JEE Main202522 Jan 2025Evening ShiftMathematicsThree Dimensional GeometryActual

The perpendicular distance, of the line x-1 2 = y+2 -1 = z+3 2 from the point P (2,-10,1) , is :

Options

A6
B5 2
C4 3
D3 5

Correct answer

D. 3 5

Step-by-step solution

aligned & . x -1 2 = y +2 -1 = z +3 2 = (let ) & (2 +1,- -2,2 -3) & PA n =0 & (2 -1) 2+(- +8)(-1)+(2 -4) 2=0 & 4 -2+ -8+4 -8=0 & 9 -18=0 =2 & ~A (5,-4,1) & AP = 3^2+6^2+0^2 = 45 =3 5 aligned

Practice Three Dimensional Geometry on Quantrex Academy →

More from Three Dimensional Geometry

Let P be the plane such that it contains the straight line x-1 2 = y-3 3 = z+2 1 and is perpendicular to the plane x + 2y + 3z = 4 . Let P₁ be the plane which passes through the po 2026 Let L be the straight line joining the points P(1, 2, -1) and Q(2, 3, 1) . Let S be the foot of the perpendicular drawn from the point R(4, -1, 5) to the line L . Another line pass 2026 Let the foot of perpendicular from the point ( , 2, 3) on the line x-4 1 = y-9 2 = z-5 1 be the point (1, , 2) . Then the distance between the lines x-1 2 = y-2 3 = z+4 6 and x- 2 2026 Let a line L₁ pass through the origin and be perpendicular to the lines L₂: r = (3+t) i + (2t-1) j + (2t+4) k and L₃: r = (3+2s) i + (3+2s) j + (2+s) k , t, s R . If (a, b, c) , a 2026 The shortest distance between the lines x-4 1 = y-3 2 = z-2 -3 and x+2 2 = y-6 4 = z-5 -5 is : 2026 Let the image of the point P(0, -5, 0) in the line x-1 2 = y 1 = z+1 -2 be the point R and the image of the point Q (0, -1 2 , 0 ) in the line x-1 -1 = y+9 4 = z+1 1 be the point S 2026 Let the image of the point P(1, 6, a) in the line L: x 1 = y-1 2 = z-a+1 b , b > 0 , be ( a 3 , 0, a+c ) . If S( , , ) , > 0 , is the point on L such that the distance of S from th 2026 Let a line L be perpendicular to both the lines L₁: x+1 3 = y+3 5 = z+5 7 and L₂: x-2 1 = y-4 4 = z-6 7 . If is the acute angle between the lines L and L₃: x - 8 7 2 = y - 4 7 1 = 2026 Full Three Dimensional Geometry list All JEE Main PYQs