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Question: Let P be a plane passing through the points (2, 1, 0), (4, 1, 1) and (5, 0, 1) and R be the point (2...

Let P be a plane passing through the points (2, 1, 0), (4, 1, 1) and (5, 0, 1) and R be the point (2, 1, 6). Then image of R in the plane P is

A

(6, 5, 2)

B

(4, 3, 2)

C

(6, 5, -2)

D

(3, 4, -2)

Answer

(6, 5, –2)

Explanation

Solution

Here's how to find the image of point R in plane P:

  1. Find two vectors in the plane: Given points A = (2, 1, 0), B = (4, 1, 1), C = (5, 0, 1):

    • AB→=B−A=(2,0,1)\overrightarrow{AB} = B - A = (2, 0, 1)
    • AC→=C−A=(3,−1,1)\overrightarrow{AC} = C - A = (3, -1, 1)

  2. Compute the normal vector (n): Using the cross-product:

    • n→=AB→×AC→=∣i^j^k^2013−11∣=(1,1,−2)\overrightarrow{n} = \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 1 \\ 3 & -1 & 1 \end{vmatrix} = (1, 1, -2)

  3. Equation of the plane: Using point A = (2, 1, 0) and the normal vector n→=(1,1,−2)\overrightarrow{n} = (1, 1, -2): 1(x−2)+1(y−1)−2(z−0)=0  ⟹  x+y−2z−3=01(x - 2) + 1(y - 1) - 2(z - 0) = 0 \implies x + y - 2z - 3 = 0

  4. Find the foot F of the perpendicular from R = (2, 1, 6) to the plane: Use the formula: t=ax0+by0+cz0+da2+b2+c2t = \frac{ax_0 + by_0 + cz_0 + d}{a^2 + b^2 + c^2} Here, (a, b, c) = (1, 1, -2) and d = -3, so: t=(2+1−12−3)1+1+4=−126=−2t = \frac{(2 + 1 - 12 - 3)}{1 + 1 + 4} = \frac{-12}{6} = -2 Then, F=R−t⋅n→=(2,1,6)−(−2)(1,1,−2)=(2,1,6)+(2,2,−4)=(4,3,2)F = R - t\cdot\overrightarrow{n} = (2, 1, 6) - (-2)(1, 1, -2) = (2, 1, 6) + (2, 2, -4) = (4, 3, 2)

  5. Determine the image Q of R by reflection in the plane: Since the foot F is the midpoint of R and its image Q: Q=2F−R=2(4,3,2)−(2,1,6)=(8,6,4)−(2,1,6)=(6,5,−2)Q = 2F - R = 2(4, 3, 2) - (2, 1, 6) = (8, 6, 4) - (2, 1, 6) = (6, 5, -2)

Therefore, the image of R in the plane P is (6, 5, -2).