Solution

If $\vec{r}$ is a unit vector in the $X Y$-plane, then $\vec{r}=\cos \theta \hat{i}+\sin \theta \hat{j}$.

Here, $\theta$ is the angle made by the unit vector with the positive direction of the $x$-axis.

Therefore, for $\theta=30^{\circ}$ :

$\vec{r}=\cos 30^{\circ} \hat{i}+\sin 30^{\circ} \hat{j}=\frac{\sqrt{3}}{2} \hat{i}+\frac{1}{2} \hat{j}$

Hence, the required unit vector is $\frac{\sqrt{3}}{2} \hat{i}+\frac{1}{2} \hat{j}$

Solution

The vector joining the points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ can be obtained by, $\overrightarrow{{}PQ}=$ Position vector of $Q-$ Position vector of $P$

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

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

Hence, the scalar components and the magnitude of the vector joining the given points

are respectively ${(x_2-x_1),(y_2-y_1),(z_2-z_1)}$ and $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}+(z_2-z_1)^{2}}$.

Solution

Let $O$ and $B$ be the initial and final positions of the girl respectively. Then, the girl’s position can be shown as:

Now, we have:

$ \begin{aligned} \overrightarrow{{}OA} & =-4 \hat{i} \\ \overrightarrow{{}AB} & =\hat{i}|\overrightarrow{{}AB}| \cos 60^{\circ}+\hat{j}|\overrightarrow{{}AB}| \sin 60^{\circ} \\ & =\hat{i} 3 \times \frac{1}{2}+\hat{j} 3 \times \frac{\sqrt{3}}{2} \\ & =\frac{3}{2} \hat{i}+\frac{3 \sqrt{3}}{2} \hat{j} \end{aligned} $

By the triangle law of vector addition, we have:

$ \begin{aligned} \overrightarrow{{}OB} & =\overrightarrow{{}OA}+\overrightarrow{{}AB} \\ & =(-4 \hat{i})+(\frac{3}{2} \hat{i}+\frac{3 \sqrt{3}}{2} \hat{j}) \\ & =(-4+\frac{3}{2}) \hat{i}+\frac{3 \sqrt{3}}{2} \hat{j} \\ & =(\frac{-8+3}{2}) \hat{i}+\frac{3 \sqrt{3}}{2} \hat{j} \\ & =\frac{-5}{2} \hat{i}+\frac{3 \sqrt{3}}{2} \hat{j} \end{aligned} $

Hence, the girl’s displacement from her initial point of departure is

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

Solution

In $\triangle ABC$, let $\overrightarrow{{}CB}=\vec{a}, \overrightarrow{{}CA}=\vec{b}$, and $\overrightarrow{{}AB}=\vec{c}$ (as shown in the following figure).

Now, by the triangle law of vector addition, we have $\vec{a}=\vec{b}+\vec{c}$.

It is clearly known that $|\vec{a}|,|\vec{b}|$, and $|\vec{c}|$ represent the sides of $\triangle ABC$.

Also, it is known that the sum of the lengths of any two sides of a triangle is greater than the third side.

$\therefore|\vec{a}|<|\vec{b}|+|\vec{c}|$

Hence, it is not true that $|\vec{a}|=|\vec{b}|+|\vec{c}|$.

Solution

$x(\hat{i}+\hat{j}+\hat{k})$ is a unit vector if $|x(\hat{i}+\hat{j}+\hat{k})|=1$.

Now,

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

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

$\Rightarrow \sqrt{3 x^{2}}=1$

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

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

Hence, the required value of $x$ is $\pm \frac{1}{\sqrt{3}}$.

Solution

We have,

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

Let $\vec{c}$ be the resultant of $\vec{a}$ and $\vec{b}$.

