NCERT Exemplar Class 12 Maths Solutions Chapter 10 Vector Algebra

Question:1

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

Answer:

We have,
$\overrightarrow{a}=2\hat{ı}-\hat{ȷ}+\hat{k}$
$\overrightarrow{\mathrm{b}}=2\hat{ȷ}+\hat{\mathrm{k}}$

Since, unit vector is needed to be found in the direction of the sum of vectors $\overrightarrow{a}$ and $\overrightarrow{b}$.
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 $\overrightarrow{\mathrm{b}}$.
$\Rightarrow \overrightarrow{c}=(2\hat{ı}-\hat{ȷ}+\hat{\mathrm{k}})+(2\hat{ȷ}+\hat{\mathrm{k}})$

$\Rightarrow \overrightarrow{\mathrm{c}}=2\hat{ı}-\hat{ȷ}+\hat{\mathrm{k}}+2\hat{ȷ}+\hat{\mathrm{k}}$

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

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:
$\begin{array}{rl}& \hat{c}=\frac{\overrightarrow{\mathrm{c}}}{|\overrightarrow{\mathrm{c}}|}\\ & \text{Substitute the value of }\overrightarrow{\mathrm{c}}\text{ . }\\ & \Rightarrow \hat{c}=\frac{2\hat{ı}+\hat{ȷ}+2\hat{\mathrm{k}}}{|2\hat{\mathrm{i}}+\hat{\mathrm{j}}+2\hat{\mathrm{k}}|}\\ & \text{ Here, }|2\hat{ı}+\hat{ȷ}+2\hat{\mathrm{k}}|=\sqrt{2^2+1^2+2^2}\end{array}$
$\Rightarrow \hat{c}=\frac{2\hat{ı}+\hat{ȷ}+2\hat{\mathrm{k}}}{\sqrt{2^2+1^2+2^2}}$

$\Rightarrow \hat{\mathrm{c}}=\frac{2\hat{ı}+\hat{ȷ}+2\hat{\mathrm{k}}}{\sqrt{4+1+4}}$

$\Rightarrow \hat{c}=\frac{2\hat{ı}+\hat{ȷ}+2\hat{\mathrm{k}}}{\sqrt{9}}$

$\Rightarrow \hat{c}=\frac{2\hat{ı}+\hat{ȷ}+2\hat{\mathrm{k}}}{3}$
Thus, unit vector in the direction of sum of vectors $\overrightarrow{a}\text{ and }\overrightarrow{b}$ 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,

$\overrightarrow{\mathrm{b}}=2\hat{ı}+\hat{ȷ}-2\hat{\mathrm{k}}$

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.
$\Rightarrow 6\overrightarrow{\mathrm{b}}=12\hat{ı}+6\hat{ȷ}-12\hat{\mathrm{k}}$
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.
$\Rightarrow 6\hat{\mathrm{b}}=\frac{12\hat{\mathrm{i}}+6\hat{\mathrm{j}}-12\hat{\mathrm{k}}}{|12\hat{\mathrm{l}}+6\hat{\mathrm{j}}-12\hat{\mathrm{k}}|}$

$\text{Here, }|12\hat{\mathrm{i}}+6\hat{\mathrm{j}}-12\hat{\mathrm{k}}|=\sqrt{{12}^2+6^2+(-12{)}^2}$

$\Rightarrow 6\hat{\mathrm{b}}=\frac{12\hat{\mathrm{i}}+6\hat{\mathrm{j}}-12\hat{\mathrm{k}}}{\sqrt{144+36+144}}$

$\Rightarrow 6\hat{\mathrm{b}}=\frac{12\hat{\mathrm{i}}+6\hat{\mathrm{j}}-12\hat{\mathrm{k}}}{\sqrt{324}}$
$\begin{array}{rl}& \Rightarrow 6\hat{\mathrm{b}}=\frac{12\hat{\mathrm{i}}+6\hat{\mathrm{j}}-12\hat{\mathrm{k}}}{18}\\ & \text{ Let us simplify. }\\ & \Rightarrow 6\hat{\mathrm{b}}=\frac{6(2\hat{\mathrm{i}}+\hat{\mathrm{j}}-2\hat{\mathrm{k}})}{18}\\ & \Rightarrow 6\hat{\mathrm{b}}=\frac{2\hat{ı}+\hat{\mathrm{j}}-2\hat{\mathrm{k}}}{3}\end{array}$
Thus, unit vector in the direction of $6\overrightarrow{\mathrm{b}}$ is $\frac{2\hat{ı}+\hat{\mathrm{j}}-2\hat{\mathrm{k}}}{3}$

