JEE Main20265 April 2026Morning ShiftMathematicsThree Dimensional GeometryActual
The square of the distance of the point of intersection of the lines r = ( i + j - k ) + (a i - j ) , a 0 and r = (4 i - k ) + (2 i + a k ) from the origin is:
Options
- A5
- B10
- C17
- D26
Correct answer
C. 17
Step-by-step solution
Equating the position vectors of the two lines for intersection: (1 + a ) i + (1 - ) j - k = (4 + 2 ) i + 0 j + (-1 + a ) k Comparing the coefficients of i , j , k : 1 - = 0 = 1 -1 = -1 + a a = 0 Since a 0 , we get = 0 . Substituting = 1 and = 0 in the i component equation: 1 + a(1) = 4 + 2(0) a = 3 The position vector of the point of intersection is obtained by substituting = 0 in the second line: r = 4 i - k The coordinates of the point of intersection are (4, 0, -1) . The square of the distance from the origin i