Aren't we just finding the distance from the point
(4,-2,k) to the plane 10x - 7y + z - 28 = 0 ?
and we know that distance is 10
|10(4) - 7(-2) + 1(k) - 28|/√(100 + 49 + 1) = 10
|k + 26| = 10√150 = 50√6
k + 26 = 50√6 OR -k-26 = 50√6
k = 50√6 - 26 OR k = - 50√- 26
Consider the lines
L1: [x,y,z] = [4,-2,k] + s[1,1-3] +t[2,3,1]
Determine a value of k that is 10 units away from the plane
with the equation 10x-7y+Z-28=0
1 answer