91Ó°ÊÓ

A vector $\stackrel{→}{A}=2\hat{i}-3\hat{j}+\hat{k}$.

So magnitude of vector is:

$$\begin{gathered} \left|A\right|=\sqrt{{\left(2\right)}^2+{\left(-3\right)}^2+{\left(1\right)}^2} \\ =\sqrt{4+9+1} \\ =\sqrt{14} \end{gathered}$$

If$\stackrel{→}{A}$ and$\stackrel{→}{B}$ are two given vectors with angle$\mathrm{θ}$ between them, then

Dot product is given as:

$\stackrel{→}{A}·\stackrel{→}{B}=\left|A\right|\left|B\right|\cos \mathrm{θ}$

Cross product is given as

$\stackrel{→}{A}×\stackrel{→}{B}=\left|A\right|\left|B\right|\sin \mathrm{θ}$

Given $\stackrel{→}{A}=2\hat{i}-3\hat{j}+\hat{k}$.

And$\stackrel{→}{A}·\stackrel{→}{B}=0$

This implies

$\left|A\right|\left|B\right|\cos \mathrm{θ}=0$

Since $\left|A\right|=\sqrt{14}≠0$

So either,$\left|B\right|=0$ or $\cos \mathrm{θ}=0$

If $\cos \mathrm{θ}=0$, then $\mathrm{θ}=90°$

This implies and are perpendicular vectors.

So any vector perpendicular to$\stackrel{→}{A}$ will make dot product zero, and$\stackrel{→}{B}$ need not be zero.

Now, find a vector perpendicular to $\stackrel{→}{A}$.

$$\begin{gathered} \left|\begin{array}{ccc}i& j& k\\ 2& -3& 1\\ 1& 2& 3\end{array}\right|=\left(-9-2\right)\hat{i}-\left(6-1\right)\hat{j}+\left(4-\left(-3\right)\right)\hat{k} \\ =-11\hat{i}-5\hat{j}+7\hat{k} \end{gathered}$$

Let $\stackrel{→}{B}=-11\hat{i}-5\hat{j}+7\hat{k}$.

$$\begin{gathered} \left|B\right|=\sqrt{{\left(-11\right)}^2+{\left(-5\right)}^2+{\left(7\right)}^2} \\ =\sqrt{121+25+49} \\ =\sqrt{195} \\ ≠0 \end{gathered}$$

Thus, dot product is zero, as vectors are perpendicular but $\stackrel{→}{B}≠0$

Given $\stackrel{→}{A}=2\hat{i}-3\hat{j}+\hat{k}$.

And $\stackrel{→}{A}·\stackrel{→}{B}=0$as well as$\stackrel{→}{A}×\stackrel{→}{B}=0$

This implies

$\left|A\right|\left|B\right|\sin \mathrm{θ}=0$

Since $\left|A\right|=\sqrt{14}≠0$

So either,$\left|B\right|=0$ or $\sin \mathrm{θ}=0$

If $\sin \mathrm{θ}=0$, then $\mathrm{θ}=0°$

This implies$\stackrel{→}{A}$ and$\stackrel{→}{B}$ are parallel vectors.

So any vector parallel to$\stackrel{→}{A}$ will make cross product zero, and$\stackrel{→}{B}$ need not be zero.

Now, find a vector parallel to $\stackrel{→}{A}$.

$-\left(2\hat{i}-3\hat{j}+\hat{k}\right)=-2\hat{i}+3\hat{j}-\hat{k}$

Let $\stackrel{→}{B}=-2\hat{i}+3\hat{j}-\hat{k}$.

$$\begin{gathered} \left|B\right|=\sqrt{{\left(-2\right)}^2+{\left(3\right)}^2+{\left(-1\right)}^2} \\ =\sqrt{4+9+1} \\ =\sqrt{14} \\ ≠0 \end{gathered}$$

Thus, cross product is zero as vectors are parallel but$\stackrel{→}{B}≠0$

Given $\stackrel{→}{A}=2\hat{i}-3\hat{j}+\hat{k}$.

And$\stackrel{→}{A}×\stackrel{→}{B}=0$

This implies

$\left|A\right|\left|B\right|\cos \mathrm{θ}=0$and $\left|A\right|\left|B\right|\sin \mathrm{θ}=0$

Since $\left|A\right|=\sqrt{14}≠0$

So either,$\left|B\right|=0$ or $\sin \mathrm{θ}=0, \cos \mathrm{θ}=0$

Since, $\sin \mathrm{θ}$and $\cos \mathrm{θ}$can not be equal to zero simultaneously, so $\left|B\right|=0$.

Hence $\stackrel{→}{B}=0$.