BITSAT 2013 Paper

Equation of the line through the given points is

x−3

5−3

=

y−4

1−4

=

z−1

6−1

⇒

x−

2

=

y−4

−3

=

z−1

5

Any point on this line can be taken as (3 + 2λ, 4 – 3λ, 1 + 5λ)
If this point lies on XY-plane then the z-coordinate is zero
⇒ 1 + 5 λ = 0 ⇒ λ = −

1

5

Thus the required coordinates of the point are
(3−

2

5

,4−3(−

1

5

),0) = (

13

5

,

23

5

,0)