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  • A unit of vector in the plane of <img alt="\overrightarrow{b}= 2\widehat{i}+\widehat{j}" src

A unit of vector\overrightarrow{a} in the plane of \overrightarrow{b}= 2\widehat{i}+\widehat{j} and \overrightarrow{c}= \widehat{i}-\widehat{j}+\widehat{k} is such that    \vec{a}\wedge \vec{b}=\vec{a}\wedge \vec{d}     where \overrightarrow{d}= \widehat{j}+2\widehat{k} is:

Option: 1

\frac{\widehat{i}+\widehat{j}+\widehat{k}}{\sqrt{3}}


Option: 2

\frac{\widehat{i}-\widehat{j}+\widehat{k}}{\sqrt{3}}


Option: 3

\frac{2\widehat{i}+\widehat{j}}{\sqrt{5}}


Option: 4

\frac{2\widehat{i}+\widehat{j}}{\sqrt{5}}


Answers (1)

 

Vectors -

Physical quantities which can be described by its magnitude and directions.
 

- wherein

Physical quantities like Displacement ,force, velocity etc. are vectors.

 

 

Let \overrightarrow{a}=\lambda \overrightarrow{b}+\mu \overrightarrow{c}{}'

Then \frac{\overrightarrow{a}.\overrightarrow{b}}{ab}=\frac{\overrightarrow{a}.\overrightarrow{d}}{ad}

i.e. \frac{\left ( \lambda \overrightarrow{b}+\mu\overrightarrow{c} \right ).\overrightarrow{b}}{b}= \frac{\left ( \lambda \overrightarrow{b}+\mu\overrightarrow{c} \right ).\overrightarrow{d}}{d}

i.e. \frac{\left [ \lambda \left ( 2\widehat{i}+\widehat{j} \right )+\mu \left ( \widehat{i}-\widehat{j}+\widehat{k} \right ) \right ].\left ( 2\widehat{i}+\widehat{j} \right )}{\sqrt{5}}

    =\frac{\left [ \lambda \left ( 2\widehat{i}+\widehat{j} \right )+\mu \left ( \widehat{i}-\widehat{j}+\widehat{k} \right ) \right ].\left ( \widehat{j}+2\widehat{k} \right )}{\sqrt{5}}

i.e \lambda (4+1)+\mu (2-1)=\lambda(1)+\mu (-1+2) i.e  4\lambda=0 , \lambda =0

\therefore \widehat{a}=\frac{\widehat{i}-\widehat{j}+\widehat{k}}{\sqrt{3}}

Posted by

Kshitij

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