First slide

Straight lines in 3D

Question

The equation of line parallel to the plane $x + y + z = 2$ and intersecting the line $x + y - z = 0 = x + 2 y - 3 z + 5$ at the point $(- 4, 1, 3)$ is

Moderate

A

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

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

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

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

Solution

The equation of plane is $x + y - z + λ (x + 2 y - 3 z + 5) = 0$
It is passing through the point $(- 4, 1, 3)$
It implies that $- 4 + 1 - 3 + λ (- 4 + 2 - 9 + 5) = 0 \Rightarrow λ = - 1$
Hence, the equation of plane is $y - 2 z + 5 = 0$
The required line is parallel to both planes $y - 2 z + 5 = 0$ and $x + y + z = 2$
Normal vectors to above planes are $n_{1} = i + j + k, n_{2} = j - 2 k$
The vector along the line is perpendicular to both the vectors, so it is along $n_{1} \times n_{2}$
Hence, $n_{1} \times n_{2} = |\begin{matrix} i & j & k \\ 1 & 1 & 1 \\ 0 & 1 & - 2 \end{matrix}| = - 3 i + 2 j + k$
Hence, the direction ratios of the required line are $⟨ - 3, 2, 1 ⟩$
Therefore, the equation of the required line is $\frac{x + 4}{- 3} = \frac{y - 1}{2} = \frac{z - 3}{1}$

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