Then,

$\vec{c}=\vec{a}+\vec{b}=(2+1) \hat{i}+(3-2) \hat{j}+(-1+1) \hat{k}=3 \hat{i}+\hat{j}$

$\therefore|\vec{c}|=\sqrt{3^{2}+1^{2}}=\sqrt{9+1}=\sqrt{10}$

$\therefore \hat{c}=\frac{\vec{c}}{|\vec{c}|}=\frac{(3 \hat{i}+\hat{j})}{\sqrt{10}}$

Hence, the vector of magnitude 5 units and parallel to the resultant of vectors $\vec{a}$ and $\vec{b}$ is

$\pm 5 \cdot \hat{c}= \pm 5 \cdot \frac{1}{\sqrt{10}}(3 \hat{i}+\hat{j})= \pm \frac{3 \sqrt{10} \hat{i}}{2} \pm \frac{\sqrt{10}}{2} \hat{j}$.

Solution

We have,

$ \begin{aligned} & \vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}-\hat{j}+3 \hat{k} \text{ and } \vec{c}=\hat{i}-2 \hat{j}+\hat{k} \\ & \begin{aligned} 2 \vec{a}-\vec{b}+3 \vec{c} & =2(\hat{i}+\hat{j}+\hat{k})-(2 \hat{i}-\hat{j}+3 \hat{k})+3(\hat{i}-2 \hat{j}+\hat{k}) \\ & =2 \hat{i}+2 \hat{j}+2 \hat{k}-2 \hat{i}+\hat{j}-3 \hat{k}+3 \hat{i}-6 \hat{j}+3 \hat{k} \\ & =3 \hat{i}-3 \hat{j}+2 \hat{k} \end{aligned} \\ & \begin{aligned} |2 \vec{a}-\vec{b}+3 \vec{c}| & =\sqrt{3^{2}+(-3)^{2}+2^{2}}=\sqrt{9+9+4}=\sqrt{22} \end{aligned} \end{aligned} $

Hence, the unit vector along $2 \vec{a}-\vec{b}+3 \vec{c}$ is

$ \frac{2 \vec{a}-\vec{b}+3 \vec{c}}{|2 \vec{a}-\vec{b}+3 \vec{c}|}=\frac{3 \hat{i}-3 \hat{j}+2 \hat{k}}{\sqrt{22}}=\frac{3}{\sqrt{22}} \hat{i}-\frac{3}{\sqrt{22}} \hat{j}+\frac{2}{\sqrt{22}} \hat{k} $

Solution

The given points are $A(1,-2,-8), B(5,0,-2)$, and $C(11,3,7)$.

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

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

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

$|\overrightarrow{{}AB}|=\sqrt{4^{2}+2^{2}+6^{2}}=\sqrt{16+4+36}=\sqrt{56}=2 \sqrt{14}$

$|\overrightarrow{{}BC}|=\sqrt{6^{2}+3^{2}+9^{2}}=\sqrt{36+9+81}=\sqrt{126}=3 \sqrt{14}$

$|\overrightarrow{{}AC}|=\sqrt{10^{2}+5^{2}+15^{2}}=\sqrt{100+25+225}=\sqrt{350}=5 \sqrt{14}$

$\therefore|\overrightarrow{{}AC}|=|\overrightarrow{{}AB}|+|\overrightarrow{{}BC}|$

Thus, the given points $A, B$, and $C$ are collinear.

Now, let point $B$ divide $AC$ in the ratio $\lambda: 1$. Then, we have:

$ \begin{aligned} & \overrightarrow{{}OB}=\frac{\lambda \overrightarrow{{}OC}+\overrightarrow{{}OA}}{(\lambda+1)} \\ & \Rightarrow 5 \hat{i}-2 \hat{k}=\frac{\lambda(11 \hat{i}+3 \hat{j}+7 \hat{k})+(\hat{i}-2 \hat{j}-8 \hat{k})}{\lambda+1} \\ & \Rightarrow(\lambda+1)(5 \hat{i}-2 \hat{k})=11 \lambda \hat{i}+3 \lambda \hat{j}+7 \lambda \hat{k}+\hat{i}-2 \hat{j}-8 \hat{k} \\ & \Rightarrow 5(\lambda+1) \hat{i}-2(\lambda+1) \hat{k}=(11 \lambda+1) \hat{i}+(3 \lambda-2) \hat{j}+(7 \lambda-8) \hat{k} \end{aligned} $

On equating the corresponding components, we get:

$5(\lambda+1)=11 \lambda+1$

$\Rightarrow 5 \lambda+5=11 \lambda+1$

$\Rightarrow 6 \lambda=4$

$\Rightarrow \lambda=\frac{4}{6}=\frac{2}{3}$

Hence, point $B$ divides $A C$ in the ratio $2: 3$.