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 $2\overrightarrow{a}-\overrightarrow{b}$

Answer:

We need to find the unit vector in the direction of $2\overrightarrow{a}-\overrightarrow{b}$
First, let us calculate .$2\overrightarrow{a}-\overrightarrow{b}$
As we have,
$\begin{array}{l}\overrightarrow{\mathrm{a}}=\hat{1}+\hat{\mathrm{j}}+2{\hat{\mathrm{k}}}_{\dots }(\mathrm{a})\\ \overrightarrow{\mathrm{b}}=2\hat{ı}+\hat{\mathrm{j}}-2{\hat{\mathrm{k}}}_{\dots (\mathrm{b})}\end{array}$
Then multiply equation (a) by 2 on both sides,
$2\overrightarrow{a}=2(\hat{ı}+\hat{ȷ}+2\hat{k})$
We can easily multiply vector by a scalar by multiplying similar components, that is, vector’s magnitude by the scalar’s magnitude.
$\Rightarrow 2\overrightarrow{a}=2\hat{ı}+2\hat{ȷ}+4\hat{k}$ \ldots $(c)$
Subtract (b) from (c). We get
$\Rightarrow 2\overrightarrow{a}-\overrightarrow{b}=2\hat{l}-2\hat{l}+2\hat{j}-\hat{j}+4\hat{k}+2\hat{k}$
$\Rightarrow 2\overrightarrow{a}-\overrightarrow{b}=\hat{ȷ}+6\hat{k}$
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:

$2\hat{a}-\hat{b}=\frac{2\overrightarrow{a}-\overrightarrow{b}}{|2\overrightarrow{a}-\overrightarrow{b}|}$

Now we know the value of $2\overrightarrow{a}-\overrightarrow{b}$, so, we just need to substitute in the above equation.
$\begin{array}{rl}& \Rightarrow 2\hat{a}-\hat{b}=\frac{\hat{ȷ}+6\hat{k}}{|\hat{ȷ}+6\hat{k}|}\\ & \text{ Here },|\hat{ȷ}+6\hat{k}|=\sqrt{1^2+6^2}\\ & \Rightarrow 2\hat{a}-\hat{b}=\frac{\hat{ȷ}+6\hat{k}}{\sqrt{1^2+6^2}}\end{array}$

$\begin{array}{rl}& \Rightarrow 2\hat{a}-\hat{b}=\frac{\hat{ȷ}+6\hat{\mathrm{k}}}{\sqrt{1+36}}\\ & \Rightarrow 2\hat{a}-\hat{b}=\frac{\hat{ȷ}+6\hat{k}}{\sqrt{37}}\\ & \text{ Thus, unit vector in the direction of }\\ & 2\overrightarrow{a}-\overrightarrow{b}is\frac{\hat{ȷ}+6\hat{k}}{\sqrt{37}}\end{array}$

Question:3

Find a unit vector in the direction of $\overline{PQ}$ , 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,
$\begin{array}{rl}& \overrightarrow{\mathrm{OP}}=5\hat{\mathrm{i}}+0\hat{\mathrm{j}}+8\hat{\mathrm{k}}\\ & \Rightarrow \overrightarrow{\mathrm{OP}}=5\hat{\mathrm{i}}+8\hat{\mathrm{k}}\\ & \text{ Position vector of }Q\text{ is given by, }\\ & \overrightarrow{\mathrm{OQ}}=3\hat{ı}+3\hat{ȷ}+2\hat{\mathrm{k}}\end{array}$
To find unit vector in the direction of PQ, we need to find position vector of PQ.
Position vector of PQ is given by,
$\overrightarrow{\mathrm{PQ}}=\text{ Position vector of }\mathrm{Q}-\text{ Position vector of }\mathrm{P}$

$\Rightarrow \overrightarrow{\mathrm{PQ}}=\overrightarrow{\mathrm{OQ}}-\overrightarrow{\mathrm{OP}}$

