First slide

Straight lines in 3D

Question

The lines $x - y + z = 1, 3 x - y - z + 1 = 0 and x + 4 y - z = 1, x + 2 y + z + 1 = 0$ are

Moderate

A

Parallel
B

intersection
C

perpendicular and skew
D

non-perpendicular and skew

Solution

The direction ratios of line of intersection of two planes $x - y + z = 1, 3 x - y - z + 1 = 0$ are proportional to the vector $|\begin{matrix} i & j & k \\ 1 & - 1 & 1 \\ 3 & - 1 & 1 \end{matrix}| = i (0) - j (- 2) + k (2)$
The direction ratios of the line $L_{1}$ is $〈0, 1, 1〉$
A point on the line $P (0, - 1, 0)$ , by substituting $z = 0$ solve the simultaneous equations.
The direction ratios of line of intersection of two planes $x + 4 y - z = 1, x + 2 y + z + 1 = 0$ are proportional to the vector $|\begin{matrix} i & j & k \\ 1 & 4 & - 1 \\ 1 & 2 & 1 \end{matrix}| = i (6) - j (2) + k (- 2)$
The direction ratios of the line $L_{2}$ is $〈3, - 1, - 1〉$

A point on the line is $Q (- 3, 1, 0)$ , by substituting $z = 0$ solve the simultaneous equations.
To check whether the lines lies in the same plane or not , find the value of $|\begin{matrix} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{matrix}|$
Hence, $|\begin{matrix} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{matrix}| = |\begin{matrix} 0 & 0 & - 2 \\ 1 & 2 & 1 \\ 3 & - 1 & - 1 \end{matrix}| = 14$ , not equal to zero.
Therefore, the lines are skew lines and not perpendicular.

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