First slide

Straight lines in 3D

Question

The symmetric form of the line $x + y - z = 1$ and $2 x - 3 y + z = 2$ is

Very Easy

A

$\frac{x}{2} = \frac{y}{3} = \frac{z}{5}$
B

$\frac{x}{2} = \frac{y}{3} = \frac{z - 1}{5}$
C

$\frac{x - 1}{2} = \frac{y}{3} = \frac{z}{5}$
D

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

Solution

The given planes are $x + y - z = 1$ and $2 x - 3 y + z = 2$
Plug in $z = 0$ in equations, and then solve for $x, y$
$x + y = 1, 2 x - 3 y = 2$
by solving the above two equations $x = 1, y = 0$
The direction ratios are proportional to the vector $n_{1} \times n_{2} = |\begin{matrix} i & j & k \\ 1 & 1 & - 1 \\ 2 & - 3 & 1 \end{matrix}| = i (- 2) - j (3) + k (- 5)$
Therefore, the direction ratios of the required line are proportional to $⟨ 2, 3, 5 ⟩$
Therefore, the equation of the line in symmetric form is $\frac{x - 1}{2} = \frac{y}{3} = \frac{z}{5}$

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