Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Find the point on the plane 4x+3y+z=2 that is closest to (1,-1,1). Â
Dear Akif Indirectly we hwve to find out foot of perpendicular from (1, - 1,1) to the given plane x -1 /4 = y +1/3 = z – 1/1 = – 2(0) / 26 x -1 /4 = y +1/3 = z – 1/1 = 0 hence x =1, y = -1, z = 1 Means (1, – 1, 1) lies on givne planeÂ
Dear student The normal vector to the plane is (4,3,1)P = (1,-1,1) + c(4,3,1) = (1+4c, -1+3c , 1+c) 4(1+4c) +3(-1+3c) + 1+c = 126c +2 = 1 c = -1/26 P = (1,-1,1) - 1/26 ( 4,3,1) P = (0.84 , -1.11 , 0.96) Â
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -