Given that a, b. p and q are four vectors such that a + b = p, b.… - Vidyadip

MCQ

Given that $\vec a,\vec b.\vec p$ and $\vec q$ are four vectors such that $\vec a + \vec b = \mu \vec p$, $\vec b.\vec q = 0$ and ${\left( {\vec b} \right)^2}=1$ then $\left| {\left( {\vec a.\vec q} \right)\vec p - \left( {\vec p.\vec q} \right)\vec a} \right|$ is equal to

A
$2\left| {\vec p.\vec q} \right|$
B
$\frac{1}{2}\left| {\vec p.\vec q} \right|$
C
$\left| {\vec p \times \vec q} \right|$
✓
$\left| {\vec p.\vec q} \right|$

✓

Answer

Correct option: D.

$\left| {\vec p.\vec q} \right|$

d
$|(\vec{a} \cdot \vec{q}) \vec{p}-(\vec{p} \cdot \vec{q}) \vec{a}|=|(\vec{a} \times \vec{p}) \times \vec{q}|$

$=|\{(\mu \vec{p}-\vec{b}) \times \vec{p}\} \times \vec{q}|=|(\vec{b} \times \vec{p}) \times \vec{q}|$

$=|0-(\vec{p} \cdot \vec{q}) \vec{b}|=|\vec{p} \cdot \vec{q}|$

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