MCQ
Let $\vec \alpha \, = \,3\hat i\, + \hat j$ and $\vec \beta \, = \,2\hat i\, - \hat j + 3\hat k.$ If $\vec \beta \, = \,{\vec \beta _1} - {\vec \beta _2},$ where ${\vec \beta _1}$ is parallel to $\vec \alpha $ and $\vec \beta_2 $ is perpendicular to $\vec \alpha ,$  then ${\vec \beta _1} \times {\vec \beta _2}$ is equal to
  • $\frac{1}{2}( - 3\hat i + 9\hat j + 5\hat k)$
  • B
    $\frac{1}{2}(  3\hat i - 9\hat j + 5\hat k)$
  • C
    $- 3\hat i + 9\hat j + 5\hat k$
  • D
    $3\hat i - 9\hat j - 5\hat k$

Answer

Correct option: A.
$\frac{1}{2}( - 3\hat i + 9\hat j + 5\hat k)$
a
$\vec{\alpha}=3 \hat{i}+\hat{j}$

$\vec{\beta}=2 \hat{i}-\hat{j}+3 \hat{k}$

$\vec{\beta}=\vec{\beta}_{1}-\vec{\beta}_{2}$

$\overrightarrow {{\beta _1}}  = \lambda (3\hat i + \hat j),\overrightarrow {{\beta _2}}  = \lambda (3\hat i + \hat j) - 2\hat i + \hat j - 3\hat k$

$\vec{\beta}_{2} \cdot \vec{\alpha}=0$

$(3 \lambda-2) \cdot 3+(\lambda+1)=0$

$9 \lambda-6+\lambda+1=0$

$\lambda=\frac{1}{2}$

$\Rightarrow \vec{\beta}_{1}=\frac{3}{2} \hat{i}+\frac{1}{2} \hat{j}$

$\Rightarrow \vec{\beta}_{2}=-\frac{1}{2} \hat{i}+\frac{3}{2} \hat{j}-3 \hat{k}$

${\rm{ Now, }}{\vec \beta _1} \times {\vec \beta _2} = \left| {\begin{array}{*{20}{c}}
{\hat i}&{\hat j}&{\hat k}\\
{\frac{3}{2}}&{\frac{1}{2}}&0\\
{ - \frac{1}{2}}&{\frac{3}{2}}&{ - 3}
\end{array}} \right|$

$=\hat{i}\left(-\frac{3}{2}-0\right)-\hat{j}\left(-\frac{9}{2}-0\right)+\hat{k}\left(\frac{9}{4}+\frac{1}{4}\right)$

$=\frac{3}{2} \hat{i}+\frac{9}{2} \hat{j}+\frac{5}{2} \hat{k}$

$=\frac{1}{2}(-3 \hat{i}+9 \hat{j}+5 \hat{k})$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions