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10. vector algebra — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths10. vector algebra1 Mark
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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