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

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