Let P( , , ) be the image of the point Q(3,-3,1) in the line (x-0)/(1)=(y-3)/(1)=(z-1)/(-1…

Question

Let $P(\alpha ,\beta ,\gamma )$ be the image of the point $Q(3,-3,1)$ in the line $\frac{x-0}{1}=\frac{y-3}{1}=\frac{z-1}{-1}$ and $R$ be the point $(2,5,-1)$. If the area of the triangle $PQR$ is $\lambda $ and ${\lambda }^{2}=14K$, then $K$ is equal to :

Accepted Answer

$L\Rightarrow \frac{x-0}{1}=\frac{y-3}{1}=\frac{z-1}{-1}=\mu $

Point $M(\mu ,\mu +3,-\mu +1)$

Direction vector of $QM=(\mu -3,\mu +6,-\mu )$

Direction vector of line $L=(1,1,-1)$

Both direction vector are perpendicular so

$\begin{gathered} 1	imes (\mu -3)+1	imes (\mu +6)-1	imes (-\mu )=0 \ \mu =-1 \end{gathered}$

Point $M(-1,2,2)$ midpoint of $PQ$

So point $P(\alpha ,\beta ,\gamma )=(-5,7,3)$

Area of $	rianglePQR=\frac{1}{2}|\overset{	o \leftarrow }{PQ}	imes \overset{	o \leftarrow }{QR}|$

$\begin{gathered} \overset{	o \leftarrow }{PQ}=(8,-10,-2) \ \overset{	o \leftarrow }{QR}=(-1,8,-2) \ \overset{	o \leftarrow }{PQ}	imes \overset{	o \leftarrow }{QR}=|\begin{array}{ccc}\hat{i} \quad  \hat{j} \quad  \hat{k} \ 8 \quad  -10 \quad  -2 \ -1 \quad  8 \quad  -2\end{array}|=36\hat{i}+18\hat{j}+54\hat{k} \end{gathered}$

Area of $	rianglePQR=\lambda $

$\lambda =\frac{1}{2}\sqrt{(36{)}^{2}+(18{)}^{2}+(54{)}^{2}}$

$\begin{gathered} {\lambda }^{2}=\frac{1}{4}(4536) \ {\lambda }^{2}=1134 \ {\lambda }^{2}=14K \ 14K=1134 \ K=\frac{1134}{14}=81 \ K=81 \end{gathered}$

Solutions · 1