Find the equation of the plane through the point whose position vector

1.	Find the equation of the plane through the point whose position vector is vector(2i - j + k) and perpendicular to the vector(4i + 2j - 3k).
Answer» The required plane is perpendicular to vector(4i + 2j - 3k) So, it is parallel to the plane 4x + 2y – 3z = k ∴ the equation of the plane is 4x + 2y – 3z = k The plane passes through the point (2, -1, 1) ⇒ (4)(2) + 2(-1) – 3(1) = λ i.e. λ = 8 – 2 – 3 = 3 So, the equation of the plane is 4x + 2y – 3z = 3.

Find the equation of the plane through the point whose position vector is vector(2i - j + k) and perpendicular to the vector(4i + 2j - 3k).

Answer»

The required plane is perpendicular to vector(4i + 2j - 3k)

So, it is parallel to the plane 4x + 2y – 3z = k

∴ the equation of the plane is 4x + 2y – 3z = k

The plane passes through the point (2, -1, 1)

⇒ (4)(2) + 2(-1) – 3(1) = λ i.e. λ = 8 – 2 – 3 = 3

So, the equation of the plane is 4x + 2y – 3z = 3.