Find the equation of plane passing through the intersection of two planes x+2y+3z-4=0 , 2x+y-z+5=0 and which is perpendicular to the plane 5x+3y-6z+8=0
Answers
Answer:
33x + 45y + 50z - 41 = 0
Step-by-step explanation:
The planes passing through the line of intersection of the two planes:
P : x + 2y + 3z - 4 = 0
and
Q : 2x + y - z + 5 = 0
are the linear combinations of these planes:
aP + bQ : (a + 2b)x + (2a + b)y + (3a - b)z + (-4a + 5b) = 0.
A normal vector to the plane 5x + 3y - 6z + 8 = 0 is (5, 3, -6).
A normal vector to the plane aP+bQ is (a+2b, 2a+b, 3a-b).
These planes are perpendicular
<=> these normal vectors are perpendicular
<=> the dot product of the vectors is zero
<=> 5(a+2b) + 3(2a+b) - 6(3a-b) = 0
<=> (5+6-18)a + (10+3+6)b = 0
<=> -7a + 19b = 0
A solution to these is a=19 and b = 7.
Therefore, a plane passing through the required line of intersection and perpendicular to 5x+3y-6z+8=0 is 19P+7Q, whose equation is:
(19+14)x + (38+7)y + (57-7)z + (-76+35) = 0
=> 33x + 45y + 50z - 41 = 0