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.