MCQ
If $\vec{a}, \vec{b}, \overrightarrow{ c }$ are three non-zero vectors and $\hat{ n }$ is a unit vector perpendicular to $\vec{c}$ such that $\overrightarrow{ a }=\alpha \overrightarrow{ b }-\hat{ n },(\alpha \neq 0) \quad$ and $\quad \overrightarrow{ b } \cdot \overrightarrow{ c }=12$, then $|\overrightarrow{ c } \times(\overrightarrow{ a } \times \overrightarrow{ b })|$ is equal to :
  • A
    $15$
  • B
    $9$
  • $12$
  • D
    $6$

Answer

Correct option: C.
$12$
c
$\hat{ n } \perp \overrightarrow{ c } \quad \overrightarrow{ a }=\alpha \overrightarrow{ b }-\overrightarrow{ n }$

$\overrightarrow{ b } \cdot \overrightarrow{ c }=12$

$\overrightarrow{ a } \cdot \overrightarrow{ c }=\alpha(\overrightarrow{ b } \cdot \overrightarrow{ c })-\overrightarrow{ n } \cdot \overrightarrow{ c }$

$\overrightarrow{ a } \cdot \overrightarrow{ c }=\alpha(\overrightarrow{ b } \cdot \overrightarrow{ c })$

$|\overrightarrow{ c } \times(\overrightarrow{ a } \times \overrightarrow{ b })|=|(\overrightarrow{ c } \cdot \overrightarrow{ b }) \overrightarrow{ a }-(\overrightarrow{ c } \cdot \overrightarrow{ a }) \overrightarrow{ b }|$

$=|(\overrightarrow{ c } \cdot \overrightarrow{ b }) \overrightarrow{ a }-\alpha(\overrightarrow{ b } \cdot \overrightarrow{ c }) \overrightarrow{ b }|$

$=|(\overrightarrow{ c } \cdot \overrightarrow{ b })||\overrightarrow{ a }-\alpha \overrightarrow{ b }|$

$=12 \times(|\overrightarrow{ n }|)$

$=12 \times 1$

$=12$

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