Prove that the lines x = py + q, z = ry + s and x = p′y + q′, z = r′y + s′ are perpendicular if pp′ + rr′ + 1 = 0.
Given: Lines x = py + q, z = ry + s and x = p’y + q’, z = r’y + s’ are perpendicular.
To Prove: pp’ + rr’ + 1 = 0
Proof: Take x = py + q and z = ry + s
From x = py + q,
⇒ py = x – q
From z = ry + s,
⇒ ry = z – s
So,
Or,
…(i)
Now, take x = p’y + q’ and z = r’y + s’
From x = p’y + q’,
⇒ p’y = x – q’
From z = r’y + s’,
⇒ r’y = z – s’
So,
Or,
…(ii)
From (i),
Line L1 is parallel to . [From the denominators of the equation (i)]
From (ii),
Line L2 is parallel to . [From the denominators of the equation (ii)]
According to the question, the lines are perpendicular.
Then, the dot product of the vectors must be equal to 0.
That is,
⇒ pp’ + 1 + rr’ = 0
[∵, by vector dot multiplication, ]
Or,
Thus, the given lines are perpendicular if pp’ + rr’ + 1 = 0.