A perpendicular is drawn from a point on the line \frac{x \textrm{ } - \textrm{ } 1}{2} \textrm{ } = \textrm{ } \frac{y \textrm{ } + \textrm{ } 1}{- 1} \textrm{ } = \textrm{ } \frac{z}{1} to the plane x + y + z =3 such that the foot of the perpendicular Q also lies on the plane x – y + z = 3. Then the co-ordinates of Q are

Engineering

Mathematics

Plane and Its Different Forms

Question

A perpendicular is drawn from a point on the line $\frac{x - 1}{2} = \frac{y + 1}{- 1} = \frac{z}{1}$ to the plane x + y + z =3 such that the foot of the perpendicular Q also lies on the plane x – y + z = 3. Then the co-ordinates of Q are

(1, 0, 2)

(– 1, 0, 4)

(4, 0, – 1)

(2, 0, 1)

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Let point P on the line is (2λ + 1, – λ –1, l) foot of perpendicular Q is given by

$\frac{x - 2 λ - 1}{1} = \frac{y + λ + 1}{1} = \frac{z - λ}{1} = \frac{- (2 λ - 3)}{3}$

$∴$ Q lies on x + y + z + 3 and x – y + z = 3

⇒ x + z = 3 and y = 0

$y = 0 \Rightarrow λ + 1 = \frac{- 2 λ + 3}{3} \Rightarrow λ = 0$

⇒ Q is (2, 0, 1)