Solution

It is given that $\overrightarrow{{}OP}=2 \vec{a}+\vec{b}, \overrightarrow{{}OQ}=\vec{a}-3 \vec{b}$.

It is given that point $R$ divides a line segment joining two points $P$ and $Q$ externally in the ratio $1: 2$. Then, on using the section formula, we get:

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

Therefore, the position vector of point $R$ is $3 \vec{a}+5 \vec{b}$.

Position vector of the mid-point of $RQ=\frac{\overrightarrow{{}OQ}+\overrightarrow{{}OR}}{2}$ $=\frac{(\vec{a}-3 \vec{b})+(3 \vec{a}+5 \vec{b})}{2}$

$=2 \vec{a}+\vec{b}$

$=\overrightarrow{{}OP}$

Hence, $P$ is the mid-point of the line segment $R Q$.

Solution

Adjacent sides of a parallelogram are given as: $\vec{a}=2 \hat{i}-4 \hat{j}+5 \hat{k}$ and $\vec{b}=\hat{i}-2 \hat{j}-3 \hat{k}$

Then, the diagonal of a parallelogram is given by $\vec{a}+\vec{b}$.

$\vec{a}+\vec{b}=(2+1) \hat{i}+(-4-2) \hat{j}+(5-3) \hat{k}=3 \hat{i}-6 \hat{j}+2 \hat{k}$

Thus, the unit vector parallel to the diagonal is

$ \frac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|}=\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{9+36+4}}=\frac{3 \hat{i}-6 \hat{j}+2 \hat{k}}{7}=\frac{3}{7} \hat{i}-\frac{6}{7} \hat{j}+\frac{2}{7} \hat{k} . $

$\therefore$ Area of parallelogram $ABCD=|\vec{a} \times \vec{b}|$

$ \begin{aligned} & \vec{a} \times \vec{b}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -4 & 5 \\ 1 & -2 & -3 \end{vmatrix} \\ &=\hat{i}(12+10)-\hat{j}(-6-5)+\hat{k}(-4+4) \\ &=22 \hat{i}+11 \hat{j} \\ &=11(2 \hat{i}+\hat{j}) \\ & \therefore|\vec{a} \times \vec{b}|=11 \sqrt{2^{2}+1^{2}}=11 \sqrt{5} \end{aligned} $

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

Solution

Let a vector be equally inclined to axes OX, OY, and OZ at angle $a$.

Then, the direction cosines of the vector are $\cos a, \cos a$, and $\cos a$.

Now,

$\cos ^{2} \alpha+\cos ^{2} \alpha+\cos ^{2} \alpha=1$

$\Rightarrow 3 \cos ^{2} \alpha=1$

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

Hence, the direction cosines of the vector which are equally inclined to the axes

are $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$.

Solution

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

Since $\vec{d}$ is perpendicular to both $\vec{a}$ and $\vec{b}$, we have:

$\vec{d} \cdot \vec{a}=0$

$\Rightarrow d_1+4 d_2+2 d_3=0$

And,

$\vec{d} \cdot \vec{b}=0$

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

Also, it is given that:

$\vec{c} \cdot \vec{d}=15$

$\Rightarrow 2 d_1-d_2+4 d_3=15$

On solving (i), (ii), and (iii), we get:

$d_1=\frac{160}{3}, d_2=-\frac{5}{3}$ and $d_3=-\frac{70}{3}$

$\therefore \vec{d}=\frac{160}{3} \hat{i}-\frac{5}{3} \hat{j}-\frac{70}{3} \hat{k}=\frac{1}{3}(160 \hat{i}-5 \hat{j}-70 \hat{k})$

Hence, the required vector is $\frac{1}{3}(160 \hat{i}-5 \hat{j}-70 \hat{k})$.

