Answer :
The angle between the two planes is 118.47 degrees.
A) To find the equation of the plane containing the points (0,0,1), (1,2,3), (2,4,5), and (4,8,11), we can use the point-normal form of the equation of a plane. This form is given by:
ax + by + cz + d = 0
where (a,b,c) is the normal vector to the plane, and d is the distance from the origin to the plane.
To find the normal vector, we can take the cross product of two nonparallel vectors in the plane. For example, we can take the cross product of the vectors (1,2,3) - (0,0,1) and (2,4,5) - (0,0,1). This gives us the normal vector:
(1,2,3) - (0,0,1) = (1,2,2)
(2,4,5) - (0,0,1) = (2,4,4)
(1,2,2) x (2,4,4) = (-8, -4, 4)
Now that we have the normal vector, we need to find the distance from the origin to the plane. We can do this by plugging the coordinates of any of the points in the plane and the normal vector into the point-normal form of the equation of the plane. For example, using the point (0,0,1):
(-8) * 0 + (-4) * 0 + 4 * 1 + d = 0
d = -4
So, the equation of the plane is:
-8x - 4y + 4z - 4 = 0
B) To find the equation of the tangent plane to the surface z = x^2 + xy - 3y at the point (2,3,1), we need to find the partial derivatives of z with respect to x and y at that point. The partial derivatives are given by:
[tex]z_x = 2x + y\\z_y = x - 3[/tex]
Evaluating these partial derivatives at (2,3,1) gives us:
[tex]z_x = 2 * 2 + 3 = 7\\z_y = 2 - 3 = -1[/tex]
So, the equation of the tangent plane is given by:
z - 1 = 7(x - 2) - 1(y - 3)
z - 1 = 7x - 14 - y + 3
z - 7x + y + 11 = 0
To find the angle of intersection between the plane from part (B) and the plane from part (A), we can use the dot product of their normal vectors. The normal vector to the plane from part (B) is (1, -1, 1). The dot product of these two normal vectors is:
(-8) * 1 + (-4) * (-1) + 4 * 1 = -4 - 4 + 4 = -4
The angle between the two planes is given by the inverse cosine of the dot product of their normal vectors divided by the product of their magnitudes. The magnitudes of the normal vectors are:
[tex]$|(-8, -4, 4)| = \sqrt{((-8)^2 + (-4)^2 + 4^2) }= \sqrt{(64 + 16 + 16)}= \sqrt96 = 8$\\$|(1, -1, 1)| = \sqrt{(1^2 + (-1)^2 + 1^2)} = \sqrt{(1 + 1 + 1)} = \sqrt3$[/tex]
So, the angle between the two planes is given by:
[tex]angle =cos\frac{-4}{8 \ * \sqrt3}\\= cos(\frac{-1}3)\\= cos(-0.33)\\= 118.47 \ degrees\\[/tex]
Note that this angle is the acute angle between the planes, which is the angle between the planes when the planes intersect and the angle between the planes is less than 90 degrees. If the angle between the planes was greater than 90 degrees, the angle returned by the arccos function would be the obtuse angle between the planes, which is the complement of the acute angle (i.e., 180 degrees - the acute angle).
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