$\Rightarrow \overrightarrow{\mathrm{PQ}}=(3\hat{\mathrm{i}}+3\hat{\mathrm{j}}+2\hat{\mathrm{k}})-(5\hat{\mathrm{l}}+8\hat{\mathrm{k}})$

$\Rightarrow \overrightarrow{\mathrm{PQ}}=3\hat{\mathrm{i}}-5\hat{\mathrm{i}}+3\hat{\mathrm{j}}+2\hat{\mathrm{k}}-8\hat{\mathrm{k}}$
$\Rightarrow \overrightarrow{\mathrm{PQ}}=-2\hat{ı}+3\hat{ȷ}-6\hat{\mathrm{k}}$
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:

$$\begin{gathered} \widehat{\mathrm{PQ}}=\frac{\overrightarrow{\mathrm{PQ}}}{|\overrightarrow{\mathrm{PQ}}|} \\ \Rightarrow \widehat{\mathrm{PQ}}=\frac{-2\hat{ı}+3\hat{ȷ}-6\hat{\mathrm{k}}}{|-2\hat{ı}+3\hat{ȷ}-6\hat{\mathrm{k}}|} \end{gathered}$$

$\text{Here, }|-2\hat{\mathrm{i}}+3\hat{\mathrm{j}}-6\hat{\mathrm{k}}|=\sqrt{(-2{)}^2+3^2+(-6{)}^2}$

$\Rightarrow \widehat{\mathrm{PQ}}=\frac{-2\hat{\mathrm{i}}+3\hat{\mathrm{j}}-6\hat{\mathrm{k}}}{\sqrt{(-2{)}^2+3^2+(-6{)}^2}}$
$$\begin{gathered} \Rightarrow \widehat{\mathrm{PQ}}=\frac{-2\hat{ı}+3\hat{ȷ}-6\hat{\mathrm{k}}}{\sqrt{4+9+36}} \\ \Rightarrow \widehat{\mathrm{PQ}}=\frac{-2\hat{\mathrm{l}}+3\hat{\mathrm{j}}-6\hat{\mathrm{k}}}{\sqrt{49}} \\ \Rightarrow \widehat{\mathrm{PQ}}=\frac{-2\hat{\mathrm{l}}+3\hat{\mathrm{j}}-6\hat{\mathrm{k}}}{7} \end{gathered}$$
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 $\overrightarrow{a}$ and $\overrightarrow{b}$ 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 $=\overrightarrow{a}$

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

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
$\overrightarrow{\mathrm{BC}}=1.5\overrightarrow{\mathrm{BA}}\dots (\mathrm{i})$

