Let c and d be vectors such that | c+ d|=√29 and c (2i+3j+4k)=(2i+3j+4k) d.

Question

Let $\overset{	o }{c}$ and $\overset{	o }{d}$ be vectors such that $|\overset{	o }{c}+\overset{	o }{d}|=\sqrt{29}$ and $\overset{	o }{c}	imes (2\hat{i}+3\hat{j}+4\hat{k})=(2\hat{i}+3\hat{j}+4\hat{k})	imes \overset{	o }{d}$. If ${\lambda }_{1},{\lambda }_{2}({\lambda }_{1}>{\lambda }_{2})$ are the possible values of $(\vec{c}+\vec{d})\cdot (-7\hat{i}+2\hat{j}+3\hat{k})$, then the equation ${K}^{2}{x}^{2}+({K}^{2}-5K+{\lambda }_{1})xy+(3K+\frac{{\lambda }_{2}}{2}){y}^{2}-8x+12y+{\lambda }_{2}=0$ represents a circle, for K equal to :

Accepted Answer

$\begin{gathered} |\vec{c}+\vec{d}|=\sqrt{29} \ \vec{c}	imes (2\hat{i}+3\hat{j}+4\hat{k})-(2\hat{i}+3\hat{j}+4\hat{k})	imes \vec{d}=0 \end{gathered}$

$\begin{gathered} (\vec{c}+\vec{d})	imes (2\hat{i}+3\hat{j}+4\hat{k})=0 \ \Rightarrow \vec{c}+\vec{d}=\lambda (2\hat{i}+3\hat{j}+4\hat{k}) \ \Rightarrow \sqrt{29}=|\lambda |\sqrt{29} \ \Rightarrow \lambda =\pm 1\Rightarrow \vec{c}+\vec{d}=\pm (2\hat{i}+3\hat{j}+4\hat{k}) \ \Rightarrow (\vec{c}+\vec{d})\cdot (-7\hat{i}+2\hat{j}+3\hat{k})=\pm (-14+6+12)=\pm 4 \ \Rightarrow {\lambda }_{1}=4,{\lambda }_{2}=-4 \end{gathered}$

Given equation represents a circle

$\begin{gathered} \Rightarrow {K}^{2}=3K+\frac{{\lambda }_{2}}{2}{K}^{2}-5K+{\lambda }_{1}=0 \ \Rightarrow {K}^{2}=3K-2{K}^{2}-5K+4=0 \ \Rightarrow K=1,2K=4,1 \ \Rightarrow K=1 \end{gathered}$

\Rightarrow  option (4) is correct.

Solutions · 1