VITEEE 2019 Paper

The given line is

x−2

2

=

2y−5

−3

,z=−1 ,
⇒

x−2

2

=

y− 5 2

− 3 2

=

z+1

0

This shows that the given line passes through the point (2,

5

2

,−1) and has direction ratios (2,−

3

2

,0) .
Thus, given line passes through the point having position vector

→

a

=2

^

i

+

5

2

^

j

−

^

k

and is parallel to the vector

→

b

=(2

^

i

+

3

2

^

j

−0

^

k

) .
So, its vector equation is

→

r

=(2

^

i

+

5

2

^

j

−

^

k

) +λ(2

^

i

−

3

2

^

j

+0

^

k

)
Hence, p = 0