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 plane passes through a fixed point (a,b,c). Show that the locus of the foot of the perpendicular to it from the origin is the sphere x^(2)+y^(2)+z^(2)-ax-by-cz=0

The length of the perpendicular drawn from the point P(a,b,c) from z -axis is sqrt(a^(2)+b^(2)) b.sqrt(b^(2)+c^(2)) c.sqrt(a^(2)+c^(2)) d.sqrt(a^(2)+b^(2)+c^(2))

The distance of the point P(a ,\ b ,\ c) from the x-axis is a. sqrt(b^2+c^2) b. sqrt(a^2+c^2) c. sqrt(a^2+b^2) d. none of these

Distance of the points (a,b,c) for the y axis is (a) sqrt(b^(2)+c^(2)) (b) sqrt(c^(2)+a^(2)) (c )sqrt(a^(2)+b^(2)) (d) sqrt(a^(2)+b^(2)+c^(2))

The shortest distance of the point (a,b,c) from x-axis (A) sqrt(a^2+b^2) (B) sqrt(b^2+c^2) (C) sqrt(c^2+a^2) (D) sqrt(a^2+b^2+c^2)

Find the value of sqrt((a+b +c)^(2) + (a +b+c)^(2) + 2(c^(2) -a^(2) -b^(2) -2ab))

Prove: |(a^2,b c, a c+c^2),(a^2+a b,b^2,a c ),(a b,b^2+b c,c^2)|=4a^2b^2c^2

A plane makes intercepts O A ,O Ba n dO C whose measurements are a,b and c on the O X ,O Ya n dO Z axes. The area of triangle A B C is a. 1/2(a b+b c+c a) b. 1/2a b c(a+b+c) c. 1/2(a^2b^2+b^2c^2+c^2a^2)^(1//2) d. 1/2(a+b+c)^2

Two systems of rectangular axes have the same origin. If a plane cuts them at distance a ,b ,cand a^prime ,b^(prime),c ' from the origin, then a. 1/(a^2)+1/(b^2)+1/(c^2)+1/(a^('2))+1/(b^('2))+1/(c^('2))=0 b. 1/(a^2)-1/(b^2)-1/(c^2)+1/(a^('2))-1/(b^('2))-1/(c^('2))=0 c. 1/(a^2)+1/(b^2)+1/(c^2)-1/(a^('2))-1/(b^('2))-1/(c^('2))=0 d. 1/(a^2)+1/(b^2)+1/(c^2)+1/(a^('2))+1/(b^('2))+1/(c^('2))=0