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Question

Mathematics Question on Vector Algebra

If x×b=c×b\overrightarrow{x}\times\overrightarrow{b} \,=\,\overrightarrow{c}\times\overrightarrow{b} and xa, \overrightarrow{x}\,\bot\, \overrightarrow{a}, then x\overrightarrow{x} is equal to

A

(b×c)×ab.a\frac{(\overrightarrow{b}\times\overrightarrow{c})\times \overrightarrow{a}}{\overrightarrow{b}.\overrightarrow{a}}

B

b×(a×c)b.c\frac{ \overrightarrow{b}\times(\overrightarrow{a}\times\overrightarrow{c})}{\overrightarrow{b}.\overrightarrow{c}}

C

(c×b)×aa.b\frac{(\overrightarrow{c}\times\overrightarrow{b})\times \overrightarrow{a}}{\overrightarrow{a}.\overrightarrow{b}}

D

none of these.

Answer

(b×c)×ab.a\frac{(\overrightarrow{b}\times\overrightarrow{c})\times \overrightarrow{a}}{\overrightarrow{b}.\overrightarrow{a}}

Explanation

Solution

x×b=c×b\vec{x} \times \vec{b}=\vec{c} \times \vec{b} a×(x×b)=a×(c×b)\Rightarrow \vec{a} \times \left(\vec{x} \times \vec{b}\right)= \vec{a} \times \left(\vec{c} \times \vec{b}\right) (ab)x(ax)b\Rightarrow \left(\vec{a}\cdot\vec{b}\right)\vec{x}-\left(\vec{a}\cdot \vec{x}\right) \vec{b} =(c×b)×a=-\left(\vec{c} \times \vec{b}\right)\times\vec{a} =(b×c)×a=\left(\vec{b} \times \vec{c}\right) \times \vec{a} (ab)x=(b×c)×a[a.x\Rightarrow \left(\vec{a}\cdot \vec{b}\right)\vec{x}=\left(\vec{b} \times \vec{c}\right) \times \vec{a}\quad[\because \vec{a} . \vec{x} as =0xa]=0\,\vec{x} \,\bot \,\vec{a}] x=(b×c)×aa.b=(b×c)×ab.a\Rightarrow \vec{x}=\frac{\left(\vec{b} \times \vec{c}\right)\times\vec{a}}{\vec{a} . \vec{b}}=\frac{\left(\vec{b} \times \vec{c}\right)\times \vec{a}}{\vec{b} . \vec{a}}