First slide

Straight lines in 3D

Question

Equation of the perpendicular line from (3, –1, 11) to the line $\frac{x}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ is

Moderate

A

$\frac{x - 3}{1} = \frac{y + 1}{- 6} = \frac{z - 11}{4}$
B

$\frac{x - 3}{2} = \frac{y + 1}{5} = \frac{z - 11}{7}$
C

$\frac{x - 3}{1} = \frac{y + 1}{11} = \frac{z - 11}{3}$
D

$\frac{x - 3}{1} = \frac{y + 1}{6} = \frac{z - 11}{4}$

Solution

Let P be the foot of the perpendicular from
A (3, –1, 11) to the given line then
P = (2r , 3r + 2, 4r + 3)
D.r’s of A.P are (2r – 3, 3r + 3, 4r – 8)
$\overset{\leftrightarrow}{A P}$ is perpendicular to the given line
$\Rightarrow$ 2(2r – 3) + 3(3r + 3) + 4(4r – 8)=0 $\Rightarrow$ r = 1 P = (2, 5, 7)
D.r’s of $\overset{\leftrightarrow}{A P}$ are (1, –6, 4)

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