7. Where does the line r = (1, 2, -5) + t (2, -3, 1) meet the plane 2x + 5y - 3z = 6?
7. Where does the line r = (1, 2, -5) + t (2, -3, 1) meet the plane 2x + 5y - 3z = 6?
Let the intersection point be P on L such that
P(1+2t, 2-3t, -5+t)
Substitute the values of x,y and z of P into the equation of the plane and solve for t.
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To find where the line and the plane intersect, we need to find the values of "t" that satisfy both the equation of the line and the equation of the plane.
The equation of the line is given as: r = (1, 2, -5) + t(2, -3, 1)
The equation of the plane is given as: 2x + 5y - 3z = 6
Step 1: Substitute the equation of the line into the equation of the plane.
Let's replace x, y, and z in the equation of the plane with the corresponding expressions from the equation of the line:
2[(1) + t(2)] + 5[(2) + t(-3)] - 3[(-5) + t(1)] = 6
Step 2: Simplify the equation.
2(1 + 2t) + 5(2 - 3t) - 3(-5 + t) = 6
2 + 4t + 10 - 15t + 15 + 3t = 6
6t + 27 = 6
Step 3: Solve for t.
6t = 6 - 27
6t = -21
t = -21/6
t = -7/2 or -3.5
Step 4: Substitute the value of t back into the equation of the line.
For t = -7/2:
r = (1, 2, -5) + (-7/2)(2, -3, 1)
r = (1, 2, -5) + (-14/2, 21/2, -7/2)
r = (1, 2, -5) + (-7, 10.5, -3.5)
r = (-6, 12.5, -8.5)
For t = -3.5:
r = (1, 2, -5) + (-3.5)(2, -3, 1)
r = (1, 2, -5) + (-7, 10.5, -3.5)
r = (-6, 12.5, -8.5)
Therefore, the line r = (1, 2, -5) + t (2, -3, 1) intersects the plane 2x + 5y - 3z = 6 at the points (-6, 12.5, -8.5) and (-6, 12.5, -8.5).