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

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

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

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

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

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

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

Position vector of C is given by,

$\overrightarrow{OC}=3\hat{i}+5\hat{j}+3\hat{k}$
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 $|\overrightarrow{AB}|$
To find : $|\overrightarrow{AB}|$
Position vector of B-Position vector of A
$\Rightarrow \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$
$\begin{array}{rl}& \Rightarrow \overrightarrow{\mathrm{AB}}=(\hat{ı}-\hat{ȷ}+3\hat{\mathrm{k}})-(\mathrm{k}\hat{ı}-10\hat{\mathrm{j}}+3\hat{\mathrm{k}})\\ & \Rightarrow \overrightarrow{\mathrm{AB}}=\hat{ı}-k\hat{ı}-\hat{ȷ}+10\hat{ȷ}+3\hat{\mathrm{k}}-3\hat{\mathrm{k}}\\ & \Rightarrow \overrightarrow{\mathrm{AB}}=(1-\mathrm{k})\hat{ı}+9\hat{\mathrm{j}}\\ & \text{ Now, }\\ & |\overrightarrow{\mathrm{AB}}|=\sqrt{(1-\mathrm{k}{)}^2+9^2}\end{array}$
$\begin{array}{rl}& \Rightarrow |\overrightarrow{\mathrm{AB}}|={\sqrt{(1-\mathrm{k}{)}^2+81}}_{\dots (\mathrm{i})}\\ & \text{ To find }|\overrightarrow{\mathrm{BC}}{|}_{:}\\ & \overrightarrow{\mathrm{BC}}=\text{ position vector of C-Position vector of }\mathrm{B}\\ & \Rightarrow \overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}}\end{array}$
$\begin{array}{rl}& \Rightarrow \overrightarrow{\mathrm{BC}}=(3\hat{ı}+5\hat{ȷ}+3\hat{\mathrm{k}})-(\hat{ı}-\hat{ȷ}+3\hat{\mathrm{k}})\\ & \Rightarrow \overrightarrow{\mathrm{BC}}=3\hat{ı}-\hat{ı}+5\hat{ȷ}+\hat{ȷ}+3\hat{\mathrm{k}}-3\hat{\mathrm{k}}\\ & \Rightarrow \overrightarrow{\mathrm{BC}}=2\hat{ı}+6\hat{j}\\ & \text{ Now, }\\ & |\overrightarrow{\mathrm{BC}}|=\sqrt{2^2+6^2}\end{array}$
$\begin{array}{rl}& \Rightarrow |\overrightarrow{\mathrm{BC}}|=\sqrt{4+36}\\ & \Rightarrow |\overrightarrow{\mathrm{BC}}|=\sqrt{40}\\ & \text{ To find }\\ & |\overrightarrow{\mathrm{AC}}|\end{array}$
$\overrightarrow{\mathrm{AC}}$ = Position vector of C-Position vector of A
$$\begin{gathered} \Rightarrow \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}} \\ \Rightarrow \overrightarrow{\mathrm{AC}}=(3\hat{\mathrm{i}}+5\hat{\mathrm{j}}+3\hat{\mathrm{k}})-(\mathrm{ki}-10\hat{\mathrm{j}}+3\hat{\mathrm{k}}) \\ \Rightarrow \overrightarrow{\mathrm{AC}}=3\hat{\mathrm{i}}-\mathrm{k}\hat{\mathrm{I}}+5\hat{\mathrm{j}}+10\hat{\mathrm{j}}+3\hat{\mathrm{k}}-3\hat{\mathrm{k}} \\ \Rightarrow \overrightarrow{\mathrm{AC}}=(3-\mathrm{k})\hat{\mathrm{i}}+15\hat{\mathrm{j}} \end{gathered}$$
$\begin{array}{rl}& \text{ Now, }\\ & |\overrightarrow{\mathrm{AC}}|=\sqrt{(3-\mathrm{k}{)}^2+{15}^2}\\ & \Rightarrow |\overrightarrow{\mathrm{AC}}|=\sqrt{(3-\mathrm{k}{)}^2+225}...(iii)\end{array}$

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
$\begin{array}{rl}& \Rightarrow \sqrt{(3-\mathrm{k}{)}^2+225}\sqrt{40}=\sqrt{(1-\mathrm{k}{)}^2+81}\\ & \Rightarrow \sqrt{(9+{\mathrm{k}}^2-6\mathrm{k})+225}-\sqrt{40}=\sqrt{(1+{\mathrm{k}}^2-2\mathrm{k})+81}\end{array}$
$[\because$ by algebraic identity, $(a-b{)}^2=a^2+b^2-2ab]$
$\begin{array}{rl}& \Rightarrow \sqrt{{\mathrm{k}}^2-6\mathrm{k}+225+9}-\sqrt{40}=\sqrt{{\mathrm{k}}^2-2\mathrm{k}+81+1}\\ & \Rightarrow \sqrt{{\mathrm{k}}^2-6\mathrm{k}+234}-\sqrt{40}=\sqrt{{\mathrm{k}}^2-2\mathrm{k}+82}\end{array}$

Squaring on both sides,
$\Rightarrow {[\sqrt{{\mathrm{k}}^2-6\mathrm{k}+234}-\sqrt{40}]}^2={\left[\sqrt{{\mathrm{k}}^2-2\mathrm{k}+82}\right]}^2$

