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)