Let _1, _2 be the values of for which the points ((5)/(2),1, ) and (-2,0,1) are at equal d…

Question

Let ${\lambda }_{1},{\lambda }_{2}$ be the values of $\lambda $ for which the points $(\frac{5}{2},1,\lambda )$ and $(-2,0,1)$ are at equal distance from the plane $2x+3y-6z+7=0$. If ${\lambda }_{1}>{\lambda }_{2}$, then the distance of the point $({\lambda }_{1}-{\lambda }_{2},{\lambda }_{2},{\lambda }_{1})$ from the line $\frac{x-5}{1}=\frac{y-1}{2}=\frac{z+7}{2}$ is \_\_\_\_\_\_\_\_\_\_\_\_.

Answer: 9

Accepted Answer

Since $(\frac{5}{2},1,\lambda )$ and $(-2,0,1)$ are equidistant

from plane $2x+3y-6z+7=0$

$\begin{gathered} 	herefore |\frac{2(\frac{5}{2})+3(1)-6(\lambda )+7}{\sqrt{{2}^{2}+{3}^{2}+{6}^{2}}}|=|\frac{2(-2)+3(0)-6(1)+7}{\sqrt{{2}^{2}+{3}^{2}+{6}^{2}}}| \ \Rightarrow |5+3-6\lambda +7|=|-4-6+7| \ \Rightarrow |15-6\lambda |=|-3| \ \Rightarrow 15-6\lambda =\pm 3 \ \Rightarrow 15-6\lambda =315-6\lambda =-3 \ \Rightarrow 6\lambda =126\lambda =18 \ \Rightarrow \lambda =2\lambda =3 \end{gathered}$

$\begin{gathered} \because {\lambda }_{1}>{\lambda }_{2} \ 	herefore {\lambda }_{1}=3{\lambda }_{2}=2 \end{gathered}$

So, point will be $(1,2,3)$

Let ${M}_{0}=(1,2,3)$

${M}_{1}$ is the point through which line passes i.e, $(5,1,-7)$

and $\vec{s}=\hat{i}+2\hat{j}+2\hat{k}$

$	herefore \overset{	o \leftarrow }{{M}_{0}{M}_{1}}=4\hat{i}-\hat{j}-10\hat{k}$

Now, required distance $=|\frac{\overset{	o \leftarrow }{{M}_{0}{M}_{1}}	imes \vec{s}}{|\vec{s}|}|$

$\begin{gathered} =\frac{|(4\hat{i}-\hat{j}-10\hat{k})	imes (\hat{i}+2\hat{j}+2\hat{k})|}{\sqrt{1+4+4}} \ =\frac{|18\hat{i}-18\hat{j}+9\hat{k}|}{3}=9 \end{gathered}$

Solutions · 1