The lines \( x=py+q,\ z=ry+s \) and \( x=p'y+q',\ z=r'y+s' \) are perpendicular if
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( pp'+rr'+1=0 \)
Taking y as the parameter, each line can be written in symmetric form.
\( \displaystyle x=py+q,\ z=ry+s \)
\( \displaystyle \ \Rightarrow\ \frac{x-q}{p}=\frac{y}{1}=\frac{z-s}{r} \)
\( \displaystyle x=p'y+q',\ z=r'y+s' \)
\( \displaystyle \ \Rightarrow\ \frac{x-q'}{p'}=\frac{y}{1}=\frac{z-s'}{r'} \)
So the direction ratios are (p, 1, r) and (p′, 1, r′). The lines are perpendicular when
\( \displaystyle p\,p'+1\cdot 1+r\,r'=0 \)
\( \displaystyle pp'+rr'+1=0 \)
Why the other options are incorrect
Option (1) \( pp'+rr'=0 \) — Incorrect. This forgets the y-components: each direction ratio has 1 in the middle, which adds 1 × 1 = 1 to the dot product.
Option (3) \( \displaystyle\frac{p}{p'}=\frac{r}{r'} \) — Incorrect. \( \frac{p}{p'}=\frac{r}{r'} \) (with the middle ratio 1 = 1) is the condition for parallel lines, not perpendicular ones.
Option (4) \( pp'=rr' \) — Incorrect. The perpendicularity condition is \( pp'+1+rr'=0 \), not \( pp'=rr' \).
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Three Dimensional Geometry
