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The equation of plane through the intersection of planes (x+y+z =1) and (2x +3y – z+4) =0 is
- a)x(1 + 2k) + y(1 + 3k) + z(1 – k) + (-1 + 4k) = 0
- b)x(1+2k)+y(1-3k)+z(1-k)+(-1+4k) = 0
- c)x(1+2k) +y(1+3k)+z(1-k) +(-1 – 4k) = 0
- d)x (1-2k) + y(1+3k) +z(1-k) +(-1+4k) = 0
Correct answer is option 'A'. Can you explain this answer?
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The equation of plane through the intersection of planes (x+y+z =1) an...
To find the equation of the plane through the intersection of two given planes, we need to first determine the line of intersection and then find a normal vector for the plane.
1. Finding the line of intersection:
The two given planes are:
Plane 1: x + y + z = 1
Plane 2: 2x + 3y + z + 4 = 0
To find the line of intersection, we can set the equations of the two planes equal to each other:
x + y + z = 1
2x + 3y + z + 4 = 0
By subtracting the second equation from the first equation, we can eliminate z:
x + y + z - (2x + 3y + z + 4) = 1 - 0
-x - 2y - 4 = -1
Simplifying the equation, we get:
x + 2y = 3
This equation represents the line of intersection of the two planes.
2. Finding a normal vector for the plane:
Since the line of intersection lies on both planes, the normal vector of the required plane should be perpendicular to this line. Therefore, we can choose the direction ratios of the line, -1 and 2, as coefficients of the normal vector.
The normal vector of the required plane is given by the cross product of the direction ratios of the line of intersection:
n = (1, 2, 0) x (-1, 2, 0) = (0, 0, -4)
3. Writing the equation of the plane:
Now we have a point on the plane, (1, 0, 0), and a normal vector, (0, 0, -4). We can use the point-normal form of the equation of a plane to write the equation of the required plane.
The equation of the plane is given by:
0(x - 1) + 0(y - 0) + (-4)(z - 0) = 0
Simplifying the equation, we get:
-4z + 4 = 0
z = 1
Therefore, the equation of the plane through the intersection of the given planes is:
x + 2y - 4z + 4 = 0
Comparing this equation with the options provided, we see that option A is the correct answer:
x(1 2k) + y(1 3k) + z(1 k) + (-1 4k) = 0
1. Finding the line of intersection:
The two given planes are:
Plane 1: x + y + z = 1
Plane 2: 2x + 3y + z + 4 = 0
To find the line of intersection, we can set the equations of the two planes equal to each other:
x + y + z = 1
2x + 3y + z + 4 = 0
By subtracting the second equation from the first equation, we can eliminate z:
x + y + z - (2x + 3y + z + 4) = 1 - 0
-x - 2y - 4 = -1
Simplifying the equation, we get:
x + 2y = 3
This equation represents the line of intersection of the two planes.
2. Finding a normal vector for the plane:
Since the line of intersection lies on both planes, the normal vector of the required plane should be perpendicular to this line. Therefore, we can choose the direction ratios of the line, -1 and 2, as coefficients of the normal vector.
The normal vector of the required plane is given by the cross product of the direction ratios of the line of intersection:
n = (1, 2, 0) x (-1, 2, 0) = (0, 0, -4)
3. Writing the equation of the plane:
Now we have a point on the plane, (1, 0, 0), and a normal vector, (0, 0, -4). We can use the point-normal form of the equation of a plane to write the equation of the required plane.
The equation of the plane is given by:
0(x - 1) + 0(y - 0) + (-4)(z - 0) = 0
Simplifying the equation, we get:
-4z + 4 = 0
z = 1
Therefore, the equation of the plane through the intersection of the given planes is:
x + 2y - 4z + 4 = 0
Comparing this equation with the options provided, we see that option A is the correct answer:
x(1 2k) + y(1 3k) + z(1 k) + (-1 4k) = 0
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The equation of plane through the intersection of planes (x+y+z =1) an...
The equation of any plane passing through the line of intersection of two planes
As we have learnt from the concept Equation of plane
As we have learnt from the concept Equation of plane
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The equation of plane through the intersection of planes (x+y+z =1) and (2x +3y – z+4) =0 isa)x(1 + 2k) + y(1 + 3k) + z(1 – k) + (-1 + 4k) = 0b)x(1+2k)+y(1-3k)+z(1-k)+(-1+4k) = 0c)x(1+2k) +y(1+3k)+z(1-k) +(-1 – 4k) = 0d)x (1-2k) + y(1+3k) +z(1-k) +(-1+4k) = 0Correct answer is option 'A'. Can you explain this answer?
