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A plane P passes through the line of intersection of the planes 2x - y + 3z = 2, x + y - z = 1 and the point (1 ,0 ,1 ).
Q. What are the direction ratios of the line of intersection of the given planes?
- a)( 2 ,- 5 ,- 3 )
- b)(1 ,-5 ,-3 )
- c)(2,5,3)
- d)(1,3,5)
Correct answer is option 'A'. Can you explain this answer?
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A plane P passes through the line of intersection of the planes 2x - y...
Hence direction ratios ofthe line of intersection of given plane <2,-5,-3 >
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A plane P passes through the line of intersection of the planes 2x - y...
To find the direction ratios of the line of intersection of the given planes, we need to find the normal vectors of both planes and then take their cross product.
Given planes:
1. 2x - y + 3z = 2
2. x + y - z = 1
Finding the normal vectors of the planes:
The coefficients of x, y, and z in the equations of the planes give us the normal vectors.
1. Plane 1: 2x - y + 3z = 2
The coefficients of x, y, and z are 2, -1, and 3 respectively.
So, the normal vector of plane 1 is (2, -1, 3).
2. Plane 2: x + y - z = 1
The coefficients of x, y, and z are 1, 1, and -1 respectively.
So, the normal vector of plane 2 is (1, 1, -1).
Taking the cross product of the normal vectors:
To find the direction ratios of the line of intersection, we take the cross product of the normal vectors.
Cross product = (2, -1, 3) x (1, 1, -1)
Using the formula for cross product: (a, b, c) x (d, e, f) = (bf - ce, cd - af, ae - bd)
= ((-1)(-1) - (3)(1), (2)(-1) - (3)(1), (2)(1) - (-1)(-1))
= (2 + 3, -2 - 3, 2 - 1)
= (5, -5, 1)
Therefore, the direction ratios of the line of intersection of the given planes are (5, -5, 1).
Answer: Option A) (5, -5, 1)
Given planes:
1. 2x - y + 3z = 2
2. x + y - z = 1
Finding the normal vectors of the planes:
The coefficients of x, y, and z in the equations of the planes give us the normal vectors.
1. Plane 1: 2x - y + 3z = 2
The coefficients of x, y, and z are 2, -1, and 3 respectively.
So, the normal vector of plane 1 is (2, -1, 3).
2. Plane 2: x + y - z = 1
The coefficients of x, y, and z are 1, 1, and -1 respectively.
So, the normal vector of plane 2 is (1, 1, -1).
Taking the cross product of the normal vectors:
To find the direction ratios of the line of intersection, we take the cross product of the normal vectors.
Cross product = (2, -1, 3) x (1, 1, -1)
Using the formula for cross product: (a, b, c) x (d, e, f) = (bf - ce, cd - af, ae - bd)
= ((-1)(-1) - (3)(1), (2)(-1) - (3)(1), (2)(1) - (-1)(-1))
= (2 + 3, -2 - 3, 2 - 1)
= (5, -5, 1)
Therefore, the direction ratios of the line of intersection of the given planes are (5, -5, 1).
Answer: Option A) (5, -5, 1)
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A plane P passes through the line of intersection of the planes 2x - y + 3z = 2, x + y - z = 1 and the point (1 ,0 ,1 ).Q. What are the direction ratios of the line of intersection of the given planes?a)( 2 ,- 5 ,- 3 )b)(1 ,-5 ,-3 )c)(2,5,3)d)(1,3,5)Correct answer is option 'A'. Can you explain this answer?
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