Question

If $\overrightarrow{\mathbf{A}}\mathbf{=}\mathbf{2}\hat{\mathbf{i}}\mathbf{-}\mathbf{3}\hat{\mathbf{j}}\mathbf{+}\hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{A}}\mathbf{·}\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{0}$, does it follow that $\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{0}$? (Either prove that it does or give a specific example to show that it doesn’t.) Answer the same question if $\overrightarrow{\mathbf{A}}\mathbf{\times }\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{0}$. And again answer the same question if$\overrightarrow{\mathbf{A}}\mathbf{·}\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{0}$ and$\overrightarrow{\mathbf{A}}\mathbf{\times }\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{0}$ .

Short answer

If$\overrightarrow{A}·\overrightarrow{B}=0$ then$\overrightarrow{B}$ need not be zero.

If$\overrightarrow{A}\times \overrightarrow{B}=0$ then$\overrightarrow{B}$ need not be zero.

If$\overrightarrow{A}·\overrightarrow{B}=0$ and$\overrightarrow{A}\times \overrightarrow{B}=0$ then $\overrightarrow{B}=0$.

Step-by-step solution

Given that

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

So magnitude of vector is:

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

Results used

If$\overrightarrow{A}$ and$\overrightarrow{B}$ are two given vectors with angle$\theta$ between them, then

Dot product is given as:

$\overrightarrow{A}·\overrightarrow{B}=\left|A\right|\left|B\right|\cos \theta$

Cross product is given as

$\overrightarrow{A}\times \overrightarrow{B}=\left|A\right|\left|B\right|\sin \theta$

If A→·B→=0

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

And$\overrightarrow{A}·\overrightarrow{B}=0$

This implies

$\left|A\right|\left|B\right|\cos \theta =0$

Since $\left|A\right|=\sqrt{14}\ne 0$

So either,$\left|B\right|=0$ or $\cos \theta =0$

If $\cos \theta =0$, then $\theta =90°$

This implies and are perpendicular vectors.

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

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

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

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

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

Thus, dot product is zero, as vectors are perpendicular but $\overrightarrow{B}\ne 0$

If A→×B→=0

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

And $\overrightarrow{A}·\overrightarrow{B}=0$as well as$\overrightarrow{A}\times \overrightarrow{B}=0$

This implies

$\left|A\right|\left|B\right|\sin \theta =0$

Since $\left|A\right|=\sqrt{14}\ne 0$

So either,$\left|B\right|=0$ or $\sin \theta =0$

If $\sin \theta =0$, then $\theta =0°$

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

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

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

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

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

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

Thus, cross product is zero as vectors are parallel but$\overrightarrow{B}\ne 0$

If A→·B→=0 and A→×B→=0

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

And$\overrightarrow{A}\times \overrightarrow{B}=0$

This implies

$\left|A\right|\left|B\right|\cos \theta =0$and $\left|A\right|\left|B\right|\sin \theta =0$

Since $\left|A\right|=\sqrt{14}\ne 0$

So either,$\left|B\right|=0$ or $\sin \theta =0, \cos \theta =0$

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

Hence $\overrightarrow{B}=0$.