Find the vector equation of the plane passing through the intersection of the plane r.(i+j+k)=6 and r.(2i+3j+4k)=-5 and the point (1,1,1).
Solution:
We know that the vector equation of plane :
a + A(b) = 0
Here a = r.(i+j+k)-6 and b = r.(2i+3j+4k)+5
r.(i+j+k)-6+A[r.(2i+3j+4k)+5]=0
r.(i+j+k)+A[r.(2i+3j+4k)-6+5A=0
Taking r common :
r.[i+j+k+A(2i+3j+4k)]-6+5A=0
r.[i(1+2A)+j(1+3A)+k(1+4A)]-6+5A=0 ...(1)
Now intersecting point (1,1,1) is given:
Therfore r = i+j+k
Putting r value in eq(1) we get
(i+j+k).[i(1+2A)+j(1+3A)+k(1+4A)]-6+5A=0
1+2A+1+3A+1+4A-6+5A = 0
3-6+9A+5A = 0
-3+14A = 0
A = 3/14
Putting a value in eq(1) we get,
r.[i(1+6/14)+j(1+9/14)+k(1+12/14)]-6+15/14 = 0
r.[(20i)+(23j)+(26k)]-69=0
Therefore the vector equation of plane is r.[(20i+23j+26k)]-69=0.
Thank you!❤️