mathrm{j}}) \\
& \Rightarrow \overrightarrow{\mathrm{a}} \times \hat{i}=-\mathrm{a}_2 \hat{k}+\mathrm{a}_3 \hat{j}
\end{aligned}$

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

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 $\begin{aligned} &\left|\begin{array}{ccc} \lambda & 1 & 2 \\ 1 & \lambda & -1 \\ 2 & -1 & \lambda \end{array}\right|=0\\ \end{aligned}$

$\begin{aligned} & \Rightarrow \lambda\left(\lambda^2-1\right)-1(\lambda+2)+2(-1-2 \lambda)=0 \\ & \Rightarrow \lambda^3-\lambda-\lambda-2-2-4 \lambda=0 \\ & \Rightarrow \lambda^3-6 \lambda-4=0 \\ & \Rightarrow(\lambda+2)\left(\lambda^2-2 \lambda-2\right)=0\end{aligned}$

$\\ \Rightarrow \lambda=-2 \text { and } \lambda=\frac{2 \pm \sqrt{(-2)^{2}-4 \times 1 \times-2}}{2}=\frac{2 \pm \sqrt{12}}{2} \\ \Rightarrow \lambda=-2 \text { and } \lambda=1 \pm \sqrt{3}$

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

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

Answer:

Answer :(A)

Let θ be the angle between $\vec {a} $ and $\vec{b}$
From figure we can see that, length 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

$\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{\imath} & \hat{\jmath} & \hat{k} \\
2 & 1 & 2 \\
0 & 1 & 1
\end{array}\right|=\hat{\imath}(1-2)-\hat{\jmath}(2-0)+\hat{\mathrm{k}}(2-0)=-\hat{\imath}-2 \hat{\jmath}+2 \hat{\mathrm{k}}$

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
$\begin{aligned}
& \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}=\frac{\vec{b} \times \vec{a}}{|\vec{b} \times \vec{a}|} \\
& \Rightarrow \frac{\vec{b} \times \vec{a}}{|\vec{b} \times \vec{a}|}=\frac{1}{3} \hat{\imath}+\frac{2}{3} \hat{\jmath}-\frac{2}{3} k
\end{aligned}$

Hence, there are two unit length perpendicular to the $\vec{a}$ 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 $\overrightarrow{\mathrm{r}}$

$\Rightarrow\vec{r}\text{ is perpendicular to } \vec{a}, \vec{b}$ and $\vec{c}$
$\Rightarrow \vec{a}, \vec{b}$ and $\vec{c}{\text { are coplanar }}$.
$\Rightarrow \vec{a} \cdot(\vec{b} \times \vec{c})=0$

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.
$\begin{aligned}
& \Rightarrow \overrightarrow{d_1}=\vec{a}+\vec{b}=(3 \hat{\imath}-2 \hat{\jmath}+2 \hat{k})+(-\hat{\imath}-2 \hat{k})=(3-1) \hat{\imath}+(-2+0) \hat{\jmath}+(2-2) \hat{k} \\
& \Rightarrow \overrightarrow{d_1}=2 \hat{\imath}-2 \hat{j} \\
& \Rightarrow \overrightarrow{d_2}=\vec{a}-\vec{b}=(3 \hat{\imath}-2 \hat{\jmath}+2 \hat{k})-(-\hat{\imath}-2 \hat{k})=(3+1) \hat{\imath}+(-2-0) \hat{\jmath}+(2+2) \hat{k} \Rightarrow \\
& \overrightarrow{d_2}=4 \hat{\imath}-2 \hat{\jmath}+4 \hat{k}
\end{aligned}$
Let $\theta$ be the angle between diagonals $\overrightarrow{d_1}$ and $\overrightarrow{d_2}$.
Then, $\overrightarrow{d_1} \cdot \overrightarrow{d_2}=\left|\overrightarrow{d_1}\right|\left|\overrightarrow{d_2}\right| \cos \theta$
$\Rightarrow \cos \theta=\frac{\overrightarrow{d_1} \cdot \overrightarrow{d_2}}{\left|\overrightarrow{d_1}\right|\left|\overrightarrow{d_2}\right|}$

$\\ \cos \theta=\frac{(2 \hat{\imath}-2 \hat{\jmath}) \cdot(4 \hat{i}-2 \hat{\jmath}+4 \hat{k})}{\sqrt{2^{2}+2^{2}} \sqrt{4^{2}+(-2)^{2}+4^{2}}}=\frac{8+4}{\sqrt{8} \sqrt{16+4+16}}(\because \hat{\imath} . \hat{\jmath}=0, \hat{\jmath} \cdot \hat{k}=0, \hat{k} \cdot \hat{\imath}=0) \\ \Rightarrow \cos \theta=\frac{12}{2 \sqrt{2} .6}=\frac{1}{\sqrt{2}} \\ \Rightarrow \theta=\frac{\pi}{4}\left(\because \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}\right)$

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.

