Find the value of λ, such that the line x-2/6 = y-1/λ = z+5/-4 is perpendicular to the plane 3x - y - 2z = 7.
Find the value of λ, such that the line x-2/6 = y-1/λ = z+5/-4 is perpendicular to the plane 3x - y - 2z = 7.
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Given, line x-2/6 = y-1/λ = z+5/-4 is perpendicular to plane 3x - y - 2z = 7
Therefore, DR's of the line are proportional to the DR's normal to the plane.
6/3 = λ/-1 = -4/-2
2 = -λ
λ = -2
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