First slide

Straight lines in 3D

Question

The equation of line passing through the points $(3, p, 0)$ and $(2, q, 4)$ where $p + q = 5$ and $p q = 6$ is $(p > q)$

Moderate

A

$\frac{x - 2}{1} = \frac{y - 2}{3} = \frac{z - 4}{2}$
B

$\frac{x - 3}{0} = \frac{y - 2}{4} = \frac{z - 1}{5}$
C

$\frac{x - 3}{- 1} = \frac{y - 3}{- 1} = \frac{z}{4}$
D

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

Solution

Given $p + q = 5$ and $p q = 6$ , it implies that $p = 3, q = 2$
The given points are $(3, 3, 0)$ and $(2, 2, 4)$
The equation of the line joining the points $A (x_{1}, y_{1}, z_{1}), B (x_{2}, y_{2}, z_{2})$ is $\frac{x - x_{1}}{x_{2} - x_{1}} = \frac{y - y_{1}}{y_{2} - y_{1}} = \frac{z - z_{1}}{z_{2} - z_{1}}$
Therefore, the equation of the line joining the points $(3, 3, 0)$ and $(2, 2, 4)$ is $\frac{x - 3}{- 1} = \frac{y - 3}{- 1} = \frac{z - 0}{4}$

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