Let the image of the point P(0,-5,0) in the line (x-1)/(2)=(y)/(1)=(z+1)/(-2) be the point…

Question

Let the image of the point $P(0,-5,0)$ in the line $\frac{x-1}{2}=\frac{y}{1}=\frac{z+1}{-2}$ be the point R and the image of the point $Q(0,\frac{-1}{2},0)$ in the line $\frac{x-1}{-1}=\frac{y+9}{4}=\frac{z+1}{1}$ be the point S . Then the square of the area of the parallelogram PQRS is $____$ .

Answer: 162

Accepted Answer

Let us find the reflected points $R$ and $S$, and then use vectors to get the area of parallelogram $PQRS$.

1. Reflection of $P(0,-5,0)$ in the line
$\frac{x-1}{2}=\frac{y}{1}=\frac{z+1}{-2}$

Write the line in parametric form:

$x=1+2t,\, y=t,\, z=-1-2t$

So, a point on the line is

$A(1,0,-1)$

and its direction vector is

${\vec{d}}_{1}=(2,1,-2)$

To reflect point $P$ in the line, first find the foot of perpendicular from $P$ to the line.

Let the foot be

$M=A+t{\vec{d}}_{1}=(1+2t,t,-1-2t)$

Since $PM\perp {\vec{d}}_{1}$,

$(\vec{PM})\cdot {\vec{d}}_{1}=0$

Now,

$\vec{PM}=M-P=(1+2t,t+5,-1-2t)$

So,

$(1+2t,\, t+5,\, -1-2t)\cdot (2,1,-2)=0$

$2(1+2t)+(t+5)+(-2)(-1-2t)=0$

$2+4t+t+5+2+4t=0$

$9+9t=0$

$t=-1$

Hence,

$M=(1+2(-1),-1,-1-2(-1))=(-1,-1,1)$

Since $M$ is the midpoint of $PR$,

$R=2M-P$

$R=2(-1,-1,1)-(0,-5,0)=(-2,3,2)$

So,

$R=(-2,3,2)$

2. Reflection of $Q(0,-\frac{1}{2},0)$ in the line
$\frac{x-1}{-1}=\frac{y+9}{4}=\frac{z+1}{1}$

Write the line in parametric form:

$x=1-u,\, y=-9+4u,\, z=-1+u$

So, a point on the line is

$B(1,-9,-1)$

and direction vector is

${\vec{d}}_{2}=(-1,4,1)$

Let the foot of perpendicular from $Q$ to this line be

$N=B+u{\vec{d}}_{2}=(1-u,-9+4u,-1+u)$

Since $QN\perp {\vec{d}}_{2}$,

$(\vec{QN})\cdot {\vec{d}}_{2}=0$

Now,

$\vec{QN}=N-Q=(1-u,-9+4u+\frac{1}{2},-1+u)=(1-u,-\frac{17}{2}+4u,-1+u)$

So,

$(1-u,-\frac{17}{2}+4u,-1+u)\cdot (-1,4,1)=0$

$-(1-u)+4(-\frac{17}{2}+4u)+(-1+u)=0$

$-1+u-34+16u-1+u=0$

$18u-36=0$

$u=2$

Hence,

$N=(1-2,-9+8,-1+2)=(-1,-1,1)$

Again, $N$ is the midpoint of $QS$, so

$S=2N-Q$

$S=2(-1,-1,1)-(0,-\frac{1}{2},0)=(-2,-2+\frac{1}{2},2)=(-2,-\frac{3}{2},2)$

So,

$S=(-2,-\frac{3}{2},2)$

3. Check the parallelogram sides
Now,

$P=(0,-5,0),\, Q=(0,-\frac{1}{2},0),\, R=(-2,3,2),\, S=(-2,-\frac{3}{2},2)$

Take adjacent sides from $P$:

$\vec{PQ}=Q-P=(0,\frac{9}{2},0)$

and

$\vec{PR}=R-P=(-2,8,2)$

Also,

$\vec{QS}=S-Q=(-2,-1,2)$

which is not equal to $\vec{PR}$, so let us check the correct order of parallelogram.

Observe:

$\vec{PS}=S-P=(-2,\frac{7}{2},2)$

and

$\vec{QR}=R-Q=(-2,\frac{7}{2},2)$

Thus, $PS\parallel QR$.

Also,

$\vec{PQ}=(0,\frac{9}{2},0),\, \vec{SR}=R-S=(0,\frac{9}{2},0)$

So the correct parallelogram order is $PQRS$, with adjacent sides

$\vec{PQ}\, \, \vec{PS}$

Hence area of parallelogram is

$|\vec{PQ}	imes \vec{PS}|$

4. Compute cross product
$\vec{PQ}=(0,\frac{9}{2},0),\, \vec{PS}=(-2,\frac{7}{2},2)$

$\vec{PQ}	imes \vec{PS}=\begin{vmatrix} \hat{i} \quad  \hat{j} \quad  \hat{k} \ 0 \quad  \frac{9}{2} \quad  0 \ -2 \quad  \frac{7}{2} \quad  2 \end{vmatrix}$

$=\hat{i}(\frac{9}{2}\cdot 2-0\cdot \frac{7}{2})-\hat{j}(0\cdot 2-0\cdot (-2))+\hat{k}(0\cdot \frac{7}{2}-\frac{9}{2}\cdot (-2))$

$=9\hat{i}+0\hat{j}+9\hat{k}$

So,

$|\vec{PQ}	imes \vec{PS}|=\sqrt{{9}^{2}+{9}^{2}}=\sqrt{162}=9\sqrt{2}$

Therefore, the square of the area is

$(9\sqrt{2}{)}^{2}=162$

$162$

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