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The shortest distance from the point (1, 2, -1) to the surface of the sphere x2 + y2 + z2 = 24 is
- a)3√6 units
- b)√6 units
- c)2√6 units
- d)2 units
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
The shortest distance from the point (1, 2, -1) to the surface of the ...
The given point (1, 2, -1) lies inside the sphere.
∵ Centre of sphere is (0, 0, 0), then the shortest distance between (1, 2, -1) and surface of sphere
∵ Centre of sphere is (0, 0, 0), then the shortest distance between (1, 2, -1) and surface of sphere
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The shortest distance from the point (1, 2, -1) to the surface of the ...
To find the shortest distance from a point to the surface of a sphere, we can use the formula:
d = |(ax + by + cz + d)| / √(a^2 + b^2 + c^2)
where (a, b, c) is the normal vector of the plane containing the point and the center of the sphere, and d is the distance from the origin to the plane.
First, let's find the center and radius of the sphere x^2 + y^2 + z^2 = 24. Since there is no constant term on the right side of the equation, we can conclude that the center of the sphere is at the origin (0, 0, 0) and the radius is √24 = 2√6.
Next, let's find the equation of the plane containing the point (1, 2, -1) and the center of the sphere (0, 0, 0). The normal vector of the plane can be obtained by subtracting the coordinates of the center from the coordinates of the point:
(a, b, c) = (1, 2, -1) - (0, 0, 0) = (1, 2, -1)
Now, we need to find the distance from the origin to the plane. We can use the formula:
d = |ax + by + cz| / √(a^2 + b^2 + c^2)
Plugging in the values, we get:
d = |(1)(0) + (2)(0) + (-1)(0)| / √(1^2 + 2^2 + (-1)^2) = 0 / √6 = 0
Finally, we can find the shortest distance from the point (1, 2, -1) to the surface of the sphere using the formula:
d = |(ax + by + cz + d)| / √(a^2 + b^2 + c^2)
Plugging in the values, we get:
d = |(1)(1) + (2)(2) + (-1)(-1) + 0| / √(1^2 + 2^2 + (-1)^2) = √6 / √6 = 1
Therefore, the shortest distance from the point (1, 2, -1) to the surface of the sphere x^2 + y^2 + z^2 = 24 is 1.
d = |(ax + by + cz + d)| / √(a^2 + b^2 + c^2)
where (a, b, c) is the normal vector of the plane containing the point and the center of the sphere, and d is the distance from the origin to the plane.
First, let's find the center and radius of the sphere x^2 + y^2 + z^2 = 24. Since there is no constant term on the right side of the equation, we can conclude that the center of the sphere is at the origin (0, 0, 0) and the radius is √24 = 2√6.
Next, let's find the equation of the plane containing the point (1, 2, -1) and the center of the sphere (0, 0, 0). The normal vector of the plane can be obtained by subtracting the coordinates of the center from the coordinates of the point:
(a, b, c) = (1, 2, -1) - (0, 0, 0) = (1, 2, -1)
Now, we need to find the distance from the origin to the plane. We can use the formula:
d = |ax + by + cz| / √(a^2 + b^2 + c^2)
Plugging in the values, we get:
d = |(1)(0) + (2)(0) + (-1)(0)| / √(1^2 + 2^2 + (-1)^2) = 0 / √6 = 0
Finally, we can find the shortest distance from the point (1, 2, -1) to the surface of the sphere using the formula:
d = |(ax + by + cz + d)| / √(a^2 + b^2 + c^2)
Plugging in the values, we get:
d = |(1)(1) + (2)(2) + (-1)(-1) + 0| / √(1^2 + 2^2 + (-1)^2) = √6 / √6 = 1
Therefore, the shortest distance from the point (1, 2, -1) to the surface of the sphere x^2 + y^2 + z^2 = 24 is 1.
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The shortest distance from the point (1, 2, -1) to the surface of the sphere x2 + y2 + z2 = 24 isa)3√6 unitsb)√6 unitsc)2√6 unitsd)2 unitsCorrect answer is option 'B'. Can you explain this answer?
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