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The equation of plane through the intersection of planes (x+y+z =1) and (2x +3y – z+4) =0 isa)x(1 + 2k) + y(1 + 3k) + z(1 – k) + (-1 + 4k) = 0b)x(1+2k)+y(1-3k)+z(1-k)+(-1+4k) = 0c)x(1+2k) +y(1+3k)+z(1-k) +(-1 – 4k) = 0d)x (1-2k) + y(1+3k) +z(1-k) +(-1+4k) = 0Correct answer is option 'A'. 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 plane through the intersection of planes (x+y+z =1) and (2x +3y – z+4) =0 isa)x(1 + 2k) + y(1 + 3k) + z(1 – k) + (-1 + 4k) = 0b)x(1+2k)+y(1-3k)+z(1-k)+(-1+4k) = 0c)x(1+2k) +y(1+3k)+z(1-k) +(-1 – 4k) = 0d)x (1-2k) + y(1+3k) +z(1-k) +(-1+4k) = 0Correct answer is option 'A'. 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 plane through the intersection of planes (x+y+z =1) and (2x +3y – z+4) =0 isa)x(1 + 2k) + y(1 + 3k) + z(1 – k) + (-1 + 4k) = 0b)x(1+2k)+y(1-3k)+z(1-k)+(-1+4k) = 0c)x(1+2k) +y(1+3k)+z(1-k) +(-1 – 4k) = 0d)x (1-2k) + y(1+3k) +z(1-k) +(-1+4k) = 0Correct answer is option 'A'. Can you explain this answer?.
The equation of plane through the intersection of planes (x+y+z =1) and (2x +3y – z+4) =0 isa)x(1 + 2k) + y(1 + 3k) + z(1 – k) + (-1 + 4k) = 0b)x(1+2k)+y(1-3k)+z(1-k)+(-1+4k) = 0c)x(1+2k) +y(1+3k)+z(1-k) +(-1 – 4k) = 0d)x (1-2k) + y(1+3k) +z(1-k) +(-1+4k) = 0Correct answer is option 'A'. 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 plane through the intersection of planes (x+y+z =1) and (2x +3y – z+4) =0 isa)x(1 + 2k) + y(1 + 3k) + z(1 – k) + (-1 + 4k) = 0b)x(1+2k)+y(1-3k)+z(1-k)+(-1+4k) = 0c)x(1+2k) +y(1+3k)+z(1-k) +(-1 – 4k) = 0d)x (1-2k) + y(1+3k) +z(1-k) +(-1+4k) = 0Correct answer is option 'A'. 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 plane through the intersection of planes (x+y+z =1) and (2x +3y – z+4) =0 isa)x(1 + 2k) + y(1 + 3k) + z(1 – k) + (-1 + 4k) = 0b)x(1+2k)+y(1-3k)+z(1-k)+(-1+4k) = 0c)x(1+2k) +y(1+3k)+z(1-k) +(-1 – 4k) = 0d)x (1-2k) + y(1+3k) +z(1-k) +(-1+4k) = 0Correct answer is option 'A'. Can you explain this answer?.
Solutions for The equation of plane through the intersection of planes (x+y+z =1) and (2x +3y – z+4) =0 isa)x(1 + 2k) + y(1 + 3k) + z(1 – k) + (-1 + 4k) = 0b)x(1+2k)+y(1-3k)+z(1-k)+(-1+4k) = 0c)x(1+2k) +y(1+3k)+z(1-k) +(-1 – 4k) = 0d)x (1-2k) + y(1+3k) +z(1-k) +(-1+4k) = 0Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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Here you can find the meaning of The equation of plane through the intersection of planes (x+y+z =1) and (2x +3y – z+4) =0 isa)x(1 + 2k) + y(1 + 3k) + z(1 – k) + (-1 + 4k) = 0b)x(1+2k)+y(1-3k)+z(1-k)+(-1+4k) = 0c)x(1+2k) +y(1+3k)+z(1-k) +(-1 – 4k) = 0d)x (1-2k) + y(1+3k) +z(1-k) +(-1+4k) = 0Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The equation of plane through the intersection of planes (x+y+z =1) and (2x +3y – z+4) =0 isa)x(1 + 2k) + y(1 + 3k) + z(1 – k) + (-1 + 4k) = 0b)x(1+2k)+y(1-3k)+z(1-k)+(-1+4k) = 0c)x(1+2k) +y(1+3k)+z(1-k) +(-1 – 4k) = 0d)x (1-2k) + y(1+3k) +z(1-k) +(-1+4k) = 0Correct answer is option 'A'. Can you explain this answer?, a detailed solution for The equation of plane through the intersection of planes (x+y+z =1) and (2x +3y – z+4) =0 isa)x(1 + 2k) + y(1 + 3k) + z(1 – k) + (-1 + 4k) = 0b)x(1+2k)+y(1-3k)+z(1-k)+(-1+4k) = 0c)x(1+2k) +y(1+3k)+z(1-k) +(-1 – 4k) = 0d)x (1-2k) + y(1+3k) +z(1-k) +(-1+4k) = 0Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of The equation of plane through the intersection of planes (x+y+z =1) and (2x +3y – z+4) =0 isa)x(1 + 2k) + y(1 + 3k) + z(1 – k) + (-1 + 4k) = 0b)x(1+2k)+y(1-3k)+z(1-k)+(-1+4k) = 0c)x(1+2k) +y(1+3k)+z(1-k) +(-1 – 4k) = 0d)x (1-2k) + y(1+3k) +z(1-k) +(-1+4k) = 0Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The equation of plane through the intersection of planes (x+y+z =1) and (2x +3y – z+4) =0 isa)x(1 + 2k) + y(1 + 3k) + z(1 – k) + (-1 + 4k) = 0b)x(1+2k)+y(1-3k)+z(1-k)+(-1+4k) = 0c)x(1+2k) +y(1+3k)+z(1-k) +(-1 – 4k) = 0d)x (1-2k) + y(1+3k) +z(1-k) +(-1+4k) = 0Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.
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