Most Important Selected Qs for JEE AdvancedMathematicsThree Dimensional Geometry

The line x=2 y=3 z meets the plane x+y+z=11 at the point P and the sphere centred at origin and radius equal to 14 , at the points R and S , then -

Options

APR + PS =28
BPR . PS = 147
CPR = PS
DPR + PS = RS

Correct answer

D. PR + PS = RS

Step-by-step solution

Any point on line x=2 y=3 z is ( , 2 , 3 ) line meets plane x+y+z=11 at P P is (6,3,2) Similarly it meets sphere given by x^2+y^2+z^2=196 at R & S ^2+ ( 2 )^2+ ( 3 )^2=196 or = 12 R (12,6,4) & S(-12,-6,-4) Now P R= 36+9+4 =7 , PS = 324+81+36 =21 & RS = 576+144+64 =28

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