BITSAT 2010
- Exam
- BITSAT
- Year
- 2010
- Subject
- Mathematics
- Chapter
- Vector Algebra
BITSAT 2010 · Mathematics · Vector Algebra
Which one of the following is the unit vector perpendicular to both...
Which one of the following is the unit vector perpendicular to both (\vec{a}=-\hat{i}+\hat{j}+\hat{k}) and (\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}) ?
Options
- A(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}})
- B(\hat{\mathrm{k}})
- C(\frac{\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{2}})
- D(\frac{\hat{\mathrm{i}}-\hat{\mathrm{j}}}{\sqrt{2}})
Correct answer
A. (\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}})
Step-by-step solution
According to question (\mathrm{a}=-\hat{i}+\hat{j}+\hat{k}) and (\mathrm{b}=\hat{i}-\hat{j}+\hat{k})
Then, (\mathrm{a} \times \mathrm{b}=\left|\right|)
(\begin{gathered}
=\hat{i}[1+1]-\hat{j}[-1-1]+\hat{k}[1-1]=2(\hat{i}+\hat{j}) \\
\quad \text{ and }|\mathrm{a} \times \mathrm{b}|=\sqrt{4+4}=2 \sqrt{2}
\end{gathered})
(\therefore \text{ Required unit vector }= \pm \frac{2(\hat{i}+\hat{j})}{2 \sqrt{2}}= \pm \frac{\hat{i}+\hat{j}}{\sqrt{2}})