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A plane passes through a fixed point `(a ,b ,c)dot` The locus of the foot of the perpendicular to it from the origin is a sphere of radius a. `1/2sqrt(a^2+b^2+c^2)` b. `sqrt(a^2+b^2+c^2)` c. `a^2+b^2+c^2` d. `1/2(a^2+b^2+c^2)`

A

`(1)/(2)sqrt(a^(2)+b^(2)+c^(2))`

B

`sqrt(a^(2)+b^(2)+c^(2))`

C

`a^(2)+b^(2)+c^(2)`

D

`(1)/(2)(a^(2)+b^(2)+c^(2))`

Text Solution

Verified by Experts

The correct Answer is:
a

Let the foot of the perpendicular from the origin on the given plane be `P (alpha, beta, gamma)`. Since the plane passes through `A(a, b, c)`,
`" "AP bot OP rArr vec(AP)*vec(OP)=0`
`rArr" "[(alpha -a)hati+(beta-b)hatj+ (gamma-c)hatk]*(alphahati+betahati+gammahatk)=0`
or `" "alpha(alpha-a)+beta(beta-b)+gamma(gamma-c)=0`
Hence, the locus of `(alpha, beta, gamma)` is
`" "x(x-a)+ y(y-b)+ z(z-c)=0`
`" "x^(2)+y^(2)+z^(2)-ax-by-cz=0`
which is a sphere of radius `(1)/(2)sqrt(a^(2)+b^(2)+c^(2))`
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