Q. 295.0( 1 Vote )

# Find the equation

Given: Planes  The equation of the planes passing through the line of intersection of these planes is given by;

2x − 3y + 4z − 1 + λ(x − y + 4) = 0

(2 + λ)x + (−3 − λ)y + (4)z − 1 + 4λ = 0

((2 + λ), (−3 − λ), (4)) ……(i)

Perpendicular plane is In Cartesian form (2, −1, 1) ……(ii)

According to the given condition product of (i) and (ii) should be zero.

2(2 + λ) − (−3 − λ) + 4 = 0 The required equation is The given line x − 1 = 2y − 4 = 3z − 12 (1, 2, 4) is a point on the line  the plane contains the line x − 1 = 2y − 4 = 3z − 12

OR

Given: The point (1, 2, −4) and lines The line perpendicular to these lines will be parallel to the line obtained by the cross product of these lines.   The vector equation of the required line is The Cartesian equation of the required line is Rate this question :

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