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in Three Dimensional Geometry by (4.0k points)
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(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.

1 Answer

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Best 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

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