First slide

Straight lines in 3D

Question

The symmetric form of the line of intersection of two planes $x - y + 2 z = 5, 3 x + y + z = 7$ is

Moderate

A

$\frac{4 x - 11}{- 3} = \frac{4 y + 9}{3} = \frac{z}{1}$
B

$\frac{x - 11}{- 3} = \frac{y + 9}{3} = \frac{z}{1}$
C

$\frac{4 x - 1}{3} = \frac{4 y - 9}{3} = \frac{z}{1}$
D

$\frac{x - 3}{3} = \frac{y + 2}{- 5} = \frac{z}{- 4}$

Solution

The given planes are $x - y + 2 z = 5, 3 x + y + z = 7$
Consider the normal vectors and then find the cross product of those two vectors to get the vector along the line of intersection of the given two planes
$n_{1} \times n_{2} = |\begin{matrix} i & j & k \\ 1 & - 1 & 2 \\ 3 & 1 & 1 \end{matrix}| = i (- 3) - j (- 5) + k (4)$
Hence the direction ratios of the required line are proportional to $⟨ 3, - 5, - 4 ⟩$
To get the point of intersection, plug in $z = 0$ in the two plane equations and then solve for $x, y$
$x - y = 5, 3 x + y = 7$
Adding these two equations and then simplify $x = 3, y = - 2$
Therefore, the equation of the line is $\frac{x - 3}{3} = \frac{y + 2}{- 5} = \frac{z}{- 4}$

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