The two lines $x=a y+b, z=c y+d$; and $x=a^{\prime} y+b^{\prime}, z=c^{\prime} y+d^{\prime}$ are perpendicular to each other if

(A) $\mathrm{aa}^{\prime}+\mathrm{cc}^{\prime}=-1$

(B) $\mathrm{aa}^{\prime}+\mathrm{cc}^{\prime}=1$

(C) $\frac{\mathrm{a}}{\mathrm{a}^{\prime}}+\frac{\mathrm{c}}{\mathrm{c}^{\prime}}=-1$

(D) $\frac{a}{a^{\prime}}+\frac{c}{c^{\prime}}=1$

JEE Main · 2006 · 30 Apr 2006 · Offline · Three Dimensional Geometry

The two lines

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

; and

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

are perpendicular to each other if

(A) $aa^{'} + cc^{'} = - 1$ ✓ Correct
(B) $aa^{'} + cc^{'} = 1$
(C) $\frac{a}{a ^{'}} + \frac{c}{c ^{'}} = - 1$
(D) $\frac{a}{a ^{'}} + \frac{c}{c ^{'}} = 1$

Answer: (A)

Solution

Read each line with

y

as the free parameter. The first,

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

, is the same as

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

, so its direction ratios are

(a, 1, c)

; in the same way the second line has direction ratios

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

. Two lines are perpendicular when the dot product of their direction vectors is zero, so

a a^{'} + (1) (1) + c c^{'} = 0

, which rearranges to

a a^{'} + c c^{'} = - 1

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option

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