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Determine the point at which the line passing through the points P(1, 0, 6) and Q(7, -1, 7) intersects the plane given by the equation x + y - 2 = 7.

User WBuck
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Final answer:

The line passing through points P(1, 0, 6) and Q(7, -1, 7) intersects the plane x + y - 2 = 7 at point Q(7, -1, 7), which is derived through parametric equations and solving for the parameter t.

Step-by-step explanation:

The question asks to find the point of intersection between a line passing through points P(1, 0, 6) and Q(7, -1, 7) with the plane represented by the equation x + y - 2 = 7. To solve this, we first need to find the parametric equations of the line by using the two given points P and Q to define the direction vector Δ(x, y, z), which will be (6, -1, 1) subtracting coordinates of P from Q. The parametric equations of the line are:


  • x = 1 + 6t

  • y = -t

  • z = 6 + t

Next, we substitute these into the equation of the plane x + y - 2 = 7 to find the value of parameter t:

1 + 6t - t - 2 = 7

Solving the above equation for t gives us t = 1. Plugging t back into the parametric equations, we get the intersection point (7, -1, 7), which is actually point Q itself. Therefore, the line intersects the plane at point Q.

User Kit Ostrihon
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