First slide

Straight lines in 3D

Question

The symmetric form of the line of intersection of two planes $3 x - y = 6, y + 2 z - 1 = 0$

Very Easy

A

$\frac{x - \frac{7}{3}}{2} = \frac{y - 1}{6} = \frac{z}{- 3}$
B

$\frac{x - \frac{7}{3}}{2} = \frac{y + 1}{6} = \frac{z}{3}$
C

$\frac{x + \frac{7}{3}}{2} = \frac{y - 1}{6} = \frac{z}{- 3}$
D

$\frac{x - 2}{2} = \frac{y}{6} = \frac{z + 0.5}{- 3}$

Solution

The given planes are $3 x - y = 6, y + 2 z - 1 = 0$
Plug in $z = 0$ to get a point on the line.
$3 x - y = 6, y = 1$
solve the above two equations, we get $x = \frac{7}{3}, y = 1$
A point on the line is $(\frac{7}{3}, 1, 0)$
Vector along the line is cross product of two normal vectors to the planes
it gives $n_{1} \times n_{2} = |\begin{matrix} i & j & k \\ 3 & - 1 & 0 \\ 0 & 1 & 2 \end{matrix}| = i (- 2) - j (6) + k (3)$
Direction ratios of a line of intersection of two planes are proportional to $⟨ 2, 6, - 3 ⟩$
Therefore, the equation of the required line is $\frac{x - \frac{7}{3}}{2} = \frac{y - 1}{6} = \frac{z}{- 3}$

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