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Mathematics Question on Vector Algebra

Let a→=i→+j→+k^,b→=i^−j^+k^\overrightarrow{a}=\overrightarrow{i}+\overrightarrow{j}+\hat{k}, \overrightarrow{b}=\hat{i}-\hat{j}+\hat{k} and c→=i→−j→−k→\overrightarrow{c}=\overrightarrow{i}-\overrightarrow{j}-\overrightarrow{k} be three vectors. A vector v^\hat{v} in the plane of a→\overrightarrow{a} and b→, \overrightarrow{b}, whose projection on c→\overrightarrow{c} is 13,\frac{1}{\sqrt{3}}, is given by

A

i^−3j^+3k^\widehat{i}-3\widehat{j}+3 \widehat{k}

B

−3i^−3j^−k^-3\widehat{i}-3\widehat{j} -\widehat{k}

C

3i^−j^+3k^3\widehat{i}-\widehat{j}+ 3\widehat{k}

D

i^+3j^−3k^\widehat{i}+3\widehat{j}- 3\widehat{k}

Answer

3i^−j^+3k^3\widehat{i}-\widehat{j}+ 3\widehat{k}

Explanation

Solution

Let v→=a→+λb→\overrightarrow{v}=\overrightarrow{a}+\lambda\overrightarrow{b}
v→=(1+λ)i^+(1+λ)j^(1+λ)k^\overrightarrow{v}=(1+\lambda)\widehat{i}+(1+\lambda)\widehat{j}(1+\lambda)\widehat{k}
Projection of v→onc→=13\overrightarrow{v} on \overrightarrow{c}=\frac{1}{\sqrt{3}}
\Rightarrow \hspace25mm \frac{\overrightarrow{v}.\overrightarrow{c}}{|\overrightarrow{c} | }=\frac{1}{\sqrt{3}}
⇒(1+λ)−(1−λ)−(1+λ)3=13\Rightarrow \frac{(1+\lambda)-(1-\lambda)-(1+\lambda)}{\sqrt{3}}=\frac{1}{\sqrt{3}}
\Rightarrow \hspace15mm 1+\lambda-1+\lambda-1-\lambda=1
\Rightarrow \hspace25mm \lambda-1=1
\Rightarrow \hspace25mm \lambda =2
\therefore \hspace25mm \overrightarrow{v}=3\widehat{i}-\widehat{j}+3\widehat{k}