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Question: Let P(3.2.6) be a point in space and Q be a point on the line F=1-1+2+(-31+] + 5k). Then the value o...

Let P(3.2.6) be a point in space and Q be a point on the line F=1-1+2+(-31+] + 5k). Then the value of u for which the vector PQ is parallel to the plane x-4y+3z = 1 is

Answer

u = 1/4

Explanation

Solution

To find the value of u for which the vector PQ is parallel to the plane x - 4y + 3z = 1, follow these steps:

  1. Express the coordinates of point Q in terms of u: Q = (1 - 3u, -1 + u, 2 + 5u)

  2. Determine the vector PQ: PQ = Q - P = ((1 - 3u) - 3, (-1 + u) - 2, (2 + 5u) - 6) = (-2 - 3u, -3 + u, -4 + 5u)

  3. Find the normal vector n to the plane x - 4y + 3z = 1: n = (1, -4, 3)

  4. Since PQ is parallel to the plane, it is perpendicular to the normal vector. Therefore, the dot product of PQ and n is 0: PQ · n = 0 (-2 - 3u)(1) + (-3 + u)(-4) + (-4 + 5u)(3) = 0

  5. Simplify and solve for u: -2 - 3u + 12 - 4u - 12 + 15u = 0 8u - 2 = 0 8u = 2 u = 1/4

Thus, the value of u for which the vector PQ is parallel to the plane is 1/4.