Grade 7Electric Current
the coordinates of foot of perpendicular (7,14,5) to the plane 2x+4y-z=2 is
the coordinates of foot of perpendicular (7,14,5) to the plane 2x+4y-z=2 is
1 Answer
Aditya Gupta
I assume you wish to know the foot of perpendicular FROM P(7, 14, 5) to the plane 2x+4y-z= 2.
Let the foot of perpendicular be (a, b, c). Clearly, the vector (a-7)i + (b-14)j + (c-15)k is parallel to the normal of the plane, viz. 2i+4j-k.
Hence (a-7)i + (b-14)j + (c-5) k = m(2i+4j-k) where m is a constant.
Comparing, we have a-7= 2m...(1)
b-14= 4m ....(2)
And c-5= -m....(3)
Also, (a,b,c) lies on the plane, and thus satisfies the eq. 2a+4b-c=2....(4)
Hence we have 4 eqns and 4 variables a, b, c, m. Solving these we get the required values of a, b, c and hence the point P's coordinates as
P(1, 2, 8).
KINDLY APPROVE :))
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