KCET 2015 Math Solved Paper

Given equation of plane
2x−3y+4z=29 → (1)
Normal vector of this plane is given by

→

N

=2

∧

i

−3

∧

j

+4

∧

k

Now, equation of line which passes through the origin and has the same direction as normal vectors N→ is
‌

x

2

=‌

y

−3

=‌

z

4

=λ → (2)
⇒x=2−x,y=−3λ,z=4λ → (3)
The coordinates of the foot of the perpendicular is
4λ+9λ+16λ=29
⇒29λ=29⇒λ=1
Putting the value of λ in Eq. (3), we have
x=2,y=−3,z=4 or (2,−3,4)