Question

Let the foot of the perpendicular from the point (1, 2, 4) on the line `(x + 2)/4 = (y - 1)/2 = (z + 1)/3` be P.

Then the distance of P from the plane 3x + 4y + 12z + 23 = 0

Options

• 5

• `50/13`

• 4

• `63/13`

Answer 1

5

Explanation:

Let the coordinates of the foot of the perpendicular be (x, y, z)

Take, `(x + 2)/4 = (y - 1)/2 = (z + 1)/3 = lambda`

(x, y, z) = (4λ − 2, 2 λ + 1, 3λ − 1)

Equation of a line from point A to point

`P = (4lambda - 3)hati + (2lambda - 1)hatj + (3lambda - 5)hatk`

Line passes through the vector `barb`. Then, `b = 4hati + 2hatj + 3hatk`

so, `bar(AP)·barb = 0`

4(4λ − 3) + 2(2λ − 1) + 3(3λ − 5) = 0

29λ = 12 + 2 + 15 = 29

= λ = 1

Now, put the value of λ in the value of point P.

Then, P = (2, 3, 2)

Distance of point P from the plane 3x + 4y + 12z + 23 = 0.

Required distance

`d = |(6 + 12 + 24 + 23)/(sqrt(9 + 16 + 144))|`

`d = |65/13|`

= 5