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Let $Q$ be the image of the point $P(-2,1,-5)$ in the plane $3 x-2 y+2 z+1=0 .$ Consider the following:
1 . The coordinates of $\mathrm{Q}$ are $(4,-3,-1)$.
2 . PQ is of length more than 8 units.
3 . The point $(1,-1,-3)$ is the mid-point of the line segment $\mathrm{PQ}$ and lies on the given plane. Which of the above statements are correct?
Options:
1 . The coordinates of $\mathrm{Q}$ are $(4,-3,-1)$.
2 . PQ is of length more than 8 units.
3 . The point $(1,-1,-3)$ is the mid-point of the line segment $\mathrm{PQ}$ and lies on the given plane. Which of the above statements are correct?
Solution:
1617 Upvotes
Verified Answer
The correct answer is:
1,2 and 3
Let $Q\left(x_{1}, y_{1}, z_{1}\right)$ be the image of the point $P$. The direction ratios of $P Q$ are $3,-2,2$
The Equation of line $P Q$ is $\frac{x+2}{3}=\frac{y-1}{-2}=\frac{z+5}{2}=r$ Coordinates of any point on the line $P Q$ is $3 r-2,-2 r+$ I and $2 r-5$.
Let $Q(3 r-2,-2 r+1,2 r-5)$ be such a point.
et $L$ be the mid point of $P Q, L=\left(\frac{3 r}{2}-2,1-r, r-5\right)$
Since $L$ lies on the plane $3 x-2 y+2 z+1=0$
So, $3\left(\frac{3 r}{2}-2\right)-2(1-r)+2(r-5)+1=0$
$\Rightarrow \frac{17}{2} r-17=0 \Rightarrow r=2$
So, coordinates of $Q$ are $(3 \times 2-2,-2 \times 2+1,2 \times 2-5)$
$=(4,-3,-1)$
Also the mid point of $P Q$ is $L=\left(\frac{3 \times 2}{2}-2,1-2,2-5\right)$ $\therefore P Q=\sqrt{(-2-4)^{2}+(1+3)^{2}+(-5+1)^{2}}=\sqrt{68}$
$\Rightarrow P Q=2 \sqrt{17}>8$
Option (d) is correct.
The Equation of line $P Q$ is $\frac{x+2}{3}=\frac{y-1}{-2}=\frac{z+5}{2}=r$ Coordinates of any point on the line $P Q$ is $3 r-2,-2 r+$ I and $2 r-5$.
Let $Q(3 r-2,-2 r+1,2 r-5)$ be such a point.
et $L$ be the mid point of $P Q, L=\left(\frac{3 r}{2}-2,1-r, r-5\right)$
Since $L$ lies on the plane $3 x-2 y+2 z+1=0$
So, $3\left(\frac{3 r}{2}-2\right)-2(1-r)+2(r-5)+1=0$
$\Rightarrow \frac{17}{2} r-17=0 \Rightarrow r=2$
So, coordinates of $Q$ are $(3 \times 2-2,-2 \times 2+1,2 \times 2-5)$
$=(4,-3,-1)$
Also the mid point of $P Q$ is $L=\left(\frac{3 \times 2}{2}-2,1-2,2-5\right)$ $\therefore P Q=\sqrt{(-2-4)^{2}+(1+3)^{2}+(-5+1)^{2}}=\sqrt{68}$
$\Rightarrow P Q=2 \sqrt{17}>8$
Option (d) is correct.
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