$\begin{array}{rl}& \Rightarrow {\left[\sqrt{{\mathrm{k}}^2-6\mathrm{k}+234}\right]}^2+[\sqrt{40}{]}^2-2\left[\sqrt{{\mathrm{k}}^2-6\mathrm{k}+234}\right][\sqrt{40}]={\mathrm{k}}^2-2\mathrm{k}+82\\ & [\because \text{ by algebraic identity, }(a-b{)}^2=a^2+b^2-2ab]\\ & \Rightarrow {\mathrm{k}}^2-6\mathrm{k}+234+40-2\left[\sqrt{{\mathrm{k}}^2-6\mathrm{k}+234}\right][\sqrt{40}]={\mathrm{k}}^2-2\mathrm{k}+82\\ & \Rightarrow {\mathrm{k}}^2-6\mathrm{k}+234+40-{\mathrm{k}}^2+2\mathrm{k}-82=2\left[\sqrt{{\mathrm{k}}^2-6\mathrm{k}+234}\right][\sqrt{40}]\end{array}$
$$\begin{gathered} \Rightarrow {\mathrm{k}}^2-{\mathrm{k}}^2-6\mathrm{k}+2\mathrm{k}+234+40-82=2\left[\sqrt{{\mathrm{k}}^2-6\mathrm{k}+234}\right][\sqrt{40}] \\ \Rightarrow -4\mathrm{k}+192=2\left[\sqrt{{\mathrm{k}}^2-6\mathrm{k}+234}\right][\sqrt{40}] \\ \Rightarrow -4\mathrm{k}+192=2\left[\sqrt{{\mathrm{k}}^2-6\mathrm{k}+234}\right].2[\sqrt{10}] \\ \Rightarrow 4(-\mathrm{k}+48)=4\left[\sqrt{{\mathrm{k}}^2-6\mathrm{k}+234}\right][\sqrt{10}] \end{gathered}$$
$\begin{array}{rl}& \Rightarrow -\mathrm{k}+48=\sqrt{{\mathrm{k}}^2-6\mathrm{k}+234}\cdot \sqrt{10}\\ & \text{ Again, squaring on both sides, we get }\\ & [48-\mathrm{k}{]}^2={[\sqrt{{\mathrm{k}}^2-6\mathrm{k}+234}\cdot \sqrt{10}]}^2\\ & \Rightarrow (48{)}^2+k^2-2(48)(k)=(k^2-6k+234)(10)[\because \text{ by algebraic identity },(a-b{)}^2=a^2+b^2-2ab]\end{array}$
$\begin{array}{rl}& \Rightarrow 2304+{\mathrm{k}}^2-96\mathrm{k}=10{\mathrm{k}}^2-60\mathrm{k}+2340\\ & \Rightarrow 10k^2-k^2-60k+96k+2340-2304=0\\ & \Rightarrow 9{\mathrm{k}}^2+36\mathrm{k}+36=0\\ & \Rightarrow 9({\mathrm{k}}^2+4\mathrm{k}+4)=0\\ & \Rightarrow {\mathrm{k}}^2+4\mathrm{k}+4=0\end{array}$
$\Rightarrow {\mathrm{k}}^2+2\mathrm{k}+2\mathrm{k}+4=0$

$\Rightarrow \mathrm{k}(\mathrm{k}+2)+2(\mathrm{k}+2)=0$

$\Rightarrow (\mathrm{k}+2)(\mathrm{k}+2)=0$

$\Rightarrow \mathrm{k}=-2\text{ or }\mathrm{k}=-2$
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 $2\sqrt{3}$ units, find $\overrightarrow{\mathrm{r}}$ .

Answer:

Given that,
Magnitude of $\overrightarrow{\mathrm{r}}$ = $2\sqrt{3}$
$\Rightarrow |\overrightarrow{r}|=2\sqrt{3}$
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,
$$\begin{gathered} {l}^2+{\mathrm{m}}^2+{\mathrm{n}}^2=1 \\ \Rightarrow l^2+l^2+l^2=1[\because \mathrm{l}=\mathrm{m}=\mathrm{n}] \\ \Rightarrow 3.l^2=1 \\ \Rightarrow l=\pm \frac{1}{\sqrt{3}} \end{gathered}$$
Also, we know that $\overrightarrow{\mathrm{r}}$ is represented in terms of direction cosines as,
$$\begin{gathered} \hat{\mathrm{r}}=l\hat{\mathrm{i}}+\mathrm{m}\hat{\mathrm{j}}+\mathrm{n}\hat{\mathrm{k}} \\ \Rightarrow \hat{\mathrm{r}}=\pm \frac{1}{\sqrt{3}}\hat{ı}\pm \frac{1}{\sqrt{3}}\hat{ȷ}\pm \frac{1}{\sqrt{3}}\hat{\mathrm{k}} \end{gathered}$$

