UPSC NDA Math Model Paper 2

Given: Equation of line is

x

6

=

y+32

2

=

z−2

3

and equation of plane is 3x+4y−12z−7=0.
As we know that the angle between the line

x−x1

a1

=

y−y1

b1

=

z1−z

c1

and the plane a2x+b2y+c2z+d=0 is given by: Here, a1=6,b1=2,c1=3,a2=3,b2=4 and c2=−12
⇒a1.a2+b1.b2+c1.c2 =18+8−36=−10
⇒√a12+b12+c12=7 and √a22+b22+c22=13
⇒sin‌θ=

10

7×13

=

10

91

⇒θ=sin−1(

10

91

)
Hence, option B is the correct answer