Find s for the given planes 2x + 2y + sz + 2 = 0 and 3x + y + z – 2 = 0, if they are perpendicular to each other.a)21b)- 7c)12d)- 8Correct answer is option 'D'. Can you explain this answer?

Question

Find s for the given planes 2x + 2y + sz + 2 = 0 and 3x + y + z &ndash; 2 = 0, if they are perpendicular to each other.a)21b)- 7c)12d)- 8Correct answer is option 'D'. Can you explain this answer?

Answer

If their normals are perpendicular to each other then a₁a₂ + b₁b₂ + c₁c₂ = 0.
2(3) + 2(1) + s(1) = 0
s(1) = - 8
k = - 8

Answer

Perpendicularity of Planes
When two planes are perpendicular to each other, the dot product of their normal vectors is zero. The normal vector of a plane is the coefficients of x, y, and z in the equation of the plane.

Finding the Normal Vectors
For the plane 2x + 2y + sz + 2 = 0, the normal vector is (2, 2, s).
For the plane 3x + y + z - 2 = 0, the normal vector is (3, 1, 1).

Dot Product Calculation
To check if the two planes are perpendicular, we need to calculate the dot product of their normal vectors and set it equal to zero.
(2, 2, s) . (3, 1, 1) = 2(3) + 2(1) + s(1) = 6 + 2 + s = 0
8 + s = 0
s = -8

Conclusion
Therefore, the value of s for the given planes to be perpendicular to each other is -8.

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