Q. 184.5( 2 Votes )

Find the equation of plane passing through the points A (3, 2, 1), B (4, 2, - 2) and C(6, 5, - 1) and hence find the value of for which A (3, 2, 1), B(4, 2, - 2), C(6, 5, - 1) and are coplanar
OR

Find the co - ordinates of the point where the line meets the plane which is perpendicular to the vector and at a distance of from origin[CBSE 2016]

Answer :

Given: A (3, 2, 1) B (4, 2, - 2) C (6, 5, - 1)


Formula: equation of plane is given by



Substituting the values in above equation



(x - 3)9 – (y - 2)7 + (z - 1)3 = 0


9x - 7y + 3z = 16


Now if A, B, C, and D are co - planar D must lie on the plane


9λ - 35 + 15 - 16 = 0


λ = 4


OR


given:


Equation of plane perpendicular to at a distance of from origin is





= (-1 + 3λ)1 + (-2 + 4λ)1 + (-3 + 3λ)3 = 4


16λ = 16


λ = 1


Hence substituting λ = 1 in


The point of intersection is ( - 1 + 3, - 2 + 2, - 3 + 3) which is


(2, 2, 0)

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