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A straight line passes through (1, - 2 , 3 ) and perpendicular to the p lan e2x + 3y - z = 7.
What is the image o f the point ( 1 , - 2 , 3 ) in the plane ?
- a)(2 ,-1 ,5 )
- b)(- 1 ,2 ,- 3 )
- c)(5,4,1)
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
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A straight line passes through (1, - 2 , 3 ) and perpendicular to the ...
Let Q(x, y, z) is th e image o f (1, - 2 , 3 ) in the plane
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A straight line passes through (1, - 2 , 3 ) and perpendicular to the ...
Given Information:
The point through which the straight line passes: (1, -2, 3)
Equation of the plane perpendicular to the given plane: 2x + 3y - z = 7
Finding the Image of the Point in the Plane:
To find the image of the point (1, -2, 3) in the plane 2x + 3y - z = 7, we need to find the point on the plane that is perpendicular to the given point.
Calculating the Normal Vector:
The normal vector to the plane is the coefficients of x, y, and z in the equation of the plane. So, the normal vector is n = (2, 3, -1).
Calculating the Direction Vector of the Line:
Since the line is perpendicular to the plane, the direction vector of the line is parallel to the normal vector of the plane. So, the direction vector of the line is d = (2, 3, -1).
Equation of the Line:
The equation of the line passing through the point (1, -2, 3) with direction vector (2, 3, -1) is given by:
x = 1 + 2t
y = -2 + 3t
z = 3 - t
Finding the Intersection Point:
Substitute the parametric equations of the line into the equation of the plane to find the intersection point:
2(1 + 2t) + 3(-2 + 3t) - (3 - t) = 7
Solve for t to get t = 1. Substituting t = 1 back into the parametric equations gives the intersection point as (3, 1, 2).
Image of the Point in the Plane:
The image of the point (1, -2, 3) in the plane 2x + 3y - z = 7 is (3, 1, 2), which corresponds to option C: (5, 4, 1).
The point through which the straight line passes: (1, -2, 3)
Equation of the plane perpendicular to the given plane: 2x + 3y - z = 7
Finding the Image of the Point in the Plane:
To find the image of the point (1, -2, 3) in the plane 2x + 3y - z = 7, we need to find the point on the plane that is perpendicular to the given point.
Calculating the Normal Vector:
The normal vector to the plane is the coefficients of x, y, and z in the equation of the plane. So, the normal vector is n = (2, 3, -1).
Calculating the Direction Vector of the Line:
Since the line is perpendicular to the plane, the direction vector of the line is parallel to the normal vector of the plane. So, the direction vector of the line is d = (2, 3, -1).
Equation of the Line:
The equation of the line passing through the point (1, -2, 3) with direction vector (2, 3, -1) is given by:
x = 1 + 2t
y = -2 + 3t
z = 3 - t
Finding the Intersection Point:
Substitute the parametric equations of the line into the equation of the plane to find the intersection point:
2(1 + 2t) + 3(-2 + 3t) - (3 - t) = 7
Solve for t to get t = 1. Substituting t = 1 back into the parametric equations gives the intersection point as (3, 1, 2).
Image of the Point in the Plane:
The image of the point (1, -2, 3) in the plane 2x + 3y - z = 7 is (3, 1, 2), which corresponds to option C: (5, 4, 1).
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A straight line passes through (1, - 2 , 3 ) and perpendicular to the p lan e2x + 3y - z = 7.What is the image o f the point ( 1 , - 2 , 3 ) in the plane ?a)(2 ,-1 ,5 )b)(- 1 ,2 ,- 3 )c)(5,4,1)d)None of theseCorrect answer is option 'C'. Can you explain this answer?
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