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A straight line passes through (1, - 2 , 3 ) and perpendicular to the p lan e2x + 3y - z = 7.
Where does the line meet the plane ?
- a)(2,3,-1)
- b)(1,2,3)
- c)(2,1,3)
- d)(3,1,2)
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A straight line passes through (1, - 2 , 3 ) and perpendicular to the ...
Equation o f line, 
Let P (2r + 1, 3r - 2, - r + 3) of the line meets the plane.
Then, 2(2r + l ) + 3 ( 3 r - 2 ) - ( - r + 3) = 0
4r + 2 + 9 r - 6 + r - 3 = 7
14r= 14
r= 1
P (3,1,2) meets the plane.
Let P (2r + 1, 3r - 2, - r + 3) of the line meets the plane.
Then, 2(2r + l ) + 3 ( 3 r - 2 ) - ( - r + 3) = 0
4r + 2 + 9 r - 6 + r - 3 = 7
14r= 14
r= 1
P (3,1,2) meets the plane.
Most Upvoted Answer
A straight line passes through (1, - 2 , 3 ) and perpendicular to the ...
Given Information:
- The line passes through the point (1, -2, 3).
- The line is perpendicular to the plane 2x + 3y - z = 7.
Find the Point of Intersection:
To find the point where the line intersects the plane, we need to first find the equation of the line passing through the given point and perpendicular to the plane.
Equation of the Line:
The direction vector of the line is the normal vector to the plane, which is (2, 3, -1).
The parametric equations of the line passing through point (1, -2, 3) with direction vector (2, 3, -1) can be written as:
x = 1 + 2t
y = -2 + 3t
z = 3 - t
Intersection with the Plane:
To find the point of intersection, substitute the parametric equations of the line into the equation of the plane:
2(1 + 2t) + 3(-2 + 3t) - (3 - t) = 7
Solve for t:
2 + 4t - 6 + 9t - 3 + t = 7
13t - 7 = 7
13t = 14
t = 14/13
Substitute t back into the parametric equations to find the point of intersection:
x = 1 + 2(14/13) = 3
y = -2 + 3(14/13) = 1
z = 3 - (14/13) = 2
Therefore, the line intersects the plane at the point (3, 1, 2), which corresponds to option 'D'.
- The line passes through the point (1, -2, 3).
- The line is perpendicular to the plane 2x + 3y - z = 7.
Find the Point of Intersection:
To find the point where the line intersects the plane, we need to first find the equation of the line passing through the given point and perpendicular to the plane.
Equation of the Line:
The direction vector of the line is the normal vector to the plane, which is (2, 3, -1).
The parametric equations of the line passing through point (1, -2, 3) with direction vector (2, 3, -1) can be written as:
x = 1 + 2t
y = -2 + 3t
z = 3 - t
Intersection with the Plane:
To find the point of intersection, substitute the parametric equations of the line into the equation of the plane:
2(1 + 2t) + 3(-2 + 3t) - (3 - t) = 7
Solve for t:
2 + 4t - 6 + 9t - 3 + t = 7
13t - 7 = 7
13t = 14
t = 14/13
Substitute t back into the parametric equations to find the point of intersection:
x = 1 + 2(14/13) = 3
y = -2 + 3(14/13) = 1
z = 3 - (14/13) = 2
Therefore, the line intersects the plane at the point (3, 1, 2), which corresponds to option 'D'.
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A straight line passes through (1, - 2 , 3 ) and perpendicular to the p lan e2x + 3y - z = 7.Where does the line meet the plane ?a)(2,3,-1)b)(1,2,3)c)(2,1,3)d)(3,1,2)Correct answer is option 'D'. Can you explain this answer?
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