Question

If $\overrightarrow{a}=\hat{i}+4\hat{j}+2\hat{k},\overrightarrow{b}=3\hat{i}-2\hat{j}+7\hat{k}$ and $\overrightarrow{c}=2\hat{i}-\hat{j}+4\hat{k}$, then find a vector $\overrightarrow{d}$ which is perpendicular to both $\overrightarrow{a}$ and $\overrightarrow{b}$ and $\overrightarrow{c}\cdot \overrightarrow{d}=15$

Answer 1

$\overrightarrow{d}$ is $\perp$ to $\overrightarrow{a}$ and $\overrightarrow{b}$

So, we take the cross product of $\overrightarrow{a}$ and $\overrightarrow{b}$

i.e. $(\hat{i}+4\hat{j}+2\hat{k})\times (3\hat{i}-2\hat{j}+7\hat{k})$

$=0-2\hat{i}\times \hat{j}+7\hat{i}\times \hat{k}+12\hat{j}\times \hat{i}-0+28\hat{j}\times \hat{k}+6\hat{k}\times \hat{i}-4\hat{k}\times \hat{j}+0$

$=-2\hat{k}-7\hat{j}-12\hat{k}+28\hat{i}+6\hat{j}+4\hat{i}$

$=32\hat{i}-\hat{j}-14\hat{k}$

$\overrightarrow{d}$ would be a multiple of the obtained cross product, such that $\overrightarrow{c}.\overrightarrow{d}=15$

$\Rightarrow (2\hat{i}-\hat{j}+4\hat{k}).(32\lambda \hat{i}-\lambda \hat{j}-14\lambda \hat{k})=15$

$\Rightarrow 64\lambda +\lambda -56\lambda =15$

$\therefore 9\lambda =15$

$\therefore \lambda =\frac{5}{3}$

$\therefore \overrightarrow{d}=\frac{160\hat{i}-5\hat{j}-70\hat{k}}{3}$