Question

Find an equation of the plane through the point (1, -4, 0) and perpendicular to the vector (5, 1, -4). Do this problem in the standard way or WebWork may not recognize a correct answer.

Answer 1

x = = -5 ; y = -1 ; z = 4
`\bigg | = 3 \bigg |`

Explanation:

The equation of the plane through (1,-4,0) is,

a(x-1) + b(y+4) + c(z+0) = 0 ....... (1)

Again the plane is perpendicular to the vector (5,1,-4)

So, the direction ratios of the normal to the plane is proportional to the direction ratios of the vector (5,1,-4)

So, a/(5) = b/(1) = c/(-4 )

Let,a/(5) = b/(1) = c/(-4 ) = k

So, a = 5k, b = k and c = -4k

So, from (1) we have,

(5k)(x-1) + (k)(y+4) + (-4k)(z+0) = 0

Thus;

-5(x-1) -(y+4) +4(z+0) = 0

So,

-5x +1 -y-4+4z+0 =0

-5x - y +4z -3 = 0

-5x - y +4z = 3

x = = -5 ; y = -1 ; z = 4
`\bigg | = 3 \bigg |`