Q.

Let the line passing through the points (–1, 2, 1) and parallel to the line x–12=y+13=z4 intersect the line x+23=y–32=z–41 at the point P. Then the distance of P from the point Q(4, –5, 1) is         [2025]

1 55  
2 56  
3 5  
4 10  

Ans.

(1)

Equation of line passing through (–1, 2, 1) and parallel to x–12=y+13=z4 is x+12=y–23=z–14.

x+12=y–23=z–14=λ (say)

Coordinates of P are (2λ–1,3λ+2,4λ+1).

As point P also lies on

x+23=y–32=z–41=μ (say)

∴   Coordinates of point P are (3μ–2,2μ+3,μ+4)

⇒ 2λ–1=3μ–2, 3λ+2=2μ+3, 4λ+1=μ+4

On solving, we get λ=μ=1

∴   The point P is (1, 5, 5)

∴   Required distance QP=(4–1)2+(–5–5)2+(1–5)2

                                                     =9+100+16=55