If the lines $x=a y+b, z=c y+d$ and $x=a^{\prime} z+b^{\prime}, y=c^{\prime} z+d^{\prime}$ are perpendicular, then:

(B) $\mathrm{aa}^{\prime}+\mathrm{c}+\mathrm{c}^{\prime}=0$

(C) $a b^{\prime}+b c{ }^{\prime}+1=0$

(D) $\mathrm{bb}^{\prime}+\mathrm{cc}^{\prime}+1=0$

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Three Dimensional Geometry

If the lines

x = a y + b, z = cy + d

and

x = a^{'} z + b^{'}, y = c^{'} z + d^{'}

are perpendicular, then:

(A) $c c^{'} + a + a^{'} = 0$
(B) $aa^{'} + c + c^{'} = 0$ ✓ Correct
(C) $a b^{'} + b c^{'} + 1 = 0$
(D) $bb^{'} + cc^{'} + 1 = 0$

Answer: (B)

Solution

Write each line in symmetric form to read off its direction. The first,

x = a y + b, z = cy + d

, becomes

\frac{x - b}{a} = \frac{y}{1} = \frac{z - d}{c}

, direction

(a, 1, c)

. The second,

x = a^{'} z + b^{'}, y = c^{'} z + d^{'}

, becomes

\frac{x - b ^{'}}{a ^{'}} = \frac{y - d ^{'}}{c ^{'}} = \frac{z}{1}

, direction

(a^{'}, c^{'}, 1)

. Perpendicular lines have zero dot product of directions:

a a^{'} + 1 \cdot c^{'} + c \cdot 1 = 0

, that is

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

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option

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