NCERT Solutions for Miscellaneous Exercise Chapter 10 Class 12 - Vector Algebra

Question 1: Write down a unit vector in XY-plane, making an angle of ${30}^{\circ }$ with the positive direction of x-axis.

Answer:

As we know

a unit vector in XY-Plane making an angle $\theta$ with x-axis :

$\overrightarrow{r}=cos\theta \hat{i}+sin\theta \hat{j}$

Hence for $\theta ={30}^0$

$\overrightarrow{r}=cos({30}^0)\hat{i}+sin({30}^0)\hat{j}$

$\overrightarrow{r}=\frac{\sqrt{3}}{2}\hat{i}+\frac{1}{2}\hat{j}$

Answer- the unit vector in XY-plane, making an angle of ${30}^{\circ }$ with the positive direction of x-axis is

$\overrightarrow{r}=\frac{\sqrt{3}}{2}\hat{i}+\frac{1}{2}\hat{j}$

Question 2: Find the scalar components and magnitude of the vector joining the points
$P(x_1,y_1,z_1)andQ(x_2,y_2,z_2).$

Answer:

Given in the question

$P(x_1,y_1,z_1)andQ(x_2,y_2,z_2).$

And we need to finrd the scalar components and magnitude of the vector joining the points P and Q

$\overrightarrow{PQ}=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}+(z_2-z_1)\hat{k}$

Magnitiude of vector PQ

$|\overrightarrow{PQ}|=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$

Scalar components are

$(x_2-x_1),(y_2-y_1),(z_2-z_1)$

Question 3: A girl walks 4 km towards west, then she walks 3 km in a direction ${30}^{\circ }$ east of north and stops. Determine the girl’s displacement from her initial point of departure.

Answer:

As the girl walks 4km towards west

Position vector = $-4\hat{i}$

Now as she moves 3km in direction 30 degree east of north.

$-4\hat{i}+3sin{30}^0\hat{i}+3cos{30}^0\hat{j}$

$-4\hat{i}+\frac{3}{2}\hat{i}+3\frac{\sqrt{3}}{2}\hat{j}$

$\frac{-5}{2}\hat{i}+3\frac{\sqrt{3}}{2}\hat{j}$

hence final position vector is;

$\frac{-5}{2}\hat{i}+3\frac{\sqrt{3}}{2}\hat{j}$

Question 4: If $\overrightarrow{a}=\overrightarrow{b}+\overrightarrow{c}$ , then is it true that $|\overrightarrow{a}|=|\overrightarrow{b}|+|\overrightarrow{c}|$ ? Justify your answer.

Answer:

No, if $\overrightarrow{a}=\overrightarrow{b}+\overrightarrow{c}$ then we can not conclude that $|\overrightarrow{a}|=|\overrightarrow{b}|+|\overrightarrow{c}|$ .

the condition $\overrightarrow{a}=\overrightarrow{b}+\overrightarrow{c}$ satisfies in the triangle.

also, in a triangle, $|\overrightarrow{a}|<|\overrightarrow{b}|+|\overrightarrow{c}|$

Since, the condition $|\overrightarrow{a}|=|\overrightarrow{b}|+|\overrightarrow{c}|$ is contradicting with the triangle inequality, if $\overrightarrow{a}=\overrightarrow{b}+\overrightarrow{c}$ then we can not conclude that $|\overrightarrow{a}|=|\overrightarrow{b}|+|\overrightarrow{c}|$

Question 5: Find the value of x for which $x(\hat{i}+\hat{j}+\hat{k})$ is a unit vector.

