Solution

We have,

$ \begin{aligned} & \vec{a}=\hat{i}-7 \hat{j}+7 \hat{k} \text{ and } \vec{b}=3 \hat{i}-2 \hat{j}+2 \hat{k} \\ & \begin{aligned} \vec{a} \times \vec{b} & = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -7 & 7 \\ 3 & -2 & 2 \end{vmatrix} \\ & =\hat{i}(-14+14)-\hat{j}(2-21)+\hat{k}(-2+21)=19 \hat{j}+19 \hat{k} \end{aligned} \\ & \therefore|\vec{a} \times \vec{b}|=\sqrt{(19)^{2}+(19)^{2}}=\sqrt{2 \times(19)^{2}}=19 \sqrt{2} \end{aligned} $

Solution

We have,

$ \begin{aligned} & \vec{a}=3 \hat{i}+2 \hat{j}+2 \hat{k} \text{ and } \vec{b}=\hat{i}+2 \hat{j}-2 \hat{k} \\ & \therefore \vec{a}+\vec{b}=4 \hat{i}+4 \hat{j}, \vec{a}-\vec{b}=2 \hat{i}+4 \hat{k} \\ & (\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{vmatrix} =\hat{i}(16)-\hat{j}(16)+\hat{k}(-8)=16 \hat{i}-16 \hat{j}-8 \hat{k} \\ & \therefore|(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})|=\sqrt{16^{2}+(-16)^{2}+(-8)^{2}} \\ & =\sqrt{2^{2} \times 8^{2}+2^{2} \times 8^{2}+8^{2}} \\ & =8 \sqrt{2^{2}+2^{2}+1}=8 \sqrt{9}=8 \times 3=24 \end{aligned} $

Hence, the unit vector perpendicular to each of the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ is given by the relation,

$= \pm \frac{(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})}{|(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})|}= \pm \frac{16 \hat{i}-16 \hat{j}-8 \hat{k}}{24}$

$= \pm \frac{2 \hat{i}-2 \hat{j}-\hat{k}}{3}= \pm \frac{2}{3} \hat{i} \mp \frac{2}{3} \hat{j} \mp \frac{1}{3} \hat{k}$

Solution

Let unit vector $\vec{a}$ have ( $.a_1, a_2, a_3)$ components.

$ \vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k} $

Since $\vec{a}$ is a unit vector, $|\vec{a}|=1$.

Also, it is given that $\vec{a}$ makes angles $\frac{\pi}{3}$ with $\hat{i}, \frac{\pi}{4}$ with $\hat{j}$, and an acute angle $\theta$ with $\hat{k}$. Then, we have:

$\cos \frac{\pi}{3}=\frac{a_1}{|\vec{a}|}$

$\Rightarrow \frac{1}{2}=a_1 \quad[|\vec{a}|=1]$

$\cos \frac{\pi}{4}=\frac{a_2}{|\vec{a}|}$

$\Rightarrow \frac{1}{\sqrt{2}}=a_2 \quad[|\vec{a}|=1]$

Also, $\cos \theta=\frac{a_3}{|\vec{a}|}$.

$\Rightarrow a_3=\cos \theta$

Now,

$|a|=1$

$\Rightarrow \sqrt{a_1^{2}+a_2^{2}+a_3^{2}}=1$

$\Rightarrow(\frac{1}{2})^{2}+(\frac{1}{\sqrt{2}})^{2}+\cos ^{2} \theta=1$

$\Rightarrow \frac{1}{4}+\frac{1}{2}+\cos ^{2} \theta=1$

$\Rightarrow \frac{3}{4}+\cos ^{2} \theta=1$

$\Rightarrow \cos ^{2} \theta=1-\frac{3}{4}=\frac{1}{4}$

$\Rightarrow \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}$

$\therefore a_3=\cos \frac{\pi}{3}=\frac{1}{2}$

Hence, $\theta=\frac{\pi}{3}$ and the components of $\vec{a}$ are $(\frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2})$.

Solution

$(\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})$

$=(\vec{a}-\vec{b}) \times \vec{a}+(\vec{a}-\vec{b}) \times \vec{b} \quad$ [By distributivity of vector product over addition]

$=\vec{a} \times \vec{a}-\vec{b} \times \vec{a}+\vec{a} \times \vec{b}-\vec{b} \times \vec{b} \quad$ [Again, by distributivity of vector product over addition]

$=\overrightarrow{{}0}+\vec{a} \times \vec{b}+\vec{a} \times \vec{b}-\overrightarrow{{}0}$

$=2 (\vec{a} \times \vec{b})$

Solution

$(2 \hat{i}+6 \hat{j}+27 \hat{k}) \times(\hat{i}+\lambda \hat{j}+\mu \hat{k})=\overrightarrow{{}0}$

