7. Where does the line r = (1, 2, -5) + t (2, -3, 1) meet the plane 2x + 5y - 3z = 6?
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
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)
thanks reiny!
welcome
To find the point where the line and plane intersect, we need to solve the system of equations formed by the equation of the line and the equation of the plane.
The equation of the line is given by:
r = (1, 2, -5) + t(2, -3, 1)
We'll substitute the values of x, y, and z from the line equation into the equation of the plane:
2x + 5y - 3z = 6
Substituting the values of x, y, and z from the line equation:
2(1 + 2t) + 5(2 - 3t) - 3(-5 + t) = 6
Simplifying the equation:
2 + 4t + 10 - 15t + 15 + 3t = 6
4t - 15t + 3t + 2 + 10 + 15 = 6
-8t + 27 = 6
Bringing the constant to the other side:
-8t = 6 - 27
-8t = -21
Dividing both sides by -8:
t = (-21) / (-8)
t = 21 / 8
Now, we have the value of t. To find the point of intersection, substitute this value of t back into the equation of the line:
r = (1, 2, -5) + (21/8)(2, -3, 1)
Evaluating this expression:
r = (1, 2, -5) + (21/8)(2, -3, 1)
r = (1, 2, -5) + (21/8)(2, -3, 1)
r = (1, 2, -5) + (21/8)(2, -3, 1)
r = (1, 2, -5) + (42/8, -63/8, 21/8)
r = (1 + 42/8, 2 - 63/8, -5 + 21/8)
r = (8/8 + 42/8, 16/8 - 63/8, -40/8 + 21/8)
r = (50/8, -47/8, -19/8)
r = (25/4, -47/8, -19/8)
Therefore, the line r = (1, 2, -5) + t(2, -3, 1) meets the plane 2x + 5y - 3z = 6 at the point (25/4, -47/8, -19/8).