We are familiar with the formula, $\hat{\mathrm{r}}=\frac{\overrightarrow{\mathrm{r}}}{|\overrightarrow{\mathrm{r}}|}$

$$\begin{gathered} \hat{\mathrm{r}}=\frac{\overrightarrow{\mathrm{r}}}{|\overrightarrow{\mathrm{r}}|} \\ \text{To find }\overrightarrow{\mathrm{r}} \\ \Rightarrow \overrightarrow{\mathrm{r}}=\hat{\mathrm{r}}|\overrightarrow{\mathrm{r}}| \end{gathered}$$Substituting values of$|\overrightarrow{\mathrm{r}}|and\hat{\mathrm{r}}$
$\begin{array}{rl}& \overrightarrow{\mathrm{r}}=(\pm \frac{1}{\sqrt{3}}\hat{ı}\pm \frac{1}{\sqrt{3}}\hat{ȷ}\pm \frac{1}{\sqrt{3}}\hat{\mathrm{k}})(2\sqrt{3})\\ & \Rightarrow \overrightarrow{\mathrm{r}}=\pm \frac{1}{\sqrt{3}}(\hat{ı}+\hat{ȷ}+\hat{\mathrm{k}})(2\sqrt{3})\\ & \Rightarrow \overrightarrow{\mathrm{r}}=\pm 2(\hat{ı}+\hat{ȷ}+\hat{\mathrm{k}})\\ & \text{ Thus, the value of }\overrightarrow{\mathrm{r}}\text{ is }\pm 2(\hat{ı}+\hat{ȷ}+\hat{\mathrm{k}})\end{array}$

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

Any vector perpendicular to both $\overrightarrow{a}$ and $\overrightarrow{b}$ can be given as,
$$\begin{gathered} \overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{ı}& \hat{ȷ}& \hat{k}\\ 2& -1& 2\\ 4& -1& 3\end{array}\right| \\ \begin{array}{r}\Rightarrow \overrightarrow{a}\times \overrightarrow{b}=\hat{i}((-1)(3)-(2)(-1))-\hat{j}((2)(3)-(2)(4))\\ +\hat{k}((2)(-1)-(-1)(4))\end{array} \\ \Rightarrow \overrightarrow{a}\times \overrightarrow{\mathrm{b}}=\hat{ı}(-3+2)-\hat{ȷ}(6-8)+\hat{\mathrm{k}}(-2+4) \\ \Rightarrow \overrightarrow{a}\times \overrightarrow{\mathrm{b}}=-\hat{ı}+2\hat{ȷ}+2\hat{\mathrm{k}} \end{gathered}$$
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 $=6\times \frac{\overrightarrow{r}}{\mid \overrightarrow{r}}$ $\Rightarrow$ vector $=6\times \frac{-\hat{ı}+2\hat{ȷ}+2\hat{k}}{|-\hat{l}+2\hat{j}+2\hat{k}|}$
Here, $|-\hat{ı}+2\hat{ȷ}+2\hat{k}|=\sqrt{(-1{)}^2+(2{)}^2+(2{)}^2}$

$$\begin{gathered} \Rightarrow \text{ vector }=6\times \frac{-\hat{1}+2\hat{ȷ}+2\hat{\mathrm{k}}}{\sqrt{(-1{)}^2+(2{)}^2+(2{)}^2}} \\ \Rightarrow \text{ vector }=6\times \frac{(-\hat{ı}+2\hat{ȷ}+2\hat{\mathrm{k}})}{\sqrt{1+4+4}} \\ \Rightarrow \text{ vector }=6\times \frac{-\hat{\mathrm{i}}+2\hat{ȷ}+2\hat{\mathrm{k}}}{\sqrt{9}} \end{gathered}$$
$\begin{array}{rl}& \Rightarrow \text{ vector }=6\times \frac{-\hat{\mathrm{i}}+2\hat{\mathrm{j}}+2\hat{\mathrm{k}}}{3}\\ & \Rightarrow \text{ vector }=2\times (-\hat{1}+2\hat{ȷ}+2\hat{\mathrm{k}})\\ & \Rightarrow \text{ vector }=-2\hat{ı}+4\hat{ȷ}+4\hat{k}\\ & \text{ Thus, required vector is }-2\hat{ı}+4\hat{ȷ}+4\hat{k}\end{array}$

Question:9

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