Search any question & find its solution
Question:
Answered & Verified by Expert
If $\vec{a}=\hat{i}+\hat{j}-\hat{k}, \vec{b}=\hat{i}-\hat{j}+\hat{k}$ and $\vec{c}$ is unit vector perpendicular to $\vec{a}$ and coplanar with $\vec{a}$ and $\vec{b}$, then unit vector $\vec{d}$ perpendicular to both $\vec{a}$ and $\vec{c}$ is
Options:
Solution:
2471 Upvotes
Verified Answer
The correct answer is:
$\pm \frac{1}{\sqrt{2}}(\hat{j}+\hat{k})$
$\because \overrightarrow{\mathrm{d}}$ is normal to the plane of $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$
$\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$ or $\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}=\pm 2(\hat{\mathrm{j}}+\hat{\mathrm{k}}) \therefore$ unit vector $\overrightarrow{\mathrm{d}}=\pm \frac{1}{\sqrt{2}}(\hat{\mathrm{j}}+\hat{\mathrm{k}})$
$\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$ or $\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}=\pm 2(\hat{\mathrm{j}}+\hat{\mathrm{k}}) \therefore$ unit vector $\overrightarrow{\mathrm{d}}=\pm \frac{1}{\sqrt{2}}(\hat{\mathrm{j}}+\hat{\mathrm{k}})$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.