Find the equation of the plane through the point (4, -3, 2) and perpendicular to the line of intersection of the planes x – y + 2z – 3 = 0 and 2x – y -3z = 0. Find the point of intersection of the line r = i+2j-k + lamda(i+3j -9k) and the plane obtained above.
Answers
The equation of the plane through the point (4, -3, 2) and perpendicular to the line of intersection of the planes x – y + 2z – 3 = 0 and 2x – y -3z = 0
1. Let equation of plane be
a(x-4) + b(y+3) + c(z-2) = 0 .... (1)
Where a, b, c are direction ratios of plane
2. Above plane is perpendicular to x-y+2z-3=0 and 2x-y-3z=0
Therefore,
a-b+2c=0 and 2a-b-3c=0
3. Solving both these equation,
a/5=b/7=c/1=k
4. Substitute values of a, b, c in equation (1) and solve
5k(x-4) + 7k(y+3) + k(z-2) =0
5. Solving further we get equation of plane,
5x + 7y + z -1 = 0
6. Writing equation of plane in vector form :-
P = (xi + yj + zk).(5i + 7j + k) - 1 = 0
7. For Point of intersection,
Equation of line, r = i + 2j - k + (lamda)(i + 3j - 9k) .....(2)
[ ( i + 2j - k) + (lamda)( i + 3j - 9k) ].( 5i + 7j + k) - 1 = 0
8. Solving for lamda, we get
Lamda = -1
Putting lamda = -1 in (2)
we get,
r = 0i - j + 8k
9. Hence, Point of intersection of line and plane is at ( 0, -1 , 8 )
