First slide

Straight lines in 3D

Question

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

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