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Let the points on the plane P be equidistant from the points (-4, 2, 1) and (2, -2, 3). Then the acute angle between the plane P and the plane 2x + y + 3z = 1 is:
- a)π/6
- b)π/4
- c)π/3
- d)5π/12
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
Let the points on the plane P be equidistant from the points (-4, 2, 1...
Let P1(x, y, z) be any point on plane P.
Then (x + 4)2 + (y - 2)2 + (z - 1)2 = (x - 2)2 + (y + 2)2 + (z - 3)2
⇒ 12x − 8y + 4z + 4 = 0
⇒ 3x - 2y + z + 1 = 0
And P2 : 2x + y + 3z = 1
∴ Angle between P1 and P2
Then (x + 4)2 + (y - 2)2 + (z - 1)2 = (x - 2)2 + (y + 2)2 + (z - 3)2
⇒ 12x − 8y + 4z + 4 = 0
⇒ 3x - 2y + z + 1 = 0
And P2 : 2x + y + 3z = 1
∴ Angle between P1 and P2
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Community Answer
Let the points on the plane P be equidistant from the points (-4, 2, 1...
Let's find the equation of the plane P first. The distance between a point (x, y, z) on the plane P and the point (-4, 2, 1) is equal to the distance between the point (x, y, z) and the point (2, -2, 3). This can be written as:
√[(x - (-4))^2 + (y - 2)^2 + (z - 1)^2] = √[(x - 2)^2 + (y - (-2))^2 + (z - 3)^2]
Simplifying this equation, we get:
(x + 4)^2 + (y - 2)^2 + (z - 1)^2 = (x - 2)^2 + (y + 2)^2 + (z - 3)^2
Expanding both sides, we get:
x^2 + 8x + 16 + y^2 - 4y + 4 + z^2 - 2z + 1 = x^2 - 4x + 4 + y^2 + 4y + 4 + z^2 - 6z + 9
Simplifying further, we get:
12x - 8y + 8z - 24 = 0
Dividing by 4, we get:
3x - 2y + 2z - 6 = 0
So, the equation of plane P is 3x - 2y + 2z - 6 = 0.
Now, let's find the equation of the plane Q, which is 2x + y + 3z = 1.
The acute angle between two planes can be found using the formula:
cosθ = (a1 * a2 + b1 * b2 + c1 * c2) / (√(a1^2 + b1^2 + c1^2) * √(a2^2 + b2^2 + c2^2))
where (a1, b1, c1) and (a2, b2, c2) are the normal vectors of the planes P and Q respectively.
The normal vector of plane P is (3, -2, 2) and the normal vector of plane Q is (2, 1, 3).
Calculating the dot product of the two normal vectors, we get:
(3 * 2) + (-2 * 1) + (2 * 3) = 6 - 2 + 6 = 10
Calculating the magnitudes of the normal vectors, we get:
√(3^2 + (-2)^2 + 2^2) = √(9 + 4 + 4) = √17
√(2^2 + 1^2 + 3^2) = √(4 + 1 + 9) = √14
Substituting these values into the formula for cosθ, we get:
cosθ = 10 / (√17 * √14)
Using a calculator, we find that cosθ ≈ 0.775.
To find the acute angle θ, we take the inverse cosine of 0.775:
θ ≈ cos^(-1)(0
√[(x - (-4))^2 + (y - 2)^2 + (z - 1)^2] = √[(x - 2)^2 + (y - (-2))^2 + (z - 3)^2]
Simplifying this equation, we get:
(x + 4)^2 + (y - 2)^2 + (z - 1)^2 = (x - 2)^2 + (y + 2)^2 + (z - 3)^2
Expanding both sides, we get:
x^2 + 8x + 16 + y^2 - 4y + 4 + z^2 - 2z + 1 = x^2 - 4x + 4 + y^2 + 4y + 4 + z^2 - 6z + 9
Simplifying further, we get:
12x - 8y + 8z - 24 = 0
Dividing by 4, we get:
3x - 2y + 2z - 6 = 0
So, the equation of plane P is 3x - 2y + 2z - 6 = 0.
Now, let's find the equation of the plane Q, which is 2x + y + 3z = 1.
The acute angle between two planes can be found using the formula:
cosθ = (a1 * a2 + b1 * b2 + c1 * c2) / (√(a1^2 + b1^2 + c1^2) * √(a2^2 + b2^2 + c2^2))
where (a1, b1, c1) and (a2, b2, c2) are the normal vectors of the planes P and Q respectively.
The normal vector of plane P is (3, -2, 2) and the normal vector of plane Q is (2, 1, 3).
Calculating the dot product of the two normal vectors, we get:
(3 * 2) + (-2 * 1) + (2 * 3) = 6 - 2 + 6 = 10
Calculating the magnitudes of the normal vectors, we get:
√(3^2 + (-2)^2 + 2^2) = √(9 + 4 + 4) = √17
√(2^2 + 1^2 + 3^2) = √(4 + 1 + 9) = √14
Substituting these values into the formula for cosθ, we get:
cosθ = 10 / (√17 * √14)
Using a calculator, we find that cosθ ≈ 0.775.
To find the acute angle θ, we take the inverse cosine of 0.775:
θ ≈ cos^(-1)(0
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Let the points on the plane P be equidistant from the points (-4, 2, 1) and (2, -2, 3). Then the acute angle between the plane P and the plane 2x + y + 3z = 1 is:a)π/6b)π/4c)π/3d)5π/12Correct answer is option 'C'. Can you explain this answer?
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