Answer:

Given in the question,

a unit vector, $\overrightarrow{u}=x(\hat{i}+\hat{j}+\hat{k})$

We need to find the value of x

$|\overrightarrow{u}|=1$

$|x(\hat{i}+\hat{j}+\hat{k})|=1$

$x\sqrt{1^2+1^2+1^2}=1$

$x\sqrt{3}=1$

$x=\frac{1}{\sqrt{3}}$

The value of x is $\frac{1}{\sqrt{3}}$

Question 6: Find a vector of magnitude 5 units, and parallel to the resultant of the vectors $\overrightarrow{a}=2\hat{i}+3\hat{j}-\hat{k}and\overrightarrow{b}=\hat{i}-2\hat{j}+\hat{k}$

Answer:

Given two vectors

$\overrightarrow{a}=2\hat{i}+3\hat{j}-\hat{k}and\overrightarrow{b}=\hat{i}-2\hat{j}+\hat{k}$

Resultant of $\overrightarrow{a}$ and $\overrightarrow{b}$ :

$\overrightarrow{R}=\overrightarrow{a}+\overrightarrow{b}$ $=2\hat{i}+3\hat{j}-\hat{k}+\hat{i}-2\hat{j}+\hat{k}=3\hat{i}+\hat{j}$

Now, a unit vector in the direction of $\overrightarrow{R}$

$\overrightarrow{u}=\frac{3\hat{i}+\hat{j}}{\sqrt{3^2+1^2}}=\frac{3}{\sqrt{10}}\hat{i}+\frac{1}{\sqrt{10}}\hat{j}$

Now, a unit vector of magnitude in direction of $\overrightarrow{R}$

$\overrightarrow{v}=5\overrightarrow{u}=5\ast \frac{3}{\sqrt{10}}\hat{i}+5\ast \frac{1}{\sqrt{10}}\hat{j}=\frac{15}{\sqrt{10}}\hat{i}+\frac{5}{\sqrt{10}}\hat{j}$

Hence the required vector is $\frac{15}{\sqrt{10}}\hat{i}+\frac{5}{\sqrt{10}}\hat{j}$

Question 7: If $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k},\overrightarrow{b}=2\hat{i}-\hat{j}+3\hat{k}and\overrightarrow{c}=\hat{i}-2\hat{j}+\hat{k}$ , find a unit vector parallel to the vector $2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}$ .

Answer:

Given in the question,

$\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k},\overrightarrow{b}=2\hat{i}-\hat{j}+3\hat{k}and\overrightarrow{c}=\hat{i}-2\hat{j}+\hat{k}$

Now,

let vector $\overrightarrow{V}=2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}$

$\overrightarrow{V}=2(\hat{i}+\hat{j}+\hat{k})-(2\hat{i}-\hat{j}+3\hat{k})+3(\hat{i}-2\hat{j}+\hat{k})$

$\overrightarrow{V}=3\hat{i}-3\hat{j}+2\hat{k}$

Now, a unit vector in direction of $\overrightarrow{V}$

$\overrightarrow{u}=\frac{3\hat{i}-3\hat{j}+2\hat{k}}{\sqrt{3^2+(-3{)}^2+2^2}}=\frac{3}{\sqrt{22}}\hat{i}-\frac{3}{\sqrt{22}}\hat{j}+\frac{2}{\sqrt{22}}\hat{k}$

Now,

A unit vector parallel to $\overrightarrow{V}$

$\overrightarrow{u}=\frac{3}{\sqrt{22}}\hat{i}-\frac{3}{\sqrt{22}}\hat{j}+\frac{2}{\sqrt{22}}\hat{k}$

OR

$-\overrightarrow{u}=-\frac{3}{\sqrt{22}}\hat{i}+\frac{3}{\sqrt{22}}\hat{j}-\frac{2}{\sqrt{22}}\hat{k}$

Question 8: Show that the points A(1, – 2, – 8), B(5, 0, –2) and C(11, 3, 7) are collinear, and find the ratio in which B divides AC.

Answer:

Given in the question,

points A(1, – 2, – 8), B(5, 0, –2) and C(11, 3, 7)

$\overrightarrow{AB}=(5-1)\hat{i}+(0-(-2))\hat{j}+(-2-(-8))\hat{k}=4\hat{i}+2\hat{j}+6\hat{k}$

$\overrightarrow{BC}=(11-5)\hat{i}+(3-0)\hat{j}+(7-(-2))\hat{k}=6\hat{i}+3\hat{j}+9\hat{k}$

