First slide

Planes in 3D

Question

The image of the line $\frac{x - 1}{3} = \frac{y - 3}{1} = \frac{z - 4}{- 5}$ in the plane $2 x - y + z + 3 = 0$ is the line

Moderate

A

$\frac{x - 3}{3} = \frac{y + 5}{1} = \frac{z - 2}{- 5}$
B

$\frac{x - 3}{- 3} = \frac{y + 5}{- 1} = \frac{z - 2}{5}$
C

$\frac{x + 3}{3} = \frac{y - 5}{1} = \frac{z - 2}{- 5}$
D

$\frac{x + 3}{- 3} = \frac{y - 5}{- 1} = \frac{z + 2}{5}$

Solution

$3 (2) + 1 (- 1) + (- 5) (1) = 0$
Given line and given plane are parallel
$∴$ Image line is also parallel to the given line
Image of A (1,3,4) w.r.to given plane lies on the image line.
Equation of the normal to the plane is $\frac{x - 1}{2} = \frac{y - 3}{- 1} = \frac{z - 4}{1}$
Any point on the line $B = (2 r + 1, - r + 3, r + 4)$
If B is the image of A (1,3,4) then mid point of AB lies on the plane
Mid point $= (\frac{2 r + 2}{2}, \frac{- r + 6}{2}, \frac{r + 8}{2})$
Mid point lies in the given plane $\Rightarrow 2 (\frac{2 r + 2}{2}) - (\frac{- r + 6}{2}) + \frac{r + 8}{2} + 3 = 0$
$\Rightarrow r = - 2, B (- 3, 5, 2)$
Image lines is $\frac{x + 3}{3} = \frac{y - 5}{1} = \frac{z - 2}{- 5}$

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