First slide

Planes in 3D

Question

Equation of a line in the plane $π≡2x−y+z−4=0$ which is perpendicular to the line whose equation is $\frac{x - 2}{1} = \frac{y - 2}{- 1} = \frac{z - 3}{- 2}$ and which passes through the point of intersection of $l and π$ is

Moderate

A

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

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

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

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

Solution

Let direction ratios of the line be (a, b, c ,) then 2a-b+c=0
$and a - b - 2 c = 0, i.e., \frac{a}{3} = \frac{b}{5} = \frac{c}{- 1}$
Therefore, direction ratios of the line are (3, 5, - 1).
Any point on the given line is $(2 + λ, 2 - λ, 3 - 2 λ)$ it lies on the given plane $π$ if
$2 (2 + λ) - (2 - λ) + (3 - 2 λ) = 4$
or $4 + 2 λ - 2 + λ + 3 - 2 λ = 4 or λ = - 1$
Therefore, the point of intersection of the line and the plane is (1, 3, 5).
Therefore, equation of the required line is
$\frac{x - 1}{3} = \frac{y - 3}{5} = \frac{z - 5}{- 1}$

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