First slide

Straight lines in 3D

Question

If the line $\frac{x - 2}{3} = \frac{y - 1}{- 5} = \frac{z + 2}{2}$ lies in the plane $x + 3 y - α z + β = 0$ then $(α, β)$ =

Moderate

A

$(6, - 17)$
B

$(- 6, 7)$
C

$(5, - 16)$
D

$(- 5, 15)$

Solution

The condition that the line $\frac{x - x_{1}}{l} = \frac{y - y_{1}}{m} = \frac{z - z_{1}}{n}$ to lies on the plane $a x + b y + c z + d = 0$ is $a l + b m + c n = 0$ and $a x_{1} + b y_{1} + c z_{1} + d = 0$
Hence,
$\begin{matrix} 1 (3) + 3 (- 5) - α (2) = 0 \\ 3 - 15 - 2 α = 0 \\ 2 α = - 12 \\ α = - 6 \end{matrix}$
And
$\begin{matrix} 1 (2) + 3 (1) + 6 (- 2) + β = 0 \\ 5 - 12 + β = 0 \\ β = 7 \end{matrix}$
Therefore, $(α, β) = (- 6, 7)$

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