Solution

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

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

Therefore, unit vector along $(2 \hat{i}+4 \hat{j}-5 \hat{k})+(\lambda \hat{i}+2 \hat{j}+3 \hat{k})$ is given as:

$\frac{(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{(2+\lambda)^{2}+6^{2}+(-2)^{2}}}=\frac{(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{4+4 \lambda+\lambda^{2}+36+4}}=\frac{(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{\lambda^{2}+4 \lambda+44}}$

Scalar product of $(\hat{i}+\hat{j}+\hat{k})$ with this unit vector is 1 .

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

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

$\Rightarrow \sqrt{\lambda^{2}+4 \lambda+44}=\lambda+6$

$\Rightarrow \lambda^{2}+4 \lambda+44=(\lambda+6)^{2}$

$\Rightarrow \lambda^{2}+4 \lambda+44=\lambda^{2}+12 \lambda+36$

$\Rightarrow 8 \lambda=8$

$\Rightarrow \lambda=1$

Hence, the value of $\lambda$ is 1 .

Solution

Since $\vec{a}, \vec{b}$, and $\vec{c}$ are mutually perpendicular vectors, we have

$\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{a}=0$.

It is given that:

$|\vec{a}|=|\vec{b}|=|\vec{c}|$

Let vector $\vec{a}+\vec{b}+\vec{c}$ be inclined to $\vec{a}, \vec{b}$, and $\vec{c}$ at angles $\theta_1, \theta_2$, and $\theta_3$ respectively.

Then, we have:

$ \begin{aligned} \cos \theta_1 & =\frac{(\vec{a}+\vec{b}+\vec{c}) \cdot \vec{a}}{|\vec{a}+\vec{b}+\vec{c}||\vec{a}|}=\frac{\vec{a} \cdot \vec{a}+\vec{b} \cdot \vec{a}+\vec{c} \cdot \vec{a}}{|\vec{a}+\vec{b}+\vec{c}||\vec{a}|} \\ & =\frac{|\vec{a}|^{2}}{|\vec{a}+\vec{b}+\vec{c}||\vec{a}|} \quad[\vec{b} \cdot \vec{a}=\vec{c} \cdot \vec{a}=0] \\ & =\frac{|\vec{a}|}{|\vec{a}+\vec{b}+\vec{c}|} \\ \cos \theta_2 & =\frac{(\vec{a}+\vec{b}+\vec{c}) \cdot \vec{b}}{|\vec{a}+\vec{b}+\vec{c}||\vec{b}|}=\frac{\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{b}+\vec{c} \cdot \vec{b}}{|\vec{a}+\vec{b}+\vec{c}| \cdot|\vec{b}|} \\ & =\frac{|\vec{b}|^{2}}{|\vec{a}+\vec{b}+\vec{c}| \cdot|\vec{b}|} \quad[\vec{a} \cdot \vec{b}=\vec{c} \cdot \vec{b}=0] \\ \cos \theta_3 & =\frac{\vec{a} \cdot \vec{c}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{c}}{|\vec{a}+\vec{b}+\vec{c}||\vec{c}|} \\ & =\frac{|\vec{c}|^{2}}{|\vec{a}+\vec{b}+\vec{c}||\vec{c}|} \\ & =\frac{|\vec{c}|}{|\vec{a}+\vec{b}+\vec{c}|} \end{aligned} $

Now, as $|\vec{a}|=|\vec{b}|=|\vec{c}|, \cos \theta_1=\cos \theta_2=\cos \theta_3$.

$\therefore \theta_1=\theta_2=\theta_3$

Hence, the vector $(\vec{a}+\vec{b}+\vec{c})$ is equally inclined to $\vec{a}, \vec{b}$, and $\vec{c}$.

