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Determine the angle between the surface z=x²+y²-1and x²+y²+z²=4 at the point (1,0,-1).

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by sarojdas1963

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by sarojdas1963

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x²+y²+z²=4 ............(ii)

Δ(x²+y²-z) = [2x, 2y, -1z]

Δ(x²+y²+z²) = [2x, 2y, 2z]

The tangent vector at the point are (1, 0, -1)

Therefore, A = (2, 0, 2)

B = (2, 0, -2)

Therefore, the product of this two vectors are,

A x B = 2 x 2 + 0 x 0 + 2 x -2

= 4 + 0 - 4

= 0

Again, A x B = ΙAΙ ΙBΙ cosθ

The magnitude of the two vectors,

ΙAΙ = √2² + 0² + 2² = √8

ΙBΙ = √2² + 0² + -2² = √8

Now, substituting the values,

A x B = ΙAΙ ΙBΙ cosθ

0 = √8 x √8 x cosθ

0 = (√8)² x cosθ

0 = 8 x cosθ

cosθ = 0

cosθ = cos90°

θ = 90°