Question

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

• 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: (C)

$\hat{a}=\pm \frac{\vec{n_{1}} \times \vec{n_{2}}}{\left|\vec{n_{1}} \times \vec{n_{2}}\right|}=\pm \frac{2 \hat{i}+5 \hat{j}+3 \hat{k}}{\sqrt{38}}$
(where $\vec{n}_{1}$ and $\vec{n}_{2}$ are normal to the planes)