$\Rightarrow \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 27 \\ 1 & \lambda & \mu\end{vmatrix} =0 \hat{i}+0 \hat{j}+0 \hat{k}$

$\Rightarrow \hat{i}(6 \mu-27 \lambda)-\hat{j}(2 \mu-27)+\hat{k}(2 \lambda-6)=0 \hat{i}+0 \hat{j}+0 \hat{k}$

On comparing the corresponding components, we have:

$6 \mu-27 \lambda=0$

$2 \mu-27=0$

$2 \lambda-6=0$

Now,

$2 \lambda-6=0 \Rightarrow \lambda=3$

$2 \mu-27=0 \Rightarrow \mu=\frac{27}{2}$

Hence, $\lambda=3$ and $\mu=\frac{27}{2}$.

Solution

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

Then,

(i) Either $|\vec{a}|=0$ or $|\vec{b}|=0$, or $\vec{a} \perp \vec{b}$ (in case $\vec{a}$ and $\vec{b}$ are non-zero)

$\vec{a} \times \vec{b}=0$

(ii) Either $|\vec{a}|=0$ or $|\vec{b}|=0$, or $\vec{a} | \vec{b}$ (in case $\vec{a}$ and $\vec{b}$ are non-zero)

But, $\vec{a}$ and $\vec{b}$ cannot be perpendicular and parallel simultaneously.

Hence, $|\vec{a}|=0$ or $|\vec{b}|=0$.

Solution

We have,

$\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}, \vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}, \vec{c}=c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}$

$(\vec{b}+\vec{c})=(b_1+c_1) \hat{i}+(b_2+c_2) \hat{j}+(b_3+c_3) \hat{k}$

Now, $\vec{a} \times(\vec{b}+\vec{c}) \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1+c_1 & b_2+c_2 & b_3+c_3\end{vmatrix} $

$=\hat{i}[a_2(b_3+c_3)-a_3(b_2+c_2)]-\hat{j}[a_1(b_3+c_3)-a_3(b_1+c_1)]+\hat{k}[a_1(b_2+c_2)-a_2(b_1+c_1)]$

$=\hat{i}[a_2 b_3+a_2 c_3-a_3 b_2-a_3 c_2]+\hat{j}[-a_1 b_3-a_1 c_3+a_3 b_1+a_3 c_1]+\hat{k}[a_1 b_2+a_1 c_2-a_2 b_1-a_2 c_1]$

$\vec{a} \times \vec{b}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3\end{vmatrix} $

$=\hat{i}[a_2 b_3-a_3 b_2]+\hat{j}[b_1 a_3-a_1 b_3]+\hat{k}[a_1 b_2-a_2 b_1]$

$\vec{a} \times \vec{c}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ c_1 & c_2 & c_3\end{vmatrix} $

$$ \begin{equation*} =\hat{i}[a_2 c_3-a_3 c_2]+\hat{j}[a_3 c_1-a_1 c_3]+\hat{k}[a_1 c_2-a_2 c_1] \tag{3} \end{equation*} $$

On adding (2) and (3), we get:

$(\vec{a} \times \vec{b})+(\vec{a} \times \vec{c})=\hat{i}[a_2 b_3+a_2 c_3-a_3 b_2-a_3 c_2]+\hat{j}[b_1 a_3+a_3 c_1-a_1 b_3-a_1 c_3]$

$+\hat{k}[a_1 b_2+a_1 c_2-a_2 b_1-a_2 c_1]$

Now, from (1) and (4), we have:

$\vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}$

Hence, the given result is proved.

Solution

Take any parallel non-zero vectors so that $\vec{a} \times \vec{b}=\overrightarrow{{}0}$.

Let $\vec{a}=2 \hat{i}+3 \hat{j}+4 \hat{k}, \vec{b}=4 \hat{i}+6 \hat{j}+8 \hat{k}$.

Then,

$\vec{a} \times \vec{b}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 4 & 6 & 8\end{vmatrix} =\hat{i}(24-24)-\hat{j}(16-16)+\hat{k}(12-12)=0 \hat{i}+0 \hat{j}+0 \hat{k}=\overrightarrow{{}0}$

It can now be observed that:

$|\vec{a}|=\sqrt{2^{2}+3^{2}+4^{2}}=\sqrt{29}$

$\therefore \vec{a} \neq \overrightarrow{{}0}$

$|\vec{b}|=\sqrt{4^{2}+6^{2}+8^{2}}=\sqrt{116}$

$\therefore \vec{b} \neq \overrightarrow{{}0}$

Hence, the converse of the given statement need not be true.

Solution

The vertices of triangle $A B C$ are given as $A(1,1,2), B(2,3,5)$, and $C(1,5,5)$.

