Question

The vector $\overline{x}$ which is perpendicular to $(2,-3,1)$ and $(1,-2,3)$ and which satisfies the condition $\overline{x}(\overline{i}+\overline{2j}-7k)=10$ is

Answer 1

Consider,

$\overline{x}=x\hat{i}+y\hat{j}+z\hat{k}$

Since, $\overline{x}$ is perpendicular to vector $(2,-3,1)$

So,

$2x-3y+z=0$ ---- $\left(1\right)$

Also,

$\overline{x}$ is $\perp$ to vector $(1,-2,3)$

So,

$x-2y+3z=0$ ----- $\left(2\right)$

And

$\overrightarrow{x}.(\hat{i}+2\hat{j}-7\hat{k})=10$

$x+2y-7z=10$ ----- $\left(3\right)$

Subtracting $2\times \left(3\right)$ from $\left(1\right)$,

$(2x-3y+z)-(2x+4y-14z)=0-20$

$7y-15z=-20$ ------ $\left(4\right)$

Subtracting $\left(3\right)$ from $\left(2\right)$

$-4y+10z=-10$

$y=\frac{10+10z}{4}$ ----- $\left(5\right)$

From $\left(4\right)$ and $\left(5\right)$

$7y-15z=20$

$7\left(\frac{10+10z}{4}\right)-15z=20$

$z=1$

From $\left(5\right)$

$y=\frac{10+10z}{4}=\frac{10+10}{4}=5$

From $\left(1\right)$

$2x-3y+z=0$

$2x-3\times 5+1=0$

$x=7$

Therefore,

$\overline{x}=7\hat{i}+5\hat{j}+\hat{k}$