JEE Main202528 Jan 2025Evening ShiftMathematicsThree Dimensional GeometryActual

The square of the distance of the point ( 15 7 , 32 7 , 7 ) from the line x+1 3 = y+3 5 = z+5 7 in the direction of the vector i +4 j +7 k is :

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Correct answer

D. 66

Step-by-step solution

aligned & L= x+1 3 = y+3 5 = z+5 7 & P Q= x- 15 7 1 = y- 32 7 4 = z-7 7 = aligned Q ( + 15 7 , 4 + 32 7 , 7 +7 ) Since Q lies on line L aligned & So, + 15 7 +1 3 = 7 +7+5 7 & 7 +22=21 +36 & =-1 & Point Q ( 8 7 , 4 7 , 0 ) & PQ = ( 15 7 - 8 7 )^2+ ( 32 7 - 4 7 )^2+(7-0) & PQ = 66 & ( PQ )^2=66 aligned

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