Let a=4i-j+k, b=11i-j+k and c be a vector such that ( a+ b) c= c (-2 a+3 b).

Question

Let $\overset{	o }{a}=4\hat{i}-\hat{j}+\hat{k},\overset{	o }{b}=11\hat{i}-\hat{j}+\hat{k}$ and $\overset{	o }{c}$ be a vector such that $(\overset{	o }{a}+\overset{	o }{b})	imes \overset{	o }{c}=\overset{	o }{c}	imes (-2\overset{	o }{a}+3\overset{	o }{b})$.
If $(2\vec{a}+3\vec{b})\cdot \vec{c}=1670$, then $|\vec{c}{|}^{2}$ is equal to:

Accepted Answer

$\begin{gathered} \begin{gathered} \vec{a}=4\hat{i}-\hat{j}+\hat{k} \ \vec{b}=11\hat{i}-\hat{j}+\hat{k} \end{gathered}]] \ \vec{a}\cdot \vec{b}=44+1+1=46 \ |\vec{a}{|}^{2}=18,|{\vec{b}}^{2}|=123 \ (\vec{a}+\vec{b})	imes \vec{c}=\vec{c}	imes (-2\vec{a}+3\vec{b}) \ (2\vec{a}+3\vec{b})\cdot \vec{c}=1670 \ (\vec{a}+\vec{b})	imes c=(2\vec{a}-3\vec{b})	imes \vec{c} \ (-\vec{a}+4\vec{b})	imes \vec{c}=0 \ \vec{c}=\lambda (4\vec{b}-\vec{a}) \ (2\vec{a}+3\vec{b})\cdot \lambda (4\vec{b}-\vec{a})=1670 \ \lambda (5\vec{a}\cdot \vec{b}-2|\vec{a}{|}^{2}+12|\vec{b}{|}^{2})=1670 \ \lambda =\frac{1670}{5	imes 46-2	imes 18+12	imes 123} \ \lambda =1 \ \vec{c}=4\vec{b}-\vec{a} \ =4(11\hat{i}-\hat{j}+\hat{k})-(4\hat{i}-\hat{j}+\hat{k}) \ =40\hat{i}-3\hat{j}+3\hat{k} \ |{\vec{c}}^{2}|=1600+9+9 \ =1618 \end{gathered}$

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