$\overrightarrow{CA}=(11-1)\hat{i}+(3-(-2))\hat{j}+(7-(-8))\hat{k}=10\hat{i}+5\hat{j}+15\hat{k}$

now let's calculate the magnitude of the vectors

$|\overrightarrow{AB}|=\sqrt{4^2+2^2+6^2}=\sqrt{56}=2\sqrt{14}$

$|\overrightarrow{BC}|=\sqrt{6^2+3^2+9^2}=\sqrt{126}=3\sqrt{14}$

$|\overrightarrow{CA}|=\sqrt{{10}^2+5^2+{15}^2}=\sqrt{350}=5\sqrt{14}$

As we see that AB = BC + AC, we conclude that three points are colinear.

we can also see from here,

Point B divides AC in the ratio 2 : 3.

Question 9: Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are $(2\overrightarrow{a}+\overrightarrow{b})and(\overrightarrow{a}-3\overrightarrow{b})$ externally in the ratio 1: 2. Also, show that P is the mid point of the line segment RQ.

Answer:

Given, two vectors $\overrightarrow{P}=(2\overrightarrow{a}+\overrightarrow{b})and\overrightarrow{Q}=(\overrightarrow{a}-3\overrightarrow{b})$

the point R which divides line segment PQ in ratio 1:2 is given by

$=\frac{2(2\overrightarrow{a}+\overrightarrow{b})-(\overrightarrow{a}-3\overrightarrow{b})}{2-1}=4\overrightarrow{a}+2\overrightarrow{b}-\overrightarrow{a}+3\overrightarrow{b}=3\overrightarrow{a}+5\overrightarrow{b}$

Hence position vector of R is $3\overrightarrow{a}+5\overrightarrow{b}$ .

Now, Position vector of the midpoint of RQ

$=\frac{(3\overrightarrow{a}+5\overrightarrow{b}+\overrightarrow{a}-3\overrightarrow{b})}{2}=2\overrightarrow{a}+\overrightarrow{b}$

which is the position vector of Point P . Hence, P is the mid-point of RQ

Question 10: The two adjacent sides of a parallelogram are $2\hat{i}-4\hat{j}+5\hat{k}and\hat{i}-2\hat{j}-3\hat{k}$ . Find the unit vector parallel to its diagonal. Also, find its area.

Answer:

Given, two adjacent sides of the parallelogram

$2\hat{i}-4\hat{j}+5\hat{k}and\hat{i}-2\hat{j}-3\hat{k}$

The diagonal will be the resultant of these two vectors. so

resultant R:

$\overrightarrow{R}=2\hat{i}-4\hat{j}+5\hat{k}+\hat{i}-2\hat{j}-3\hat{k}=3\hat{i}-6\hat{j}+2\hat{k}$

Now unit vector in direction of R

$\overrightarrow{u}=\frac{3\hat{i}-6\hat{j}+2\hat{k}}{\sqrt{3^2+(-6{)}^2+2^2}}=\frac{3\hat{i}-6\hat{j}+2\hat{k}}{\sqrt{49}}=\frac{3\hat{i}-6\hat{j}+2\hat{k}}{7}$

Hence unit vector along the diagonal of the parallelogram

$\overrightarrow{u}=\frac{3}{7}\hat{i}-\frac{6}{7}\hat{j}+\frac{2}{7}\hat{k}$

Now,

Area of parallelogram

$A=(2\hat{i}-4\hat{j}+5\hat{k})\times (\hat{i}-2\hat{j}-3\hat{k})$

$A=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -4& 5\\ 1& -2& -3\end{array}\right|=|\hat{i}(12+10)-\hat{j}(-6-5)+\hat{k}(-4+4)|=|22\hat{i}+11\hat{j}|$

$A=\sqrt{{22}^2+{11}^2}=11\sqrt{5}$

Hence the area of the parallelogram is $11\sqrt{5}$ .

