UPSC NDA Math Model Paper 10

Given, the equation of the line is

x−1

2

=

y−2

1

=

z−3

2

⇒

x−1

2

=

y−2

1

=

z−3

2

=k
Consider A = (x, y, z) be any point on the line.

⇒

x−1

2

=k,

y−2

1

=k,

z−3

2

=k
⇒x=2k+1,y=k+2,z=2k+3
Hence, A = (2k + 1, k + 2, 2k + 3 ) be any point on the line.
Consider the point B = (1, 2, 1)
Direction ratios of AB = 2k, k, 2k + 2
Direction ratios of line

x−1

2

=

y−2

1

=

z−3

2

=2,1,2
Since, AB is perpendicular to the line
The sum of the product of direction ratios is zero.
⇒ 2(2k) + (1)(k) + (2)(2k + 2) = 0
⇒k=

−4

9

The point A becomes A=(

1

9

,

14

9

,

19

9

)
By distance formula, we have
AB=√

64

81

+

16

81

+

100

81

AB=

2√5

3

Hence, The distance of the point (1, 2, 1) from the line

x−1

2

=

y−2

1

=

z−3

2

is

2√5

3