Find the radius and the centre of the circular section of the sphere | r| = 4 cut off by the plane r.(2i-j+4k) = 3.
Ni malum
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To find the radius and center of the circular section of the sphere |r| = 4 cut off by the plane r.(2i-j+4k) = 3, we can follow these steps:
Step 1: Rewrite the equation of the plane in vector form:
r.(2i-j+4k) = 3
Step 2: Rewrite the equation of the plane in scalar form:
2x - y + 4z = 3
Step 3: Determine the center of the circular section:
The center of the circular section is obtained by finding the point where the plane intersects the origin. To find this point, set x, y, and z to 0 in the scalar equation:
2(0) - (0) + 4(0) = 3
0 - 0 + 0 = 3
0 = 3
Since 0 ≠ 3, there is no point of intersection at the origin, which means the center of the circular section is not at the origin (0, 0, 0).
Step 4: Determine the radius of the circular section:
The radius of the circular section can be found by substituting the coordinates (x, y, z) = (0, 0, 0) into the equation |r| = 4:
|0, 0, 0| = 4
√(0^2 + 0^2 + 0^2) = 4
0 ≠ 4
Since 0 ≠ 4, the radius of the circular section is not 4.
In conclusion, based on the given equation of the plane and the radius condition, there doesn't seem to be a circular section of the sphere |r| = 4 cut off by the plane r.(2i-j+4k) = 3.
To find the radius and center of the circular section of the sphere |r| = 4 cut off by the plane r.(2i-j+4k) = 3, we need to work with the equation of the sphere and the equation of the plane.
The equation of the sphere |r| = 4 represents a sphere centered at the origin (0,0,0) with a radius of 4 units.
To address the equation of the plane, let's start by rewriting it in vector form:
r.(2i-j+4k) = 3
Expanding this equation, we get:
2x - y + 4z = 3
Now, let's find the point of intersection between the sphere and the plane. We can substitute the equation of the plane into the equation of the sphere:
2x - y + 4z = 3 [equation of the plane]
|<x, y, z>| = 4 [equation of the sphere]
Substituting, we get:
2x - y + 4z = 3
sqrt(x^2 + y^2 + z^2) = 4
To solve this system of equations, we can use the method of substitution or elimination. Let's use the method of substitution:
From the equation of the plane, we have:
y = 2x + 4z - 3
Substituting this value into the equation of the sphere, we get:
sqrt(x^2 + (2x + 4z - 3)^2 + z^2) = 4
Squaring both sides, we have:
x^2 + (2x + 4z - 3)^2 + z^2 = 16
Expanding and simplifying the equation, we get:
5x^2 + 20xz + 20z^2 - 20x - 24z = 0
x^2 + 4xz + 4z^2 - 4x - 4z = 0
Rearranging the terms, we have:
x^2 + 4xz - 4x + 4z^2 - 4z = 0
Now, we have an equation that represents a circle on the xz-plane. We can complete the square to find the center and radius of this circle.
To complete the square, we need to group the x-terms and z-terms together:
(x^2 - 4x) + (4xz - 4z) + 4z^2 = 0
Now, let's complete the square for the x-terms:
(x^2 - 4x + 4) + (4xz - 4z) + 4z^2 = 4
(x - 2)^2 + (4xz - 4z) + 4z^2 = 4
Next, complete the square for the z-terms:
(x - 2)^2 + [4(z^2 - z)] + 4z^2 = 4
(x - 2)^2 + 4(z^2 - z + 1/4) = 4 + 1
(x - 2)^2 + 4(z - 1/2)^2 = 5
Now, the equation represents a circle in the form:
(x - h)^2 + (z - k)^2 = r^2
Comparing the equations, we have:
(x - 2)^2 + 4(z - 1/2)^2 = 5
This implies that the center of the circular section is at (2, 1/2, 0) and the radius is sqrt(5).
Therefore, the radius and the center of the circular section of the sphere |r| = 4 cut off by the plane r.(2i-j+4k) = 3 are sqrt(5) and (2, 1/2, 0), respectively.