Let u ;be a vector coplanar with the vector a = 2 i + 3 j - k and… - Vidyadip

MCQ

Let $\vec u\;$be a vector coplanar with the vector $\vec a = 2\hat i + 3\hat j - \hat k$ and $\vec b = \hat j + \hat k$ . If $\vec u$ is perpendicular to $\vec a$ and $\vec u \cdot \vec b = 24$ ,then ${\left| {\vec u} \right|^2} = $ . . . .

A
$315$
B
$256$
C
$84$
✓
$336$

✓

Answer

Correct option: D.

$336$

d
$\because \overrightarrow{\mathrm{u}}, \overrightarrow{\mathrm{a}} \& \overrightarrow{\mathrm{b}}$ are coplanar

$\therefore \overrightarrow{\mathrm{u}}=\lambda(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{a}}=\lambda\left\{\overrightarrow{\mathrm{a}}^{2} \cdot \overrightarrow{\mathrm{b}}-(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}) \overrightarrow{\mathrm{a}}\right\}$

$=\lambda\{-4 \hat{\hat{\imath}}+8 \hat{\jmath}+16 \hat{k}\}=\lambda^{\prime}\{-\hat{i}+2 \hat{j}+4 \hat{k}\}$

Also, $\overrightarrow{\mathrm{u}} . \overrightarrow{\mathrm{b}}=24 \Rightarrow \lambda^{\prime}=4$

$\overrightarrow{\mathrm{u}}=-4 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}+16 \hat{\mathrm{k}} \Rightarrow \quad|\overrightarrow{\mathrm{u}}|^{2}=336$

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