Question

A unit vector parallel to the intersection of the planes $\overrightarrow{r}\cdot (\hat{i}-\hat{j}+\hat{k})=5$ and $\overrightarrow{r}\cdot (2\hat{i}+\hat{j}-3\hat{k})=4$ can be

• A. $\frac{2\hat{i}+5\hat{j}+3\hat{k}}{\sqrt{38}}$
• B. $\frac{2\hat{i}-5\hat{j}+3\hat{k}}{\sqrt{38}}$
• C. $\frac{-2\hat{i}-5\hat{j}-3\hat{k}}{\sqrt{38}}$
• D. $\frac{-2\hat{i}+5\hat{j}-3\hat{k}}{\sqrt{38}}$

Answer 1

Answer: (A), (C)

$\frac{2\hat{i}+5\hat{j}+3\hat{k}}{\sqrt{38}}$
$\frac{-2\hat{i}-5\hat{j}-3\hat{k}}{\sqrt{38}}$

Vector parallel to intersection of planes

$\overrightarrow{p}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& 1\\ 2& 1& -3\end{array}\right|$

$=\hat{i}(3-1)-\hat{j}(-3-2)+\hat{k}(1+2)$

$=2\hat{i}+5\hat{j}+3\hat{k}$

$|\overrightarrow{p}|=\sqrt{2^2+5^2+3^2}=\sqrt{38}$

unit vector parallel to intersection of planes

$=\pm \frac{(2\hat{i}+5\hat{j}+3\hat{k})}{\sqrt{2^2+5^2+3^2}}$

$=\pm \frac{(2\hat{i}+5\hat{j}+3\hat{k})}{\sqrt{38}}$