Question

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

Answer 1

Given $\overrightarrow{a}=\hat{i}+4\hat{j}+2\hat{k}$

$\overrightarrow{b}=3\hat{i}+2\hat{j}+7\hat{k}$

$\overrightarrow{c}=2\hat{i}-\hat{j}+4\hat{k}$

$\overrightarrow{d}$ is perpendicular to both $\overrightarrow{a}$ and $\overrightarrow{b}$

Let $\overrightarrow{d}=d_1\hat{i}+d_2\hat{j}+d_3\hat{k}$

$\therefore \overrightarrow{d}.\overrightarrow{a}=0$

$\therefore (d_1\hat{i}+d_2\hat{j}+d_3\hat{k}).(\hat{i}+4\hat{j}+2\hat{k})=0$

$\therefore d_1+4d_2+2d_3=0$ (1)

Similarly, $\overrightarrow{d}.\overrightarrow{b}=0$

$\therefore (d_1\hat{i}+d_2\hat{j}+d_3\hat{k}).(3\hat{i}+2\hat{j}+7\hat{k})=0$

$\therefore 3d_1+2d_2+7d_3=0$ (2)

Given $\overrightarrow{c}.\overrightarrow{d}=15$

$\therefore (2\hat{i}-\hat{j}+4\hat{k}).(d_1\hat{i}+d_2\hat{j}+d_3\hat{k})=15$

$\therefore 2d_1-d_2+4d_3=15$ (3)

Solving (1), (2) and (3), we get,

$d_1=40$

$d_2=-1.667$

$d_3=-16.667$

Thus, required vector is,

$\overrightarrow{d}=40\hat{i}-1.667\hat{j}-16.667\hat{k}$