First slide

Straight lines in 3D

Question

The two lines $x = a y + b, z = c y + d and x = a^{'} y + b^{'}, Z = c^{'} y + d^{'}$ will be perpendicular if and only if

Moderate

A

$a a^{'} + b b^{'} + c c^{'} + 1 = 0$
B

$a a^{'} + b b^{'} + c c^{'} = 0$
C

$a a^{'} + b b^{'} + c c^{'} - 1 = 0$
D

$a a^{'} + c c^{'} + 1 = 0$

Solution

The given lines in symmetric forms are $\frac{x - b}{a} = y = \frac{z - d}{c}, \frac{x - b^{'}}{a^{'}} = y = \frac{z - d^{'}}{c^{'}}$
Two lines $\frac{x - x_{1}}{a_{1}} = \frac{y - y_{1}}{b_{1}} = \frac{z - z_{1}}{c_{1}}$ and $\frac{x - x_{2}}{a_{2}} = \frac{y - y_{2}}{b_{2}} = \frac{z - z_{2}}{c_{2}}$ are to be perpendicular if and only if $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$
Hence, the condition that the given two lines are to be perpendicular is $a a^{'} + c c^{'} + 1 = 0$

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