VITEEE 2015 Paper

Given, lines are

x−7

3

=

y+4

−16

=

z−6

7

and

x−10

3

=

y−30

8

=

4−z

5

The vector form of given lines are
r = 7

^

i

−4

^

j

+6

^

k

+ λ (3

^

i

−16

^

j

+7

^

k

)
and r = 10

^

i

+30

^

j

+4

^

k

+ µ (3

^

i

+8

^

j

−5

^

k

)
On comparing these equations with
r = a1+λb1 and r = a2+µb2 we get

→

a1

= 7

^

i

−4

^

j

+6

^

k

→

a2

= 10

^

i

+30

^

j

+4

^

k

→

b1

= 3

^

i

−16

^

j

+7

^

k

and

→

b2

= 3

^

i

+8

^

j

−5

^

k

Shortest distance = |

( → a2 − → a1 ).( → b1 × → b2 )

|b1×b2|

|
= |

72+1224−144

84

| = |

1152

84

| =

288

21

units