Let $\overrightarrow{a}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overrightarrow{b}$ be two vectors such that $\overrightarrow{a}·\overrightarrow{b}=1,\cos \left(\overrightarrow{a}·\overrightarrow{b}\right)=\frac{1}{3}$ and the components of $\overrightarrow{b}$ w.r.t $\left(\hat{\mathrm{i}}, \hat{\mathrm{j}}, \hat{\mathrm{k}}\right)$ be integers. Then the number of possible vectors that represent $\overrightarrow{b}$ is

Given,

$\overrightarrow{a}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overrightarrow{b}$ be two vectors such that $\overrightarrow{a}·\overrightarrow{b}=1 \& \cos \left(\overrightarrow{a}·\overrightarrow{b}\right)=\frac{1}{3}$

So, applying formula of dot product we get,

$\overrightarrow{a}·\overrightarrow{b}=1\Rightarrow \left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\times \frac{1}{3}=1$

$\Rightarrow \sqrt{3}\left|\overrightarrow{b}\right|\times \frac{1}{3}=1$

$\Rightarrow \left|\overrightarrow{b}\right|=\sqrt{3}$

Also given that the components of $\overrightarrow{b}$ w.r.t $\left(\hat{\mathrm{i}}, \hat{\mathrm{j}}, \hat{\mathrm{k}}\right)$ be integers,

Then the possible vectors that represent $\overline{b}$ will be $\left\{\left(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\right), \left(-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\right), \left(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}\right)\right\}$

Then the number of possible vectors that represent $\overrightarrow{b}$ is $3$