Let P be the image of the point ( 3, 1, 7 ) with respect to the...

Let

P

be the image of the point

( 3, 1, 7 )

with respect to the plane

x - y + z = 3 .

Then the equation of the plane passing through

P

and containing the straight line

\dfrac{x}{1} = \dfrac{y}{2} = \dfrac{z}{1}

Let P be the image of the point ( 3, 1, 7 ) with respect to the plane x - y + z = 3 . Then the equation of the plane

Options

A $x + y - 3z = 0$
B $3x + z = 0$
C $x - 4y + 7z = 0$
D $2x - y = 0$

Correct answer

C. $x - 4y + 7z = 0$

Step-by-step solution

Let P be the image of the point ( 3, 1, 7 ) with respect to the plane x - y + z = 3 . Then the equation of the plane

Let $F$ be mid point of $AP$
Let P be the image of the point ( 3, 1, 7 ) with respect to the plane x - y + z = 3 . Then the equation of the plane
Line $AP: \dfrac{x - 3}{1} = \dfrac{y - 1}{- 1} = \dfrac{z - 7}{1} = \lambda$

$⇒ F\left(3 + \lambda, 1 - \lambda, \lambda + 7\right)$ lies in the plane

$∴ 3 + \lambda - \left(1 - \lambda\right) + \lambda + 7 = 3$

$3\lambda = - 6 ⇒ \lambda = - 2$

$⇒ F\left(1, 3, 5\right)$

$⇒ P\left(- 1, 5, 3\right)$ (using mid point $F$ )

So required plane (containing line $\dfrac{x}{1} = \dfrac{y}{2} = \dfrac{z}{1}$ & passing through $P\left(- 1, 5, 3\right)$ ) is

\left| $\begin{cases}x - 0 & y - 0 & z - 0 \\ 1 & 2 & 1 \\ - 1 & 5 & 3\end{cases}$ \right| = 0 ∴ x - 4y + 7z = 0

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