Manipal Entrance Test 2015 Math Paper

Let P(3,8,2) be the given point.
Let the equation of a line through P(3,8,2) and parallel to the plane 3x+2y−2z+15=0 be

x−3

a

=

y−8

b

=

z−2

c

....(i)
Then, normal to the plane is perpendicular to the parallel line.
Then, 3a+2b−2c=0 .....(ii)
Line (i) intersects the line

x−1

2

=

y−3

4

=

z−2

3

at Q.
∴ |

3−1	8−3	2−2
a	b	c
2	4	3

|=0
⇒ |

2	5	0
a	b	c
2	4	3

|=0
⇒ 2(3b−4c)−5(3a−2c)=0
⇒ 15a−6b−2c=0 .......(iii)
On solving Eqs. (ii) and (iii) by cross-multiplication, we get

a

−4−12

=

b

−30+6

=

c

−18−30

⇒

a

−16

=

b

−24

=

c

−48

⇒

a

2

=

b

3

=

c

6

Substituting a, b and c in Eq. (i), we get

x−3

2

=

y−8

3

=

z−2

6

=λ [say]
Let the coordinates of Q be (2λ+3,3λ+8,6λ+2). It lies on the line

x−1

2

=

y−3

4

=

z−2

3

∴ λ+1=

3λ+5

4

=2λ
⇒ λ=1
∴ Coordinate of Q = (2 + 3, 3 + 8, 6 + 2) i.e. Q =(5,11, 8).
Hence, PQ=√(5−3)2+(11−8)2+(8−2)2
=√4+9+36=7