Question 11: Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are $\pm (\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$

Answer:

Let a vector $\overrightarrow{a}$ is equally inclined to axis OX, OY and OZ.

let direction cosines of this vector be

$cos\alpha ,cos\alpha andcos\alpha$

Now

$co{s}^2\alpha +co{s}^2\alpha +co{s}^2\alpha =1$

$co{s}^2\alpha =\frac{1}{3}$

$cos\alpha =\frac{1}{\sqrt{3}}$

Hence direction cosines are:

$(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$

Question 12: Let $\overrightarrow{a}=\hat{i}+4\hat{j}+2\hat{k},\overrightarrow{b}=3\hat{i}-2\hat{j}+7\hat{k}and\overrightarrow{c}=2\hat{i}-\hat{j}+4\hat{k}$ . Find a vector $\overrightarrow{d}$ which is perpendicular to both $\overrightarrow{a}and\overrightarrow{b}and\overrightarrow{c}.\overrightarrow{d}=15$

Answer:

Given,

$\overrightarrow{a}=\hat{i}+4\hat{j}+2\hat{k},\overrightarrow{b}=3\hat{i}-2\hat{j}+7\hat{k}and\overrightarrow{c}=2\hat{i}-\hat{j}+4\hat{k}$

Let $\overrightarrow{d}=d_1\hat{i}+d_2\hat{j}+d_3\hat{k}$

now, since it is given that d is perpendicular to $\overrightarrow{a}$ and $\overrightarrow{b}$ , we got the condition,

$\overrightarrow{b}.\overrightarrow{d}=0$ and $\overrightarrow{a}.\overrightarrow{d}=0$

$(\hat{i}+4\hat{j}+2\hat{k})\cdot (d_1\hat{i}+d_2\hat{j}+d_3\hat{k})=0$
And
$(3\hat{i}-2\hat{j}+7\hat{k})\cdot (d_1\hat{i}+d_2\hat{j}+d_3\hat{k})=0$

$d_1+4d_2+2d_3=0$ And $3d_1-2d_2+7d_3=0$

here we got 2 equation and 3 variable. one more equation will come from the condition:

$\overrightarrow{c}.\overrightarrow{d}=15$

$(2\hat{i}-\hat{j}+4\hat{k})\cdot (d_1\hat{i}+d_2\hat{j}+d_3\hat{k})=15$

$2d_1-d_2+4d_3=15$

so now we have three equations and three variables,

$d_1+4d_2+2d_3=0$

$3d_1-2d_2+7d_3=0$

$2d_1-d_2+4d_3=15$

On solving this three equation we get,

$d_1=\frac{160}{3},d_2=-\frac{5}{3}and{d}_3=-\frac{70}{3}$ ,

Hence required vector :

$\overrightarrow{d}=\frac{160}{3}\hat{i}-\frac{5}{3}\hat{j}-\frac{70}{3}\hat{k}$

Question 13: The scalar product of the vector $\hat{i}+\hat{j}+\hat{k}$ with a unit vector along the sum of vectors $2\hat{i}+4\hat{j}-5\hat{k}$ and $\lambda \hat{i}+2\hat{j}+3\hat{k}$ is equal to one. Find the value of $\lambda$.

Answer:

Let, the sum of vectors $2\hat{i}+4\hat{j}-5\hat{k}$ and $\lambda \hat{i}+2\hat{j}+3\hat{k}$ be

$\overrightarrow{a}=(\lambda +2)\hat{i}+6\hat{j}-2\hat{k}$

unit vector along $\overrightarrow{a}$

$\overrightarrow{u}=\frac{(\lambda +2)\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{(\lambda +2{)}^2+6^2+(-2{)}^2}}=\frac{(\lambda +2)\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{{\lambda }^2+4\lambda +44}}$

Now, the scalar product of this with $\hat{i}+\hat{j}+\hat{k}$

$\overrightarrow{u}.(\hat{i}+\hat{j}+\hat{k})=\frac{(\lambda +2)\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{{\lambda }^2+4\lambda +44}}.(\hat{i}+\hat{j}+\hat{k})$

$\overrightarrow{u}.(\hat{i}+\hat{j}+\hat{k})=\frac{(\lambda +2)+6-2}{\sqrt{{\lambda }^2+4\lambda +44}}=1$

$\frac{(\lambda +2)+6-2}{\sqrt{{\lambda }^2+4\lambda +44}}=1$

$\frac{(\lambda +6)}{\sqrt{{\lambda }^2+4\lambda +44}}=1$

$\lambda =1$

Question 14: If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are mutually perpendicular vectors of equal magnitudes, show that the vector $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}$ is equally inclined to $\overrightarrow{a},\overrightarrow{b}and\overrightarrow{c}$ .

