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The angle between the plane 2x – y + z = 6 and a plane perpendicular to the planes x + y + 2z = 7 and x – y = 3 is
- a)π/4
- b)π/3
- c)π/6
- d)π/2
Correct answer is option 'D'. Can you explain this answer?
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The angle between the plane 2x – y + z = 6 and aplane perpendicu...
We are given the equation of two planes:
2x - y + z = 4 ... (1)
4x - 2y + 3z = 6 ... (2)
To find the angle between them, we first need to find the normal vectors of each plane. The normal vector of a plane is the vector perpendicular to the plane, and can be found by taking the coefficients of x, y, and z from the equation of the plane.
For plane (1), the normal vector is <2, -1,="" 1="">, and for plane (2), the normal vector is <4, -2,="" 3="">.
Next, we can use the dot product formula to find the angle between the normal vectors (and therefore the angle between the planes):
cos(theta) = (n1 dot n2) / (|n1| * |n2|)
where n1 and n2 are the normal vectors of the two planes.
Plugging in the values, we get:
cos(theta) = ((2)(4) + (-1)(-2) + (1)(3)) / (√(2^2 + (-1)^2 + 1^2) * √(4^2 + (-2)^2 + 3^2))
= 13 / (√6 * √29)
≈ 0.895
To find the angle itself, we take the inverse cosine of cos(theta):
theta = acos(0.895)
≈ 26.5 degrees
Therefore, the angle between the planes 2x - y + z = 4 and 4x - 2y + 3z = 6 is approximately 26.5 degrees.
2x - y + z = 4 ... (1)
4x - 2y + 3z = 6 ... (2)
To find the angle between them, we first need to find the normal vectors of each plane. The normal vector of a plane is the vector perpendicular to the plane, and can be found by taking the coefficients of x, y, and z from the equation of the plane.
For plane (1), the normal vector is <2, -1,="" 1="">, and for plane (2), the normal vector is <4, -2,="" 3="">.
Next, we can use the dot product formula to find the angle between the normal vectors (and therefore the angle between the planes):
cos(theta) = (n1 dot n2) / (|n1| * |n2|)
where n1 and n2 are the normal vectors of the two planes.
Plugging in the values, we get:
cos(theta) = ((2)(4) + (-1)(-2) + (1)(3)) / (√(2^2 + (-1)^2 + 1^2) * √(4^2 + (-2)^2 + 3^2))
= 13 / (√6 * √29)
≈ 0.895
To find the angle itself, we take the inverse cosine of cos(theta):
theta = acos(0.895)
≈ 26.5 degrees
Therefore, the angle between the planes 2x - y + z = 4 and 4x - 2y + 3z = 6 is approximately 26.5 degrees.
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The angle between the plane 2x – y + z = 6 and aplane perpendicular to the planes x + y + 2z = 7 and x – y = 3 isa)π/4b)π/3c)π/6d)π/2Correct answer is option 'D'. Can you explain this answer?
Question Description
The angle between the plane 2x – y + z = 6 and aplane perpendicular to the planes x + y + 2z = 7 and x – y = 3 isa)π/4b)π/3c)π/6d)π/2Correct answer is option 'D'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about The angle between the plane 2x – y + z = 6 and aplane perpendicular to the planes x + y + 2z = 7 and x – y = 3 isa)π/4b)π/3c)π/6d)π/2Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The angle between the plane 2x – y + z = 6 and aplane perpendicular to the planes x + y + 2z = 7 and x – y = 3 isa)π/4b)π/3c)π/6d)π/2Correct answer is option 'D'. Can you explain this answer?.
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