Question
Question: The plane 2x – y + 3z + 5 = 0 is rotated through 90<sup>0</sup> about its line of intersection with ...
The plane 2x – y + 3z + 5 = 0 is rotated through 900 about its line of intersection with the plane 5x – 4y + 2z + 1 = 0. The equation of the plane in the new position is
A
6x – 9y – 29z – 31 = 0
B
27x – 24y – 26z – 13 = 0
C
43x – 32y – 2z + 27 = 0
D
26x – 43y – 151z – 165 = 0
Answer
27x – 24y – 26z – 13 = 0
Explanation
Solution
Equation of a plane passing through the line of intersection of the given planes is
2x – y + 3z + 5 + l (5x – 4y – 2z + 1) = 0
or (2 + 5l)x – (1 + 4l)y + (3 – 2l)z + 5 + l = 0
This will be perpendicular to the plane
2x – y + 3z + 5 = 0
if 2(2 + 5l) + (1 + 4l) + 3(3 – 2l) = 0
Ž l = –7/4 and the required equation of the plane is
4(2x – y + 3z + 5) – 7(5x – 4y – 2z + 1) = 0
27x – 24y – 26z – 13 = 0