Answer:

Given

$|\overrightarrow{a}|=|\overrightarrow{b}|=|\overrightarrow{c}|$ and

$\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{c}=\overrightarrow{c}.\overrightarrow{a}=0$

Now, let vector $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}$ is inclined to $\overrightarrow{a},\overrightarrow{b}and\overrightarrow{c}$ at ${\theta }_1,{\theta }_{2}and{\theta }_3$ respectively.

Now,

$cos{\theta }_1=\frac{(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}).\overrightarrow{a}}{|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}||\overrightarrow{a}|}=\frac{\overrightarrow{a}.\overrightarrow{a}+\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{c}.\overrightarrow{a}}{|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}||\overrightarrow{a}|}=\frac{\overrightarrow{a}.\overrightarrow{a}}{|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}||\overrightarrow{a}|}=\frac{|\overrightarrow{a}|}{|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|}$

$cos{\theta }_2=\frac{(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}).\overrightarrow{b}}{|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}||\overrightarrow{b}|}=\frac{\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{b}+\overrightarrow{c}.\overrightarrow{b}}{|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}||\overrightarrow{b}|}=\frac{\overrightarrow{b}.\overrightarrow{b}}{|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}||\overrightarrow{b}|}=\frac{|\overrightarrow{b}|}{|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|}$

$cos{\theta }_3=\frac{(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}).\overrightarrow{c}}{|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}||\overrightarrow{c}|}=\frac{\overrightarrow{a}.\overrightarrow{c}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{c}}{|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}||\overrightarrow{c}|}=\frac{\overrightarrow{c}.\overrightarrow{c}}{|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}||\overrightarrow{c}|}=\frac{|\overrightarrow{c}|}{|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|}$

Now, Since, $|\overrightarrow{a}|=|\overrightarrow{b}|=|\overrightarrow{c}|$

$cos{\theta }_1=cos{\theta }_2=cos{\theta }_3$

${\theta }_1={\theta }_2={\theta }_3$

Hence vector $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}$ is equally inclined to $\overrightarrow{a},\overrightarrow{b}and\overrightarrow{c}$ .

Question 15: Prove that $(\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}+\overrightarrow{b})=|{\overrightarrow{a}}^2|+|\overrightarrow{b}{|}^2$ , if and only if $\overrightarrow{a},\overrightarrow{b}$ are perpendicular, given $\overrightarrow{a}\ne 0,\overrightarrow{b}\ne 0$

Answer:

Given in the question,

$\begin{array}{rl}& \overrightarrow{a},\overrightarrow{b}\text{are perpendicular and we need to prove that}\\ & (\overrightarrow{a}+\overrightarrow{b})\cdot (\overrightarrow{a}+\overrightarrow{b})=\left|{\overrightarrow{a}}^2\right|+|\overrightarrow{b}{|}^2\\ & \text{LHS}=(\overrightarrow{a}+\overrightarrow{b})\cdot (\overrightarrow{a}+\overrightarrow{b})=\overrightarrow{a}\cdot \overrightarrow{a}+\overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{b}\cdot \overrightarrow{a}+\overrightarrow{b}\cdot \overrightarrow{b}\\ & =\overrightarrow{a}\cdot \overrightarrow{a}+2\overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{b}\cdot \overrightarrow{b}\\ & =|\overrightarrow{a}{|}^2+2\overrightarrow{a}\cdot \overrightarrow{b}+|\overrightarrow{b}{|}^2\\ & \text{if}\overrightarrow{a},\overrightarrow{b}\text{are perpendicular,}\overrightarrow{a}\cdot \overrightarrow{b}=0\\ & (\overrightarrow{a}+\overrightarrow{b})\cdot (\overrightarrow{a}+\overrightarrow{b})=|\overrightarrow{a}{|}^2+2\overrightarrow{a}\cdot \overrightarrow{b}+|\overrightarrow{b}{|}^2\\ & =|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2\end{array}$

= RHS

LHS ie equal to RHS

Hence proved.

