A line L : x - 2 1 = y - 3 2 = z - 1 - 1 is perpendicular to a plane (P), which passing through the point (4, 3, 9). If the mirror image of point ' S ' on the line ( L ) in the given plane ( P ) is (2, 3, 1), then co-ordinates of point S, is:

Question

A line $L : \frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 1}{- 1}$ is perpendicular to a plane (P), which passing through the point (4, 3, 9). If the mirror image of point $' S'$ on the line $(L)$ in the given plane $(P)$ is (2, 3, 1), then co-ordinates of point S, is:

Smart Achievers · Accepted Answer

(4)

Ans. $(4, 7, - 1)$

Explanation:

$Line L$

$\frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 1}{- 1} = λ (let)$

$x = λ + 2, y = 2 λ + 3, z = - λ + 1$

$direction ratio of plane (2, 0, 8)$

$Than$

$2 (λ + 2) + 0 (2 λ + 3) + 8 (- λ + 1) = 0$

$2 λ + 4 + 0 - 8 λ + 8 = 0$

$- 6 λ = - 12$

$λ = 2$

$Co-ordinate of point S = (2 + 2, 2 \times 2 + 3, - 2 + 1)$

$S (4, 7, - 1)$

Solution

Q.
A line $L : \frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 1}{- 1}$ is perpendicular to a plane (P), which passing through the point (4, 3, 9). If the mirror image of point $' S'$ on the line $(L)$ in the given plane $(P)$ is (2, 3, 1), then co-ordinates of point S, is:

1 $(1, 0, 3)$

2 $(0, - 1, 3)$

3 $(- 2, - 3, - 1)$

4 $(4, 7, - 1)$

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Solution

Q. A line L:x-21=y-32=z-1-1 is perpendicular to a plane (P), which passing through the point (4, 3, 9). If the mirror image of point 'S' on the line (L) in the given plane (P) is (2, 3, 1), then co-ordinates of point S, is:

1 (1, 0, 3)

2 (0, -1, 3)

3 (-2, -3, -1)

4 (4, 7, -1)