find the angel between the surface x^2 +y^2 +z^2 and 3x^2 −y^2 +2z=1 at (1,−2,1)


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Question Number 47190 by 23kpratik last updated on 06/Nov/18

$f i n d t h e a n g e l b e t w e e n t h e s u r f a c e x^{2} + y^{2} + z^{2} a n d 3 x^{2} - y^{2} + 2 z = 1 a t (1, - 2, 1)$
	Answered by tanmay.chaudhury50@gmail.com last updated on 06/Nov/18

	$\vec{▽} (x^{2} + y^{2} + z^{2})$ $(i \frac{\partial}{\partial x} + j \frac{\partial}{\partial y} + k \frac{\partial}{\partial z}) (x^{2} + y^{2} + z^{2})$ $= i 2 x + j 2 y + k 2 z$ $= i 2 (1) + j 2 (- 2) + k 2 (1)$ $= 2 i - 4 j + 2 k$ $\vec{▽} (3 x^{2} - y^{2} + 2 z - 1)$ $= i (6 x) + j (- 2 y) + k (2)$ $= i (6 \times 1) + j (- 2 \times - 2) + k (2)$ $= 6 i + 4 j + 2 k$ $c o s θ = \frac{6 \times 2 - 4 \times 4 + 2 \times 2}{\sqrt{2^{2} + {(- 4)}^{2} + 2^{2}} \times \sqrt{6^{2} + 4^{2} + 2^{2}}} = 0$ $θ = \frac{π}{2}$
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