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The equation of a plane containing the line of intersection of the planes 2x - y - 4 = 0 and y + 2z - 4 = 0 and passing through the point (1, 1, 0) is:
- a)x + 3y + z = 4
- b)x - y - z = 0
- c)x - 3y - 2z = -2
- d)2x - z = 2
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
The equation of a plane containing the line of intersection of the pla...
The required plane is:
(2x - y - 4) + λ(y + 2z - 4) = 0 ... (1)
It passes through (1, 1, 0).
⇒(2−1−4) +λ(1 − 4) = 0
⇒ −3 − 3λ = 0 ⇒ λ = −1
Substitute the value of λ=−1 in equation (1).
The required equation of plane is:
x−y−z=0
(2x - y - 4) + λ(y + 2z - 4) = 0 ... (1)
It passes through (1, 1, 0).
⇒(2−1−4) +λ(1 − 4) = 0
⇒ −3 − 3λ = 0 ⇒ λ = −1
Substitute the value of λ=−1 in equation (1).
The required equation of plane is:
x−y−z=0
Most Upvoted Answer
The equation of a plane containing the line of intersection of the pla...
Given Information:
- The equation of the first plane is 2x - y - 4 = 0
- The equation of the second plane is y + 2z - 4 = 0
- The point (1, 1, 0) lies on the plane we need to find
Step-by-Step Solution:
Finding the Normal Vector:
1. To find the normal vector to the plane containing the line of intersection of the given planes, we take the cross product of the normal vectors of the two planes.
2. The normal vectors are (2, -1, 0) and (0, 1, 2) for the given planes.
3. The cross product of these vectors is (2, -4, 3).
Forming the Equation of the Plane:
4. Using the point (1, 1, 0) and the normal vector (2, -4, 3), we can form the equation of the plane in point-normal form.
5. The equation becomes 2(x-1) - 4(y-1) + 3z = 0.
6. Simplifying the equation gives us 2x - 4y + 3z - 2 = 0.
7. Rearranging the terms, we get 2x - 4y + 3z = 2.
8. Dividing the entire equation by 2, we finally get x - 2y + 1.5z = 1.
Comparing with Given Options:
9. Comparing the equation obtained with the given options, we see that the closest match is option 'B': x - y - z = 0.
Therefore, the correct equation of the plane containing the line of intersection of the given planes and passing through the point (1, 1, 0) is x - y - z = 0.
- The equation of the first plane is 2x - y - 4 = 0
- The equation of the second plane is y + 2z - 4 = 0
- The point (1, 1, 0) lies on the plane we need to find
Step-by-Step Solution:
Finding the Normal Vector:
1. To find the normal vector to the plane containing the line of intersection of the given planes, we take the cross product of the normal vectors of the two planes.
2. The normal vectors are (2, -1, 0) and (0, 1, 2) for the given planes.
3. The cross product of these vectors is (2, -4, 3).
Forming the Equation of the Plane:
4. Using the point (1, 1, 0) and the normal vector (2, -4, 3), we can form the equation of the plane in point-normal form.
5. The equation becomes 2(x-1) - 4(y-1) + 3z = 0.
6. Simplifying the equation gives us 2x - 4y + 3z - 2 = 0.
7. Rearranging the terms, we get 2x - 4y + 3z = 2.
8. Dividing the entire equation by 2, we finally get x - 2y + 1.5z = 1.
Comparing with Given Options:
9. Comparing the equation obtained with the given options, we see that the closest match is option 'B': x - y - z = 0.
Therefore, the correct equation of the plane containing the line of intersection of the given planes and passing through the point (1, 1, 0) is x - y - z = 0.
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The equation of a plane containing the line of intersection of the planes 2x - y - 4 = 0 and y + 2z - 4 = 0 and passing through the point (1, 1, 0) is:a)x + 3y + z = 4b)x - y - z = 0c)x - 3y - 2z = -2d)2x - z = 2Correct answer is option 'B'. Can you explain this answer?
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The equation of a plane containing the line of intersection of the planes 2x - y - 4 = 0 and y + 2z - 4 = 0 and passing through the point (1, 1, 0) is:a)x + 3y + z = 4b)x - y - z = 0c)x - 3y - 2z = -2d)2x - z = 2Correct answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The equation of a plane containing the line of intersection of the planes 2x - y - 4 = 0 and y + 2z - 4 = 0 and passing through the point (1, 1, 0) is:a)x + 3y + z = 4b)x - y - z = 0c)x - 3y - 2z = -2d)2x - z = 2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation of a plane containing the line of intersection of the planes 2x - y - 4 = 0 and y + 2z - 4 = 0 and passing through the point (1, 1, 0) is:a)x + 3y + z = 4b)x - y - z = 0c)x - 3y - 2z = -2d)2x - z = 2Correct answer is option 'B'. Can you explain this answer?.
The equation of a plane containing the line of intersection of the planes 2x - y - 4 = 0 and y + 2z - 4 = 0 and passing through the point (1, 1, 0) is:a)x + 3y + z = 4b)x - y - z = 0c)x - 3y - 2z = -2d)2x - z = 2Correct answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The equation of a plane containing the line of intersection of the planes 2x - y - 4 = 0 and y + 2z - 4 = 0 and passing through the point (1, 1, 0) is:a)x + 3y + z = 4b)x - y - z = 0c)x - 3y - 2z = -2d)2x - z = 2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation of a plane containing the line of intersection of the planes 2x - y - 4 = 0 and y + 2z - 4 = 0 and passing through the point (1, 1, 0) is:a)x + 3y + z = 4b)x - y - z = 0c)x - 3y - 2z = -2d)2x - z = 2Correct answer is option 'B'. Can you explain this answer?.
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