(i) Find the vector equation of the Plane Passing through the intersec

1.	(i) Find the vector equation of the Plane Passing through the intersection of the Planes \(\bar{r}\).(i + j + k) = 6 and \(\bar{r}\) (2i + 3j + 4k) = – 5 and through the point (1, 1, 1). (ii) Express the vector equation \(\bar{r}\).(5i + 3j + 4k) = 0 of a Plane in Cartesian form and hence find its perpendicular distance from the origin.
Answer» (i) The Cartesian equation of the given planes are x + y + z – 6 = 0 and 2x + 3 – y + 4z + 5 = 0 The family of such planes is x + y + z – 6 + λ(2x + 3y+ 4z + 5) = 0 …..(1) Since it passes through (1, 1, 1) \(1+1+1-6+λ(2+3+4+5)=0\\⇒-3+λ(14)=0\\⇒ λ =\frac{3}{14}\) (1) ⇒ x + y + z - 6 + \(\frac{3}{14}\)(2x + 3y +4z +5) = 0 ⇒ 14x + 14y + 14z - 84 +6x +9y + 12z + 15 = 0 ⇒ 20x + 23y + 26z - 69 = 0 Vector Equation is \(\bar{r}\).(20i + 23j + 26k) = 69 (ii) Cartesian Equation is ⇒ 5x + 3y + 4z = 0 Perpendicular distance from origin is \(\frac{5\times0 + 3\times0+4\times0}{\sqrt{25+9+16}}\) = 0

(i) Find the vector equation of the Plane Passing through the intersection of the Planes \(\bar{r}\).(i + j + k) = 6 and \(\bar{r}\) (2i + 3j + 4k) = – 5 and through the point (1, 1, 1). (ii) Express the vector equation \(\bar{r}\).(5i + 3j + 4k) = 0 of a Plane in Cartesian form and hence find its perpendicular distance from the origin.

Answer»

(i) The Cartesian equation of the given planes are

x + y + z – 6 = 0 and 2x + 3 – y + 4z + 5 = 0

The family of such planes is x + y + z – 6 + λ(2x + 3y+ 4z + 5) = 0 …..(1)

Since it passes through (1, 1, 1)

\(1+1+1-6+λ(2+3+4+5)=0\\⇒-3+λ(14)=0\\⇒ λ =\frac{3}{14}\)

(1) ⇒ x + y + z - 6 + \(\frac{3}{14}\)(2x + 3y +4z +5) = 0

⇒ 14x + 14y + 14z - 84 +6x +9y + 12z + 15 = 0

⇒ 20x + 23y + 26z - 69 = 0

Vector Equation is \(\bar{r}\).(20i + 23j + 26k) = 69

(ii) Cartesian Equation is ⇒ 5x + 3y + 4z = 0

Perpendicular distance from origin is

\(\frac{5\times0 + 3\times0+4\times0}{\sqrt{25+9+16}}\)

= 0