Question

Find the angle of intersection of the sphere x2+y2+z2=29 and x2+y2+z2+4x-6y-8z-47=0 at the point (4,-3,2).​

Answer 1

Step-by-step explanation:

x2+y2+z2=29 and x2+y2+z2+4x-6y-8z-47=x2+y2+z2=29 and x2+y2+z2+4x-6y-8z-47=x2+y2+z2=29 and x2+y2+z2+4x-6y-8z-47=x2+y2+z2=29 and x2+y2+z2+4x-6y-8z-47

Answer 2

Answer:

120°.

Step-by-step explanation:

Given:- Equation of the two spheres are $x^{2} + y^{2}+ z^{2} = 29$ and $x^{2}+ y^{2} + z^{2} + 4x - 6y - 8z - 47 = 0$

To Find:- Angle of intersection of the two given spheres at the point (4, -3, 2).

Solution:-

Derivative of 1st sphere = 2xi + 2yj + 2zk

At point (4, -3, 2) the derivative will be 8i - 6j + 4k.

$n_{1} = \frac{(8i - 6j + 4k)}{\sqrt{8^{2} +(-6)^{2} +4^{2} } }$ $= \frac{8i -6j+4k}{\sqrt{64+36+16} }$

Now, $n_{1}$  will be $\frac{2(4i-3j+2k)}{\sqrt{116} } = \frac{(4i-3j+2k)}{\sqrt{29} }$.

Now let's find the derivative of the 2nd equation of sphere, we get

2xi + 2yj + 2zk + i - j - k = (2x + 1)i + (2y - 1)j + (2z - 1)k

At point (4, -3, 2) the derivative will be 9i - 7j + 3k.

$n_{2} = \frac{9i-7j+3k}{\sqrt{81+49+9}} = \frac{1}{\sqrt{139}}(9i-7j+3k)$

Now, angle between the two sphere at point (4, -3, 2) will be

cos θ = $\frac{n_{1} - n_{2} }{|n_{1}||n_{2}|}$

         = $\frac{1}{\sqrt{29} \sqrt{139} }((16 + 9 + 4)-(36 + 21 + 6))$

         = $\frac{29-63}{\sqrt{29} \sqrt{139} }$

θ = $cos^{-1} (\frac{-34}{\sqrt{29} \sqrt{139} } )$

   ≈ $cos^{-1} (-0.5)$

   = 120°

Therefore, angle of intersection for the two spheres at point (4, -3, 2) is  120°.

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