How do I find the vector parametric equation for this line, where The planes x−4y+4z=−24 and y−5x−2z=10 are not parallel, so they must intersect along a line that is common to both of them.?

1 Answer
Somebody N.
Jul 18, 2018

See below.

Explanation:

We are looking for the line of intersection of the two planes. To find this we first find the normals to the two planes:

#x-4y+4z=-24 \ \ \ \[1]#

#-5x+y-2z=10 \ \ \ \ \ [2]#

Normal to #[1]# is:

#[(1),(-4),(4)]#

Normal to #[2]# is:

#[(-5),(1),(-2)]#

Since these are perpendicular to each plane, the vector product of the normals will give us a vector that is perpendicular to the direction of both of these. This will therefore be parallel to the line of intersection.

We can find the cross product by finding the determinant :

#|(bbi,bbj,bbk),(1,-4,4),(-5,1,-2)|#

This is:

#4bbi-18bbj-19bbk#

We now need to find a point that is on both planes. Using equations #[1] and [2]# and setting #z=0#

#x-4y+0=-24 \ \ \ \[1]#

#-5x+y+0=10 \ \ \ \ \ [2]#

Solving these simultaneously, we get:

#x=-16/19 , y=110/19, z=0#

So our vector equation is:

#[(-16/19),(110/19),(0)]+t[(4),(-18),(-10)]#

Or:

#x=-16/19+4t, y=110/19-18t, z=-10t#

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