Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12

# Find the Vector Equation of the Plane Through the Line of Intersection of the Planes X + Y + Z = 1 and 2x + 3y + 4z = 5 Which is Perpendicular to the Plane X − Y + Z = 0. - Mathematics

Sum

Find the vector equation of the plane through the line of intersection of the planes x + yz = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x − y + z = 0.

#### Solution

\text{ The equation of the plane passing through the line of intersection of the given planes is}
$x + y + z - 1 + \lambda \left( 2x + 3y + 4z - 5 \right) = 0$
$\left( 1 + 2\lambda \right)x + \left( 1 + 3\lambda \right)y + \left( 1 + 4\lambda \right)z - 1 - 5\lambda = 0 . . . \left( 1 \right)$
\text{ This plane is perpendicular to x - y + z = 0 . So,}
$1 + 2\lambda - 1 \left( 1 + 3\lambda \right) + 1 + 4\lambda = 0 (\text{ Because a}_1 a_2 + b_1 b_2 + c_1 c_2 = 0)$
$\Rightarrow 1 + 2\lambda - 1 - 3\lambda + 1 + 4\lambda = 0$
$\Rightarrow 3\lambda + 1 = 0$
$\Rightarrow \lambda = \frac{- 1}{3}$
\text{ Substituting this in (1), we get }
$\left( 1 + 2 \left( \frac{- 1}{3} \right) \right)x + \left( 1 + 3 \left( \frac{- 1}{3} \right) \right)y + \left( 1 + 4 \left( \frac{- 1}{3} \right) \right)z - 1 - 5 \left( \frac{- 1}{3} \right) = 0$
$\Rightarrow x - z + 2 = 0$

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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 29 The Plane
Exercise 29.8 | Q 16 | Page 40