The adjacent sides $\overrightarrow{{}AB}$ and $\overrightarrow{{}BC}$ of $\triangle ABC$ are given as:

$ \begin{aligned} & \overrightarrow{{}AB}=(2-1) \hat{i}+(3-1) \hat{j}+(5-2) \hat{k}=\hat{i}+2 \hat{j}+3 \hat{k} \\ & \overrightarrow{{}BC}=(1-2) \hat{i}+(5-3) \hat{j}+(5-5) \hat{k}=-\hat{i}+2 \hat{j} \end{aligned} $

Area of $\triangle ABC=\frac{1}{2}|\overrightarrow{{}AB} \times \overrightarrow{{}BC}|$

$\overrightarrow{{}AB} \times \overrightarrow{{}BC}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -1 & 2 & 0\end{vmatrix} =\hat{i}(-6)-\hat{j}(3)+\hat{k}(2+2)=-6 \hat{i}-3 \hat{j}+4 \hat{k}$

$\therefore|\overrightarrow{{}AB} \times \overrightarrow{{}BC}|=\sqrt{(-6)^{2}+(-3)^{2}+4^{2}}=\sqrt{36+9+16}=\sqrt{61}$

Hence, the area of $\triangle A B C$ is $\frac{\sqrt{61}}{2}$ square units.

Solution

The area of the parallelogram whose adjacent sides are $\vec{a}$ and $\vec{b}$ is $|\vec{a} \times \vec{b}|$.

Adjacent sides are given as:

$\vec{a}=\hat{i}-\hat{j}+3 \hat{k}$ and $\vec{b}=2 \hat{i}-7 \hat{j}+\hat{k}$

$\therefore \vec{a} \times \vec{b}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 3 \\ 2 & -7 & 1\end{vmatrix} =\hat{i}(-1+21)-\hat{j}(1-6)+\hat{k}(-7+2)=20 \hat{i}+5 \hat{j}-5 \hat{k}$

$|\vec{a} \times \vec{b}|=\sqrt{20^{2}+5^{2}+5^{2}}=\sqrt{400+25+25}=15 \sqrt{2}$

Hence, the area of the given parallelogram is $15 \sqrt{2}$ square units .

Solution

It is given that $|\vec{a}|=3$ and $|\vec{b}|=\frac{\sqrt{2}}{3}$.

We know that $\vec{a} \times \vec{b}=|\vec{a}||\vec{b}| \sin \theta \hat{n}$, where $\hat{n}$ is a unit vector perpendicular to both $\vec{a}$ and $\vec{b}$ and $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.

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

$|\vec{a} \times \vec{b}|=1$

$\Rightarrow|| \vec{a}|| \vec{b}|\sin \theta \hat{n}|=1$

$\Rightarrow|\vec{a}||\vec{b}||\sin \theta|=1$

$\Rightarrow 3 \times \frac{\sqrt{2}}{3} \times \sin \theta=1$

$\Rightarrow \sin \theta=\frac{1}{\sqrt{2}}$

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

Hence, $\vec{a} \times \vec{b}$ is a unit vector if the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{4}$.

The correct answer is $B$.

Solution

The position vectors of vertices $A, B, C$, and $D$ of rectangle $A B C D$ are given as:

$\overrightarrow{{}OA}=-\hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}, \overrightarrow{{}OB}=\hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}, \overrightarrow{{}OC}=\hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}, \overrightarrow{{}OD}=-\hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}$

The adjacent sides $\overrightarrow{{}AB}$ and $\overrightarrow{{}BC}$ of the given rectangle are given as:

$ \begin{aligned} & \overrightarrow{{}AB}=(1+1) \hat{i}+(\frac{1}{2}-\frac{1}{2}) \hat{j}+(4-4) \hat{k}=2 \hat{i} \\ & \overrightarrow{{}BC}=(1-1) \hat{i}+(-\frac{1}{2}-\frac{1}{2}) \hat{j}+(4-4) \hat{k}=-\hat{j} \\ & \therefore \overrightarrow{{}AB} \times \overrightarrow{{}BC}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 0 \\ 0 & -1 & 0 \end{vmatrix} =\hat{k}(-2)=-2 \hat{k} \\ & |\overrightarrow{{}AB} \times \overrightarrow{{}AC}|=\sqrt{(-2)^{2}}=2 \end{aligned} $

Now, it is known that the area of a parallelogram whose adjacent sides are $\vec{a}$ and $\vec{b}$ is $|\vec{a} \times \vec{b}|$.

Hence, the area of the given rectangle is $|\overrightarrow{{}AB} \times \overrightarrow{{}BC}|=2$ square units. The correct answer is $C$.