Let a=i-2j+3k,b=2i+j-k,c= i+j+k and v=a b.

Question

Let $\vec{a}=\hat{i}-2\hat{j}+3\hat{k},\vec{b}=2\hat{i}+\hat{j}-\hat{k},\vec{c}=\lambda \hat{i}+\hat{j}+\hat{k}$ and $\vec{v}=\vec{a}	imes \vec{b}$. If $\vec{v}\cdot \vec{c}=11$ and the length of the projection of $\vec{b}$ on $\vec{c}$ is $p$, then $9{p}^{2}$ is equal to

Accepted Answer

$\begin{gathered} \begin{gathered} \vec{a}=\hat{i}-2\hat{j}+3\hat{k} \ \vec{b}=2\hat{i}+\hat{j}-\hat{k} \ \vec{c}=\lambda \hat{i}+\hat{j}+\hat{k} \ \vec{v}=\vec{a}	imes \vec{b} \ \vec{v}=|\begin{array}{ccc}\hat{i} \quad \hat{j} \quad \hat{k} \ 1 \quad -2 \quad 3 \ 2 \quad 1 \quad -1\end{array}| \ \vec{v}=\hat{i}(2-3)-\hat{j}(-1-6)+\hat{k}(1+4) \ \vec{v}=-\hat{i}+7\hat{j}+5\hat{k} \ \vec{v}\cdot \vec{c}=11 \end{gathered} \end{gathered}$

$\begin{gathered} (-\hat{i}+7\hat{j}+5\hat{k})\cdot (\lambda \hat{i}+\hat{j}+\hat{k})=11 \ -\lambda +7+5=11 \ -\lambda +12=11 \ \lambda =1 \ \vec{c}=\hat{i}+\hat{j}+\hat{k} \ p=\frac{|\vec{b}\cdot \vec{c}|}{|\vec{c}|} \ \vec{b}\cdot \vec{c}=(2\hat{i}+\hat{j}-\hat{k})\cdot (\hat{i}+\hat{j}+\hat{k}) \ =2+1-1=2 \ |\vec{c}|=\sqrt{{1}^{2}+{1}^{2}+{1}^{2}}=\sqrt{3} \ p=\frac{2}{\sqrt{3}} \ {p}^{2}=\frac{4}{3} \ 9{p}^{2}=12 \end{gathered}$

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