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10. vector algebra — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths10. vector algebra1 Mark
MCQ
The vectors $\vec a$ and $\vec b$ are not perpendicular and $\vec c$ and $\vec d$ are two vector satisfying $\vec b \times \vec c = \vec b \times \vec d$ and $\vec a \cdot \vec d = 0$ then the vector  $\vec d$  is equal to 
  • A
    $\vec c + \left( {\frac{{\vec a \cdot \vec c}}{{\vec a \cdot \vec b}}} \right)\vec b$
  • B
    $\;\vec b + \left( {\frac{{\vec b \cdot \vec c}}{{\vec a \cdot \vec b}}} \right)\vec c\;$
  • ✓
    $\;\vec c - \left( {\frac{{\vec a \cdot \vec c}}{{\vec a \cdot \vec b}}} \right)\vec b$
  • D
    $\;\vec b - \left( {\frac{{\vec b \cdot \vec c}}{{\vec a \cdot \vec b}}} \right)\vec c$
✓

Answer

Correct option: C.
$\;\vec c - \left( {\frac{{\vec a \cdot \vec c}}{{\vec a \cdot \vec b}}} \right)\vec b$
c
$\vec{a} \cdot \vec{b} \neq 0 \quad(\text {given})$

$\vec{a} \cdot \vec{b}=0$

$\vec{b} \times \vec{c}=\vec{b} \times \vec{d} \Rightarrow \vec{a} \times(\vec{b} \times \vec{c})=\vec{a} \times(\vec{b} \times \vec{d})$

$\Rightarrow(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}=(\vec{a} \cdot \vec{d}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{d}$

$\Rightarrow(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}=0-(\vec{a} \cdot \vec{b}) \vec{d}$

$\Rightarrow(\vec{a} \cdot \vec{b}) \vec{d}=-(\vec{a} \cdot \vec{c}) \vec{b}+(\vec{a} \cdot \vec{b}) \vec{c}$

$\vec{d}=-\frac{(\vec{a} \cdot \vec{c}) \vec{b}}{(\vec{a} \cdot \vec{b})}+\vec{c}$

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