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Let x/a + y/b + z/c = 1 be a plane, then...

Let `x/a + y/b + z/c = 1` be a plane, then the image of origin with respect to the plane is :

A

`[(2)/(a ((1)/(a ^(2))+(1)/(b ^(2) ) + (1 )/(c ^(2)))), (2)/(b((1)/(a ^(2))+ (1)/(b ^(2))+ (1)/(c ^(2)))),(2)/(c ((1)/(a ^(2))+(1)/(b ^(2)) + (1)/(c ^(2))))]`

B

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

C

`(-2, -b, -2c)`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
A
Equation of plane is `x/a+y/b+z/c=1` (Intercept from of plane)
`(x -0)/((1)/(a))= (y-0)/((1)/(b))=(z-0)/((1)/(c))=(-2(-1))/((1)/(a ^(2))+ (1)/(b^(2))+ (1)/(c ^(2)))`
`implies x = (2)/(a((1)/(a ^(2))+ (a)/(b^(2))+ (1)/(c ^(2))))`
`y = (2)/(b((1)/(a ^(2))+ (a)/(b^(2))+ (1)/(c ^(2))))," "z= (2)/(c((1)/(a ^(2))+ (a)/(b^(2))+ (1)/(c ^(2))))`
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