First slide

Straight lines in 3D

Question

Let $L$ be the line passing through the point $P (1, 2, 0)$ and perpendicular to the plane $\bar{r} \cdot (3 \bar{i} + 4 \bar{k}) = 0$ if line intersects the plane $\bar{r} \cdot (\bar{i} - \bar{j} + \bar{k}) = 13$ at point $Q$ , then $Q$ is

Moderate

A

$(3, - 6, 4)$
B

$(- 2, - 4, 1)$
C

$(7, 2, 8)$
D

$(4, 0, 9)$

Solution

The equation of the line passing through the point $P (1, 2, 0)$ and perpendicular to the plane $\bar{r} \cdot (3 \bar{i} + 4 \bar{k}) = 0$ is $\frac{x - 1}{3} = \frac{y - 2}{0} = \frac{z - 0}{4} = k$
The general point on the above line is $(3 k + 1, 2, 4 k)$
This point lies on the plane $x - y + z = 13$
It implies that
$\begin{matrix} 3 k + 1 - 2 + 4 k = 13 \\ 7 k = 14 \\ k = 2 \end{matrix}$
Therefore, the point $Q (7, 2, 8)$

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