Solution

$(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^{2}+|\vec{b}|^{2}$

$\Leftrightarrow \vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}=|\vec{a}|^{2}+|\vec{b}|^{2} \quad$ [Distributivity of scalar products over addition]

$\Leftrightarrow|\vec{a}|^{2}+2 \vec{a} \cdot \vec{b}+|\vec{b}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2} \quad[\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}$ (Scalar product is commutative)

$\Leftrightarrow 2 \vec{a} \cdot \vec{b}=0$

$\Leftrightarrow \vec{a} \cdot \vec{b}=0$

$\therefore \vec{a}$ and $\vec{b}$ are perpendicular. $\quad[\vec{a} \neq \overrightarrow{{}0}, \vec{b} \neq \overrightarrow{{}0}$ (Given) $]$

Solution

Let $\theta$ be the angle between two vectors $\vec{a}$ and $\vec{b}$.

Then, without loss of generality, $\vec{a}$ and $\vec{b}$ are non-zero vectors so

that $|\vec{a}|$ and $|\vec{b}|$ are positive.

It is known that $\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta$.

$\therefore \vec{a} \cdot \vec{b} \geq 0$

$\Rightarrow|\vec{a}||\vec{b}| \cos \theta \geq 0$

$\Rightarrow \cos \theta \geq 0 \quad[|\vec{a}|$ and $|\vec{b}|$ are positive $]$

$\Rightarrow 0 \leq \theta \leq \frac{\pi}{2}$

Hence, $\vec{a} \cdot \vec{b} \geq 0$ when $0 \leq \theta \leq \frac{\pi}{2}$.

The correct answer is $B$.

Solution

Let $\vec{a}$ and $\vec{b}$ be two unit vectors and $\theta$ be the angle between them.

Then, $|\vec{a}|=|\vec{b}|=1$.

Now, $\vec{a}+\vec{b}$ is a unit vector if $|\vec{a}+\vec{b}|=1$.

$ \begin{aligned} & |\vec{a}+\vec{b}|=1 \\ & \Rightarrow(\vec{a}+\vec{b})^{2}=1 \\ & \Rightarrow(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=1 \\ & \Rightarrow \vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}=1 \\ & \Rightarrow|\vec{a}|^{2}+2 \vec{a} \cdot \vec{b}+|\vec{b}|^{2}=1 \\ & \Rightarrow 1^{2}+2|\vec{a}||\vec{b}| \cos \theta+1^{2}=1 \\ & \Rightarrow 1+2 \cdot 1 \cdot 1 \cos \theta+1=1 \\ & \Rightarrow \cos \theta=-\frac{1}{2} \\ & \Rightarrow \theta=\frac{2 \pi}{3} \end{aligned} $

Hence, $\vec{a}+\vec{b}$ is a unit vector if $\theta=\frac{2 \pi}{3}$.

The correct answer is D.

Solution

$ \begin{aligned} & \hat{i} \cdot(\hat{j} \times \hat{k})+\hat{j} \cdot(\hat{i} \times \hat{k})+\hat{k} \cdot(\hat{i} \times \hat{j}) \\ & =\hat{i} \cdot \hat{i}+\hat{j} \cdot(-\hat{j})+\hat{k} \cdot \hat{k} \\ & =1-\hat{j} \cdot \hat{j}+1 \\ & =1-1+1 \\ & =1 \end{aligned} $

The correct answer is C.

Solution

Let $\theta$ be the angle between two vectors $\vec{a}$ and $\vec{b}$.

Then, without loss of generality, $\vec{a}$ and $\vec{b}$ are non-zero vectors, so

that $|\vec{a}|$ and $|\vec{b}|$ are positive

$|\vec{a} \cdot \vec{b}|=|\vec{a} \times \vec{b}|$

$\Rightarrow|\vec{a}||\vec{b}| \cos \theta=|\vec{a}||\vec{b}| \sin \theta$

$\Rightarrow \cos \theta=\sin \theta \quad[|\vec{a}|$ and $|\vec{b}|$ are positive $]$

$\Rightarrow \tan \theta=1$

$\Rightarrow \theta=\frac{\pi}{4}$

Hence, $|\vec{a} \vec{b}|=|\vec{a} \times \vec{b}|$ when $\theta$ isequal to $\frac{\pi}{4}$.

The correct answer is $B$.