Let a=i+2j+3k,b=3i+j-k and c be three vectors such that c is coplanar with a and b.

Question

Let $\vec{a}=\hat{i}+2\hat{j}+3\hat{k},\vec{b}=3\hat{i}+\hat{j}-\hat{k}$ and $\vec{c}$ be three vectors such that $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$. If the vector $\vec{C}$ is perpendicular to $\vec{b}$ and $\vec{a}\cdot \vec{c}=5$, then $|\vec{c}|$ is equal to

Accepted Answer

$\begin{gathered} \overset{	o }{c}=\lambda (\overset{	o }{b}	imes (\overset{	o }{a}	imes \overset{	o }{b})) \ =\lambda ((\overset{	o }{b}\cdot \overset{	o \leftarrow }{b})\overset{	o }{a}-(\overset{	o }{a}\cdot \overset{	o \leftarrow }{b})\overset{	o }{b}) \ =\lambda (11\overset{	o }{a}-2\overset{	o \leftarrow }{b})=\lambda (11i+22j+33k-6i-2j+2k) \ =\lambda (5i+20j+35k) \ =5\lambda (5i+4j+7k) \ =\overset{	o }{c}\cdot \overset{	o }{a}=5 \ =5\lambda (1+8+21)=5=\lambda =\frac{1}{30} \ \Rightarrow \overset{	o }{c}=\frac{1}{6}(i+4j+7k) \ |\overset{	o }{c}|=\frac{\sqrt{1+16+49}}{6}=\sqrt{\frac{11}{6}} \end{gathered}$

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