Let a=2i-j+k and b= j+2k, Z be two vectors.

Question

Let $\vec{a}=2\hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=\lambda \hat{j}+2\hat{k},\lambda \in Z$ be two vectors. Let $\vec{c}=\vec{a}	imes \vec{b}$ and $\vec{d}$ be a vector of magnitude 2 in $yz$-plane. If $|\vec{c}|=\sqrt{53}$, then the maximum possible value of $(\vec{c}\cdot \vec{d}{)}^{2}$ is equal to :

Accepted Answer

$\begin{gathered} \vec{c}=\vec{a}	imes \vec{b}=|\begin{array}{ccc}\vec{i} \quad  \vec{j} \quad  \vec{k} \ 2 \quad  1 \quad  1 \ 0 \quad  \lambda  \quad  2\end{array}|=(2-\lambda )\vec{i}-\vec{j}(4)+(2\lambda )\vec{k} \ |\vec{c}{|}^{2}=53\Rightarrow (2-\lambda {)}^{2}+16+(2\lambda {)}^{2}=53=5{\lambda }^{2}-4\lambda -33=0 \ \Rightarrow 5{\lambda }^{2}-15\lambda +11\lambda -33=0 \ \Rightarrow 5\lambda (\lambda -3)+11(\lambda -3)=0 \ \Rightarrow (\lambda -3)(5\lambda +11)=0 \ \Rightarrow \lambda =3,\lambda 
e -\frac{11}{5}\, \, \, \, \, (\because \lambda \in Z) \ 	herefore \vec{c}=(-1,-4,6) \end{gathered}$

Let $	herefore \vec{d}=(0,y,z)$

Given $|\vec{d}|=2\Rightarrow {y}^{2}+{z}^{2}=4$

$\begin{gathered} \vec{c}.\vec{d}=0-4y+6z \ (\vec{c}.\vec{d}{)}^{2}=(-4y+6z{)}^{2} \ =16{y}^{2}+36{z}^{2}-48yz \ =16{y}^{2}+36(4-{y}^{2})-48y\cdot \sqrt{4-{y}^{2}} \end{gathered}$

Let $f(y)=-20{y}^{2}+144-48y\sqrt{4-{y}^{2}}$

$\begin{gathered} {f}^{1}(y)=-40y-48(\sqrt{4-{y}^{2}})-48y(\frac{1}{2\sqrt{4-{y}^{2}}})(-2y) \ =-40y-48(\frac{4-{y}^{2}-{y}^{2}}{\sqrt{4-{y}^{2}}}) \ {f}^{1}(y)=0\Leftarrow 40y=96(\frac{{y}^{2}-2}{\sqrt{4-{y}^{2}}}) \ \Rightarrow 5y\sqrt{4-{y}^{2}}=12({y}^{2}-2) \ \Rightarrow 25{y}^{2}(4-{y}^{2})=144({y}^{4}+4-4{y}^{2}) \ =169{y}^{4}-676{y}^{2}+576=0 \ =169{y}^{4}-208{y}^{2}-468{y}^{2}+576=0 \ \Rightarrow 13{y}^{2}(13{y}^{2}-16)-36(13{y}^{2}-16)=0 \ \Rightarrow {y}^{2}=\frac{16}{13}()\frac{36}{13} \ \Rightarrow y=\pm \frac{4}{\sqrt{13}}()\pm \frac{6}{\sqrt{13}} \ 	herefore (\vec{c}.\vec{d}{)}^{2}=208 \end{gathered}$

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