NCERT Exemplar Class 12 Maths Solutions Chapter 10 Vector Algebra

NCERT Exemplar Class 12 Maths Solutions Chapter 10 Vector Algebra

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CBSE Class 12th Exam Date:17 Feb' 26 - 17 Feb' 26

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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  1. Sub-topics Covered in NCERT exemplar solutions for Class 12 Maths chapter 10 Vector Algebra
  2. Importance of Solving NCERT Exemplar Class 12 Maths Chapter `10 Solutions:
  3. Other NCERT Exemplar Class 12 Mathematics Chapters
  4. NCERT Exemplar Class 12 Solutions
  5. NCERT solutions for class 12 maths - Chapter-wise

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.

1
Here, https://gradeup-question-images.grdp.co/liveData/PROJ28891/1554370624968879.png and https://gradeup-question-images.grdp.co/liveData/PROJ28891/1554370625718817.png .
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,
2
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,
3
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
$\\ \sin \mathrm{A}=\frac{\text { perpendicular }}{\text { hypotenuse }} \text { in } \Delta \mathrm{BAD} \\ \qquad \because \sin \mathrm{A}=\frac{\mathrm{BD}}{\mathrm{c}} \\ \Rightarrow \mathrm{BD}=\mathrm{c} \sin \mathrm{A}$
$\\ \text { And, } A D=c \cos A \\ \qquad \cos A=\frac{\text { base }}{\text { hypotenuse }} \text { in } \Delta B A D \\ \because \\ \Rightarrow \cos A=\frac{A D}{C} \\ \Rightarrow A D=c \cos A$
4
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

If $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}$ determine the vertices of a triangle, 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 area of the triangle. Hence deduce the condition that the three points a, b, c are collinear. Also find the unit vector normal to the plane of the triangle.

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,
5
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)
6
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.
7
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:

8
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

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 solutions for class 12 maths - Chapter-wise

Must Read NCERT Solution subject-wise

Read more NCERT Notes subject-wise

Also, check the NCERT Books and the NCERT Syllabus here

Frequently Asked Questions (FAQs)

Q: Who can benefit from these solutions?
A:

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?
A:

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?
A:

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?
A:

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 ?

Hello,

The date of 12 exam is depends on which board you belongs to . You should check the exact date of your exam by visiting the official website of your respective board.

Hope this information is useful to you.

Hello,

Class 12 biology questions papers 2023-2025 are available on cbseacademic.nic.in , and other educational website. You can download PDFs of questions papers with solution for practice. For state boards, visit the official board site or trusted education portal.

Hope this information is useful to you.

Hello Pruthvi,

Taking a drop year to reappear for the Karnataka Common Entrance Test (KCET) is a well-defined process. As a repeater, you are fully eligible to take the exam again to improve your score and secure a better rank for admissions.

The main procedure involves submitting a new application for the KCET through the official Karnataka Examinations Authority (KEA) website when registrations open for the next academic session. You must pay the required application fee and complete all formalities just like any other candidate. A significant advantage for you is that you do not need to retake your 12th board exams. Your previously secured board marks in the qualifying subjects will be used again. Your new KCET rank will be calculated by combining these existing board marks with your new score from the KCET exam. Therefore, your entire focus during this year should be on preparing thoroughly for the KCET to achieve a higher score.

For more details about the KCET Exam preparation, CLICK HERE.

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

Thank you, and I wish you all the best in your bright future.

Yes, you can switch from Science in Karnataka State Board to Commerce in CBSE for 12th. You will need a Transfer Certificate from your current school and meet the CBSE school’s admission requirements. Since you haven’t studied Commerce subjects like Accountancy, Economics, and Business Studies, you may need to catch up before or during 12th. Not all CBSE schools accept direct admission to 12th from another board, so some may ask you to join Class 11 first. Make sure to check the school’s rules and plan your subject preparation.



Hello

For the 12th CBSE Hindi Medium board exam, important questions usually come from core chapters like “Madhushala”, “Jhansi ki Rani”, and “Bharat ki Khoj”.
Questions often include essay writing, letter writing, and comprehension passages. Grammar topics like Tenses, Voice Change, and Direct-Indirect Speech are frequently asked.
Students should practice poetry questions on themes and meanings. Important questions also cover summary writing and translation from Hindi to English or vice versa.
Previous years’ question papers help identify commonly asked questions.
Focus on writing practice to improve handwriting and presentation. Time management during exams is key to answering all questions effectively.