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)
how do you solve inequality problems?
example 16> 20 + g
Sure! I can help you with these explanations.
a) To determine the angle theta between the given planes, we need to find the normal vectors of both planes. The normal vector of a plane is the vector that is perpendicular (orthogonal) to the plane itself.
For the plane 2x+3y-z+9 = 0, we can identify the coefficients of x, y, and z as the components of the normal vector. Hence, the normal vector of this plane is N1 = [2, 3, -1].
Similarly, for the plane x+2y+4 = 0, the normal vector N2 = [1, 2, 0].
Now, we can use the dot product formula to find the angle between the normal vectors:
cos(theta) = (N1 · N2) / (|N1| * |N2|)
where · represents the dot product, and |N1| and |N2| represent the magnitudes of the vectors N1 and N2, respectively.
N1 · N2 = (2 * 1) + (3 * 2) + (-1 * 0) = 7
|N1| = sqrt(2^2 + 3^2 + (-1)^2) = sqrt(14)
|N2| = sqrt(1^2 + 2^2 + 0^2) = sqrt(5)
Substituting these values into the formula, we can find cos(theta) = 7 / (sqrt(14) * sqrt(5)).
To find the angle theta, we take the inverse cosine (arccos) of this value: theta = arccos(7 / (sqrt(14) * sqrt(5))).
This will give you the angle theta between the two planes in the range of 0 to 90 degrees.
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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 their directional vectors.
The directional vector of the line is given by [0, 1, 2].
The directional vector of the plane is the coefficients of the variables x, y, and z, which are [2, -10, 5].
Now, if two vectors are parallel, their directions are proportional. In other words, if we multiply the components of the line's directional vector by a scalar, it should give us the plane's directional vector.
Let's check:
[0, 1, 2] = k[2, -10, 5]
This gives us the equations 0 = 2k, 1 = -10k, and 2 = 5k.
From the first equation, we can see that k = 0. This means that the line and the plane have proportional directional vectors, and hence, they are parallel.
To check if the line is below the plane or not, we can substitute the coordinates of a point on the line into the equation of the plane. Let's choose t = 0 as an example:
For x = 0, y = 0, and z = 0, the equation of the plane becomes:
2(0) - 10(0) + 5(0) - 1 = -1.
Since -1 is below 0, we can conclude that the line x=0, y=t, z=2t is both parallel to and below the plane 2x-10y+5z-1 = 0.
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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 definition of equidistant points.
The distance between a point P(x, y, z) and point A(1, 2, 3) is given by the formula:
distance_PA = sqrt((x - 1)^2 + (y - 2)^2 + (z - 3)^2).
Similarly, the distance between point P(x, y, z) and point B(4, 0, 1) is:
distance_PB = sqrt((x - 4)^2 + (y - 0)^2 + (z - 1)^2).
For P(x, y, z) to be equidistant from A and B, we require distance_PA = distance_PB.
Therefore, we need to equate the above two distance equations and solve for the values of x, y, and z.
So, we have:
sqrt((x - 1)^2 + (y - 2)^2 + (z - 3)^2) = sqrt((x - 4)^2 + y^2 + (z - 1)^2).
Squaring both sides of the equation to eliminate the square roots:
(x - 1)^2 + (y - 2)^2 + (z - 3)^2 = (x - 4)^2 + y^2 + (z - 1)^2.
Expanding and rearranging the terms:
x^2 - 2x + 1 + y^2 - 4y + 4 + z^2 - 6z + 9 = x^2 - 8x + 16 + y^2 + z^2 - 2z + 1.
Canceling out like terms, we get:
-2x - 4y - 8z + 14 = 0.
So, the 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.
I hope these explanations help! Let me know if you have any more questions.