Calculus points

Question

Find the equation for the plane through the points P0(−5,−3,−4)​, Q0(0,5,2)​, and 26 for x R0(4,−1,4) using coefficient 26 for x

Answer 1

In order to define a plane, you need 2 things: A normal vector to the plane, and a point on the plane.

The equation of a plane is

a(x-x_0) + b(y-y_0) + c(z-z_0) = 0, where <a,b,c> is a normal vector, and (x_0, y_0, z_0) is a point on the plane. This definition comes from the construction where the dot product of any vector in the plane, with the normal vector is 0.

To find a normal vector, we need to find two vectors in the plane, and to take their cross product, which is, by definition, perpendicular to each initial vector, and therefore, normal to the plane.

The vector between two points is found by subtracting their components.

vec{PQ} = <0-(-5),5-(-3),2-(-4)> = <5,8,6>

vec{PR} = <4-(-5),-1-(-3),4-(-4)> = <9,2,8>

Both of these vectors are in the plane, so we can find their cross product

PQ x PR =

| i j k |

| 5 8 6 |

| 9 2 8 |

=i*(8*8-2*6) - j*(5*8-6*9) + k*(5*2-8*9)

= 52i + 14*j - 62*k = <52,14,-62>

Choosing point Q to be the reference point, we get

a(x-x_0) + b(y-y_0) + c(z-z_0) = 0

52*(x-0) + 14*(x-5) - 62*(x-2) = 0

Simplifying, we get

52x + 14*(x-5) - 62*(x-2) = 0

26x + 7*(x-5) - 31*(x-2) = 0

This is the equation of the plane.