the line can be written as
x = 1 + 2t
y = 2 - 3t
z = -5 + t
sub that into your plane equation and see if it satisfies the equation
7. Where does the line r = (1, 2, -5) + t (2, -3, 1) meet the plane 2x + 5y - 3z = 6?
4 answers
slightly mis-read the question.
continue from above
2(1+2t) + 5(2-3t) - 3(-5+t) = 6
2 + 4t + 10 - 15t + 15 - 3t = 6
-14t = -21
t = -21/-14 = 3/2
x = 1 + 2(3/2) = 4
y = 2 - 3(3/2) = -5/2
z = -5 + 3/2 = -7/2
the line intersects at (4, -5/2, -7/2)
continue from above
2(1+2t) + 5(2-3t) - 3(-5+t) = 6
2 + 4t + 10 - 15t + 15 - 3t = 6
-14t = -21
t = -21/-14 = 3/2
x = 1 + 2(3/2) = 4
y = 2 - 3(3/2) = -5/2
z = -5 + 3/2 = -7/2
the line intersects at (4, -5/2, -7/2)
thanks reiny!
welcome