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The Cartesian equation of the plane perpendicular to the line (x - 1)/2 = (y - 3)/-1 = (z - 4)/2 and passing through the origin is
- a)2x - y + 2z - 7 = 0
- b)2x + y + 2z = 0
- c)2x - y - z = 0
- d)2x - y + 2z = 0
Correct answer is option 'D'. Can you explain this answer?
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The Cartesian equation of the plane perpendicular to the line (x - 1)/...
Given information:
- The line (x - 1)/2 = (y - 3)/-1 = (z - 4)/2
- The plane passes through the origin
To find the equation of the plane, we can first find the normal vector to the plane. The normal vector is perpendicular to the plane, and therefore perpendicular to any vector in the plane. We can use the direction ratios of the given line to find a vector in the plane, and then find a vector perpendicular to it.
Finding a vector in the plane:
- Let's take any two points on the line. For example, (1, 3, 4) and (3, 1, 6).
- The vector joining these two points is parallel to the line. So, a vector in the direction of the line is (3-1, 1-3, 6-4) = (2, -2, 2).
- Any point on the line can be written as (1, 3, 4) + t(2, -2, 2), where t is any real number.
- Let's take t = 0 to get a point on the line: (1, 3, 4) + 0(2, -2, 2) = (1, 3, 4).
- So, (1, 3, 4) is a point in the plane.
Finding a vector perpendicular to the plane:
- Let's take any two non-parallel vectors in the plane. One of them is the vector we just found: (1, 3, 4).
- Another vector in the plane can be obtained by taking the cross product of the direction ratios of the line: (2, -1, 2) × (1, 0, 2) = (-2, -4, -1).
- Note that we could have chosen any other two non-parallel vectors in the plane, and they would be proportional to these two vectors. So, we can take any linear combination of these two vectors to get a vector perpendicular to the plane.
- For example, (1, 3, 4) × (-2, -4, -1) = (8, -9, -10) is perpendicular to the plane.
The normal vector to the plane is therefore (8, -9, -10). The equation of the plane can be written as:
2x - y - z = 0 (since the plane passes through the origin, the constant term is zero)
Therefore, the correct answer is option D.
- The line (x - 1)/2 = (y - 3)/-1 = (z - 4)/2
- The plane passes through the origin
To find the equation of the plane, we can first find the normal vector to the plane. The normal vector is perpendicular to the plane, and therefore perpendicular to any vector in the plane. We can use the direction ratios of the given line to find a vector in the plane, and then find a vector perpendicular to it.
Finding a vector in the plane:
- Let's take any two points on the line. For example, (1, 3, 4) and (3, 1, 6).
- The vector joining these two points is parallel to the line. So, a vector in the direction of the line is (3-1, 1-3, 6-4) = (2, -2, 2).
- Any point on the line can be written as (1, 3, 4) + t(2, -2, 2), where t is any real number.
- Let's take t = 0 to get a point on the line: (1, 3, 4) + 0(2, -2, 2) = (1, 3, 4).
- So, (1, 3, 4) is a point in the plane.
Finding a vector perpendicular to the plane:
- Let's take any two non-parallel vectors in the plane. One of them is the vector we just found: (1, 3, 4).
- Another vector in the plane can be obtained by taking the cross product of the direction ratios of the line: (2, -1, 2) × (1, 0, 2) = (-2, -4, -1).
- Note that we could have chosen any other two non-parallel vectors in the plane, and they would be proportional to these two vectors. So, we can take any linear combination of these two vectors to get a vector perpendicular to the plane.
- For example, (1, 3, 4) × (-2, -4, -1) = (8, -9, -10) is perpendicular to the plane.
The normal vector to the plane is therefore (8, -9, -10). The equation of the plane can be written as:
2x - y - z = 0 (since the plane passes through the origin, the constant term is zero)
Therefore, the correct answer is option D.
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The Cartesian equation of the plane perpendicular to the line (x - 1)/2 = (y - 3)/-1 = (z - 4)/2 and passing through the origin isa)2x - y + 2z - 7 = 0b)2x + y + 2z = 0c)2x - y - z = 0d)2x - y + 2z = 0Correct answer is option 'D'. Can you explain this answer?
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