Question

In the following cases find the c9ordinates of foot of perpendicular from the origin `2x + 3y + 4z - 12` = 0

Options

• `(24/29, 36/29, 48/29)`

• `(0, 18/25, 24/25)`

• `(1/3, 1/3, 1/3)`

• `(0, (-8)/5, 0)`

Answer 1

`(24/29, 36/29, 48/29)`

Explanation:

Let N `(x , y, z) be the foot of the perpendicular from the origin to the plane

`2x + 3y + 4z - 12`= 0

Direction ratios of the normal are 2, 3, 4 also the direction ratios of on are `(x_1, y_1 z_1)`

∴ `x+1 = 2k, y_1 = 3k, z = 4k`  ......(1)

The point `(x_1, y_1 z_1)` lies on the plane,

∴ `2(2k) + 3(3k) + 4(4k) - 12` = 0

`(4 + 9 + 16)k` = 12

∴ `k = 12/29`

Putting value of k in (1)

`x_1 = 2, 12/29 = 24/29, y_1 = 3k = 12/29 = 36/29`

∴ `z_1 = 4k = 4 * 12/29 = 48/29`

As a result, the normal foot.