Question 16: Choose the correct answer If $\theta$ is the angle between two vectors $\overrightarrow{a}and\overrightarrow{b}$ , then $\overrightarrow{a}\cdot \overrightarrow{b}\ge 0$ only when
$$\begin{gathered} A)0<\theta <\frac{\pi }{2} \\ B)0\le \theta \le \frac{\pi }{2} \\ C)0<\theta <\pi \\ D)0\le \theta \le \pi \end{gathered}$$

Answer:

Given in the question

$\theta$ is the angle between two vectors $\overrightarrow{a}and\overrightarrow{b}$

$\overrightarrow{a}\cdot \overrightarrow{b}\ge 0$

$|\overrightarrow{a}||\overrightarrow{b}|cos\theta \ge 0$

this will satisfy when

$cos\theta \ge 0$

$0\le \theta \le \frac{\pi }{2}$

Hence option B is the correct answer.

Question 17: Choose the correct answer. Let $\overrightarrow{a}and\overrightarrow{b}$ be two unit vectors and $\theta$ is the angle between them. Then $\overrightarrow{a}+\overrightarrow{b}$ is a unit vector if

$$\begin{gathered} A)\theta =\frac{\pi }{4} \\ B)\theta =\frac{\pi }{3} \\ C)\theta =\frac{\pi }{2} \\ D)\theta =\frac{2\pi }{3} \end{gathered}$$

Answer:

Gicen in the question

$\overrightarrow{a}and\overrightarrow{b}$ be two unit vectors and $\theta$ is the angle between them

$|\overrightarrow{a}|=1,and|\overrightarrow{b}|=1$

also

$|\overrightarrow{a}+\overrightarrow{b}|=1$

$|\overrightarrow{a}+\overrightarrow{b}{|}^2=1$

$|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2+2\overrightarrow{a}.\overrightarrow{b}=1$

$1+1+2\overrightarrow{a}.\overrightarrow{b}=1$

$\overrightarrow{a}.\overrightarrow{b}=-\frac{1}{2}$

$|\overrightarrow{a}||\overrightarrow{b}|cos\theta =-\frac{1}{2}$

$cos\theta =-\frac{1}{2}$

$\theta =\frac{2\pi }{3}$

Then $\overrightarrow{a}+\overrightarrow{b}$ is a unit vector if $\theta =\frac{2\pi }{3}$

Hence option D is correct.

Question 18: The value of $\hat{i}(\hat{j}\times \hat{k})+\hat{j}(\hat{i}\times \hat{k})+\hat{k}(\hat{i}\times \hat{j})$ is

(A) 0

(B) –1

(C) 1

(D) 3

Answer:

To find the value of $\hat{i}(\hat{j}\times \hat{k})+\hat{j}(\hat{i}\times \hat{k})+\hat{k}(\hat{i}\times \hat{j})$

$$\begin{gathered} \hat{i}(\hat{j}\times \hat{k})+\hat{j}(\hat{i}\times \hat{k})+\hat{k}(\hat{i}\times \hat{j}) \\ =\hat{i}.\hat{i}+\hat{j}(-\hat{j})+\hat{k}.\hat{k} \\ =1-1+1 \\ =1 \end{gathered}$$

Hence option C is correct.

Question 19: Choose the correct. If $\theta$ is the angle between any two vectors $\overrightarrow{a}and\overrightarrow{b}$ , then $|\overrightarrow{a}\cdot \overrightarrow{b}|=|\overrightarrow{a}\times \overrightarrow{b}|$ when $\theta$
is equal to

(A) 0
(B) $\pi /4$
( C )$\pi /2$
( D ) $\pi$

Answer:

Given in the question

$\theta$ is the angle between any two vectors $\overrightarrow{a}and\overrightarrow{b}$ and $|\overrightarrow{a}\cdot \overrightarrow{b}|=|\overrightarrow{a}\times \overrightarrow{b}|$

To find the value of $\theta$

Hence, option D is correct.