First slide

Straight lines in 3D

Question

The symmetrical form of the equation of the line $x + y + z - 1=0$ and $4x + y - 2z + 2 = 0$ is

Difficult

A

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

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

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

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

Solution

D .R of the line = $= |\begin{matrix} \bar{i} & \bar{j} & \bar{k} \\ 1 & 1 & 1 \\ 4 & 1 & - 2 \end{matrix}| = (- 3, 6, - 3)$ $= (1, - 2, 1)$
Let Z = K. Then x = K -1, y = 2 - 2 K.
$∴ (K - 1, 2 - 2 k, k)$ is any point on the line
Hence $(\frac{- 1}{2}, 1, \frac{1}{2}) (0, 0, 1) (- 1, 2, 0)$ are points on the line,

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