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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Importance of Solving NCERT Exemplar Class 12 Maths Chapter `10 Solutions:

Class 12 Maths NCERT exemplar solutions chapter 10 explores the real-world concepts and applications of Vectors. Vectors have several real-world uses in Aerospace, Fluid concepts, Complex calculations, sports, etc.
These solutions will help you understand the types of problems and their approach to solving.
These solutions will help you to prepare for your board exams as well as entrance exams.

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Other NCERT Exemplar Class 12 Mathematics Chapters

NCERT Exemplar for Class 12 Mathematics Chapter 1 Relations and Functions
NCERT Exemplar for Class 12 Mathematics Chapter 2 Inverse Trigonometric Functions
NCERT Exemplar for Class 12 Mathematics Chapter 3 Matrices
NCERT Exemplar for Class 12 Mathematics Chapter 4 Determinants
NCERT Exemplar for Class 12 Mathematics Chapter 5 Continuity and Differentiability
NCERT Exemplar for Class 12 Mathematics Chapter 6 Applications of Derivatives
NCERT Exemplar for Class 12 Mathematics Chapter 7 Integrals
NCERT Exemplar for Class 12 Mathematics Chapter 8 Applications of Integrals
NCERT Exemplar for Class 12 Mathematics Chapter 9 Differential Equations
NCERT Exemplar for Class 12 Mathematics Chapter 11 Three-Dimensional Geometry
NCERT Exemplar for Class 12 Mathematics Chapter 12 Linear Programming
NCERT Exemplar for Class 12 Mathematics Chapter 13 Probability

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NCERT solutions for class 12 maths - Chapter-wise

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

Must Read NCERT Solution subject-wise

Also, check the NCERT Books and the NCERT Syllabus here

Frequently Asked Questions (FAQs)

Q: 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.

Q: 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

Q: 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.

Q: 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.

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Questions related to CBSE Class 12th

On Question asked by student community

Have a question related to CBSE Class 12th ?

Can i improve my 12 result this year in PCM

Hello,

If you want to improve the Class 12 PCM results, you can appear in the improvement exam. This exam will help you to retake one or more subjects to achieve a better score. You should check the official website for details and the deadline of this exam.

I hope it will clear your query!!

Read Complete Answer

Rahul Mandal

8 Sep'25

how may marks should i get in my 12th board exams to get fee concesion in sastra university

Hello Aspirant,

SASTRA University commonly provides concessions and scholarships based on merit in class 12 board exams and JEE Main purposes with regard to board merit you need above 95% in PCM (or on aggregate) to get bigger concessions, usually if you scored 90% and above you may get partial concessions. I suppose the exact cut offs may change yearly on application rates too.

Read Complete Answer

gopi

4 Sep'25

Which course is best for computer science students after 12th standard

Hello,

After 12th, if you are interested in computer science, the best courses are:

B.Tech in Computer Science Engineering (CSE) – most popular choice.
BCA (Bachelor of Computer Applications) – good for software and IT jobs.
B.Sc. Computer Science / IT – good for higher studies and research.
B.Tech in Information Technology (IT) – focuses on IT and networking.

All these courses have good career scope. Choose based on your interest in coding, software, hardware, or IT field.

Hope it helps !

Read Complete Answer

Prachi Kumari

4 Sep'25

I am a CBSE class 12 private candidate will my maksheet of CBSE supplement exam will be delivered on main exam form address for supplementary exam form address ,my both addresses are different please tell

Hello Vanshika,

CBSE generally forwards the marksheet for the supplementary exam to the correspondence address as identified in the supplementary exam application form. It is not sent to the address indicated in the main exam form. Addresses that differ will use the supplementary exam address.

Read Complete Answer

gopi

3 Sep'25

12 arts previous years board paper 2025

To find Class 12 Arts board papers, go to the official website of your education board, then click on the Sample Papers, Previous Years Question Papers(PYQ) or Model Papers section, and select the Arts stream. You will find papers for the various academic year. You can then select the year of which you want to solve and do your practice. There are many other educational websites that post pyqs on their website you can also visit that.

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

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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
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NCERT Exemplar Class 12 Maths Solutions Chapter 10 Vector Algebra

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

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