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The direction ratios of the normal to the plane passing through origin and the line of intersection of the planes x + 2y + 3z = 4 and 4x + 3y + 2z = 1 are .......
- a)2, 3, 1
- b)1, 2, 3
- c)3, 1, 2
- d)3, 2, 1
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
The direction ratios of the normal to the plane passing through origi...
We have, line of intersection of the planes x + 2y+ 3z = 4 and 4x + 3y + 2z = 1
∴ Equation of plane passing through the given planes is (x +2y + 3z - 4) +λ (4x + 3y + 2z - 1) = 0
⇒ (1 + 4λ)x + (2 + 3λ)y + (3 + 2λ) + (-4 - λ) = 0
Since, a plane passing through the origin.
∴ -4 - λ = 0 ⇒ λ = -4
Now, equation of plane is
(1 - 16)x + (2 - 12)y + (3 - 8)z + 0 = 0
⇒ -15x - 10y - 5z = 0
⇒ 3x + 2y + z = 0
∴ Direction ratios of the normal to the plane are 3, 2, 1.
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The direction ratios of the normal to the plane passing through origi...
Solution:
To find the direction ratios of the normal to the plane, we need to find the normal vector of the plane.
Given planes are:
x + 2y + 3z = 4 ...(1)
4x + 3y + 2z = 1 ...(2)
Line of intersection of the planes (1) and (2) can be found by solving them simultaneously.
On solving, we get:
x = 1, y = -1, z = 1
Hence, the line of intersection of the planes is:
r = (1, -1, 1) + λ(4, 3, -2)
The direction ratios of the line of intersection are 4, 3, -2.
Now, let N be the normal vector of the plane passing through the origin and the line of intersection of the given planes.
Since the plane passes through the origin, any point on the plane can be written as λ(4, 3, -2) for some scalar λ.
Also, since the plane is perpendicular to the line of intersection, the normal vector of the plane must be perpendicular to the direction vector of the line of intersection.
Therefore, N . (4, 3, -2) = 0
This gives us:
4N1 + 3N2 - 2N3 = 0
Hence, the direction ratios of the normal vector N are:
N1 : N2 : N3 = 4 : 3 : 2
Therefore, the correct option is (D) 3, 2, 1.
To find the direction ratios of the normal to the plane, we need to find the normal vector of the plane.
Given planes are:
x + 2y + 3z = 4 ...(1)
4x + 3y + 2z = 1 ...(2)
Line of intersection of the planes (1) and (2) can be found by solving them simultaneously.
On solving, we get:
x = 1, y = -1, z = 1
Hence, the line of intersection of the planes is:
r = (1, -1, 1) + λ(4, 3, -2)
The direction ratios of the line of intersection are 4, 3, -2.
Now, let N be the normal vector of the plane passing through the origin and the line of intersection of the given planes.
Since the plane passes through the origin, any point on the plane can be written as λ(4, 3, -2) for some scalar λ.
Also, since the plane is perpendicular to the line of intersection, the normal vector of the plane must be perpendicular to the direction vector of the line of intersection.
Therefore, N . (4, 3, -2) = 0
This gives us:
4N1 + 3N2 - 2N3 = 0
Hence, the direction ratios of the normal vector N are:
N1 : N2 : N3 = 4 : 3 : 2
Therefore, the correct option is (D) 3, 2, 1.
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The direction ratios of the normal to the plane passing through origin and the line of intersection of the planes x + 2y + 3z = 4 and 4x + 3y + 2z = 1 are .......a)2, 3, 1b)1, 2, 3c)3, 1, 2d)3, 2, 1Correct answer is option 'D'. Can you explain this answer?
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The direction ratios of the normal to the plane passing through origin and the line of intersection of the planes x + 2y + 3z = 4 and 4x + 3y + 2z = 1 are .......a)2, 3, 1b)1, 2, 3c)3, 1, 2d)3, 2, 1Correct answer is option 'D'. 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 direction ratios of the normal to the plane passing through origin and the line of intersection of the planes x + 2y + 3z = 4 and 4x + 3y + 2z = 1 are .......a)2, 3, 1b)1, 2, 3c)3, 1, 2d)3, 2, 1Correct answer is option 'D'. 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 direction ratios of the normal to the plane passing through origin and the line of intersection of the planes x + 2y + 3z = 4 and 4x + 3y + 2z = 1 are .......a)2, 3, 1b)1, 2, 3c)3, 1, 2d)3, 2, 1Correct answer is option 'D'. Can you explain this answer?.
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