Show that the points (2, 0, 1), (0,4,-3) and (-2, 5, 0) are non-collinear. Hence find the equation of plane passing through them ?

Please provide guideline with answer.

1 Answer
May 17, 2017

The scalar equation of the plane that contains all 3 points is:

8x+7y+3z = 19

Explanation:

Let vecv = the vector from (2, 0, 1) to (0,4,-3):

vecv = (0-2)hati+(4-0)hatj+(-3-1)hatk

vecv = -2hati+4hatj-4hatk

Let vecu = the vector from (2, 0, 1) to (-2, 5, 0):

vecu = (-2-2)hati+(5-0)hatj+(0-1)hatk

vecu = -4hati+5hatj-1hatk

Compute the dot-product by multiplying the respective components:

vecv*vecu = (-2)(-4) + (4)(5)+(-4)(-1)

vecv*vecu = 32

Compute the magnitudes:

|vecv| = sqrt((-2)^2+4^2+ (-4)^2)

|vecv| = 6

|vecu| = sqrt((-4)^2+5^2+(-1)^2)

|vecu| = sqrt(42)

Use the equation, vecv*vecu = |vecv||vecu|cos(theta), to find the cosine of the angle between the two vectors:

32= 6sqrt42cos(theta)

cos(theta) = 32/(6sqrt42)

If the points were co-linear, then the cosine of the angle between their vectors would have been +-1; it is not, therefore, the points are not co-linear.

Use the cross-product to find a vector that is perpendicular to both vectors. I use a 5 column determinant to compute the cross-product.

First the major diagonals:

|(color(red)(hati),color(green)(hatj),color(blue)(hatk),hati,hatj), (-2,color(red)(4),color(green)(-4),color(blue)(-2),4), (-4,5,color(red)(-1),color(green)(-4),color(blue)(5)) | =

color(red)((4)(-1)hati)+color(green)((-4)(-4)hatj) + color(blue)((-2)(5)hatk)

Subtract from that the minor diagonals:

|(hati,hatj,color(blue)(hatk),color(red)(hati),color(green)(hatj)), (-2,color(blue)(4),color(red)(-4),color(green)(-2),4), (color(blue)(-4),color(red)(5),color(green)(-1),-4,5) | =

color(red)({(4)(-1)-(-4)(5)}hati)+color(green)({(-4)(-4)-(-2)(-1)}hatj) + color(blue)({(-2)(5)-(4)(-4)}hatk)=

16hati+14hatj+6hatk

Divide the vector by 2:

8hati+7hatj+3hatk

These coefficients match the coefficients for x, y, and z, respectively in the scalar equation for a plane:

8x+7y+3z = c

To determine the value of c, substitute in one of the points. I shall use the first one:

8(2)+7(0)+3(1) = c

c = 19

The scalar equation of the plane that contains all 3 points is:

8x+7y+3z = 19