First slide

Straight lines in 3D

Question

The parametric form of the equation of the line $3 x - 6 y - 2 z - 15 = 0 = 2 x + y - 2 z - 5$ is

Moderate

A

$\frac{x - 5}{14} = \frac{y}{2} = \frac{z}{15}$
B

$\frac{x - 1}{14} = \frac{y - 5}{2} = \frac{z - 1}{15}$
C

$\frac{x - 3}{14} = \frac{y + 1}{2} = \frac{z}{15}$
D

$\frac{x + 5}{14} = \frac{y}{2} = \frac{z}{15}$

Solution

The given planes are $3 x - 6 y - 2 z - 15 = 0 = 2 x + y - 2 z - 5$
To get a point on the line of intersection of the above two planes, substitute $z = 0$
it gives $3 x - 6 y = 15, 2 x + y = 5$
solving the above two equations, we get $P (3, - 1, 0)$
Suppose that the direction ratios of line are $⟨ a, b, c ⟩$
Hence, $3 a - 6 b - 2 c = 0, 2 a + b - 2 c = 0$ it implies that $\frac{a}{14} = \frac{b}{2} = \frac{c}{15}$ $(b y c r o s s m u l t i p l i c a t i o n m e t h o d)$
Therefore, the equation of the line $\frac{x - 3}{14} = \frac{y + 1}{2} = \frac{z}{15}$

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