Glet p be the plane that goes through the points a(1, 3, 2), b(2, 3, 0), and c(0, 5, 3). let ` be the line through the point q(1, 2, 0) and parallel to the line x = 5, y = 3−t, z = 6+2t. find the (x, y, z) point of intersection of the line ` and the plane p.

  (x, y, z) = (1, 1/3, 3 1/3)

Step-by-step explanation:

The normal to plane ABC can be found as the cross product ...

  AB×BC = (1, 0, 2)×(2, -2, -3) = (4, 1, 2)

Then the equation of the plane is ...

  4x +y +2z = 4·0 +5 +2·3 . . . . using point C to find the constant

  4x +y +2z = 11

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The direction vector of the reference line is the vector of coefficients of t: (0, -1, 2). Then the line through point Q is ...

  (x, y, z) = (1, 2, 0) +t(0, -1, 2) = (1, 2-t, 2t)

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The value of t that puts a point on this line in plane ABC can be found by substituting these values for x, y, and z in the plane's equation.

  4(1) +(2 -t) +2(2t) = 11

Solving for t gives ...

  t = 5/3

so the point of intersection of the plane and the line is

  (x, y, z) = (1, 2-t, 2t) = (1, 2-5/3, 2·5/3) = (1, 1/3, 3 1/3)

answer: about 22.67

step-by-step explanation: