Someone please explain these!
The angle between two planes is defined as the angle between their normals. Determine the angle theta between 0 and 90 degrees between the given planes.
a) 2x+3y-z+9 = 0 and x+2y+4 = 0
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If the positive z-axiz points up, show that the line x=0, y=t, z=2t is parallel to and below the plane 2x-10y+5z-1 = 0.
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Find an equation for the set of points P(x,y,z) that are equidistant from the points A(1,2,3) and B(4,0,1)
what is the answer to 2+3+6+7+9+n=100?
n=73
73
Sure, I'd be happy to explain these problems to you.
a) To determine the angle theta between two planes, we first need to find the normals of the planes.
The first plane is given by the equation 2x+3y-z+9 = 0. To find its normal, we can look at the coefficients of x, y, and z. The normal vector can be represented as (2, 3, -1).
The second plane is given by the equation x+2y+4 = 0. Similarly, its normal vector is (1, 2, 0).
Now that we have the normals of the planes, we can find the angle between them using the dot product formula:
cos(theta) = (n1 . n2) / (||n1|| ||n2||)
Here, n1 and n2 are the normal vectors of the planes, "." represents the dot product, and ||n1|| and ||n2|| represent the magnitudes of the vectors.
Calculating the dot product and magnitudes:
n1 . n2 = (2 * 1) + (3 * 2) + (-1 * 0) = 8
||n1|| = sqrt((2^2) + (3^2) + (-1^2)) = sqrt(14)
||n2|| = sqrt((1^2) + (2^2) + (0^2)) = sqrt(5)
Plugging these values into the equation:
cos(theta) = (8) / (sqrt(14) * sqrt(5))
Now, to find theta, we can take the inverse cosine (arccos) of this value:
theta = arccos((8) / (sqrt(14) * sqrt(5)))
This will give you the angle theta between the two planes.
b) To show that the line x=0, y=t, z=2t is parallel to and below the plane 2x-10y+5z-1 = 0, we can compare the direction vector of the line to the normal vector of the plane.
The direction vector of the line is (0, 1, 2) since y=t and z=2t.
The normal vector of the plane is (2, -10, 5) since it can be read from the coefficients of x, y, and z.
If the direction vector of the line is parallel to the normal vector of the plane, then their dot product should be zero (perpendicular to each other).
Calculating the dot product:
(0, 1, 2) . (2, -10, 5) = (0*2) + (1*-10) + (2*5) = -10 + 10 = 0
Since the dot product is zero, we can conclude that the line x=0, y=t, z=2t is parallel to the plane 2x-10y+5z-1 = 0.
To show that the line is below the plane, we can substitute the coordinates (0, t, 2t) into the equation of the plane and check if the result is less than zero.
For example, substituting x=0, y=t, and z=2t into the equation 2x-10y+5z-1 = 0:
2(0) - 10(t) + 5(2t) - 1 = -10t + 10t - 1 = -1
Since the result is less than zero (-1 < 0), we can conclude that the line is below the plane.
c) To find an equation for the set of points P(x,y,z) that are equidistant from the points A(1,2,3) and B(4,0,1), we can use the distance formula.
The distance between a point (x,y,z) and point A is given by:
distance = sqrt((x - 1)^2 + (y - 2)^2 + (z - 3)^2)
Similarly, the distance between the point (x,y,z) and point B is given by:
distance = sqrt((x - 4)^2 + (y - 0)^2 + (z - 1)^2)
Since we want the point P to be equidistant from A and B, we can set these two expressions equal to each other:
sqrt((x - 1)^2 + (y - 2)^2 + (z - 3)^2) = sqrt((x - 4)^2 + (y - 0)^2 + (z - 1)^2)
To simplify this equation, we can square both sides:
(x - 1)^2 + (y - 2)^2 + (z - 3)^2 = (x - 4)^2 + (y - 0)^2 + (z - 1)^2
Expanding and combining like terms:
(x^2 - 2x + 1) + (y^2 - 4y + 4) + (z^2 - 6z + 9) = (x^2 - 8x + 16) + (y^2 + 0) + (z^2 - 2z + 1)
Rearranging terms:
-2x - 4y - 8z + 14 = 0
Therefore, an equation for the set of points P(x,y,z) that are equidistant from A(1,2,3) and B(4,0,1) is -2x - 4y - 8z + 14 = 0.