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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 :

Option 1)
$-3\hat{i}+9\hat{j}+5\hat{k}$

Option 2)
$-3\hat{i}-9\hat{j}-5\hat{k}$

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

Option 4)
$\frac{1}{2}\left ( 3\hat{i}-9\hat{j}+5\hat{k} \right )$

Answers (1)

best_answer

$\vec{\beta _{1}}=\frac{\vec{\alpha \cdot \vec{\beta }}}{\left | \vec{\alpha } \right |^{2}}\cdot \vec{\alpha }=\frac{5}{10}\vec{\alpha }=\frac{3}{2}\hat{i}+\frac{1}{2}\hat{j}$

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

$\therefore \vec{\beta _{1}}\times \vec{\beta _{2}}=\begin{vmatrix} \hat{i}& \hat{j}& \hat{k}\\ \frac{3}{2}& \frac{1}{2}& 0\\ \frac{-1}{2}& \frac{3}{2}& -3 \end{vmatrix}$

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

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

Option 1)

$-3\hat{i}+9\hat{j}+5\hat{k}$

Option 2)

$-3\hat{i}-9\hat{j}-5\hat{k}$

Option 3)

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

Option 4)

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

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