Question:
The point of intersection of the line $\frac{x-1}{3}=\frac{y+2}{4}=\frac{z-3}{-2}$ and plane $2 x-y+3 z-1= 0$ is.
The point of intersection of the line $\frac{x-1}{3}=\frac{y+2}{4}=\frac{z-3}{-2}$ and plane $2 x-y+3 z-1= 0$ is.
Updated On: Jan 30, 2025
- (10, -10, 3)
- (10, 10, -3)
- (-10, 10, 3)
- none of these
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The Correct Option is B
Solution and Explanation
We have line, $\frac{x-1}{3}=\frac{y+2}{4}=\frac{z-3}{-2}=\lambda$ ...(1)
General point $P$ on the line is,
$P =(3 \lambda+1,4 \lambda-2,-2 \lambda+3)$
since line (1) intersect plane $2 x-y+3 z-1=0$,
Assume it intersects at a point $P$
Therefore,
$2(3 \lambda+1)-(4 \lambda-2)+3(-2 \lambda+3)-1=0 $
$6 \lambda+2-4 \lambda+2-6 \lambda+9-1=0 $
$4 \lambda=12$
$\lambda=3$
Therefore,
$P =(10,10,-3))$
General point $P$ on the line is,
$P =(3 \lambda+1,4 \lambda-2,-2 \lambda+3)$
since line (1) intersect plane $2 x-y+3 z-1=0$,
Assume it intersects at a point $P$
Therefore,
$2(3 \lambda+1)-(4 \lambda-2)+3(-2 \lambda+3)-1=0 $
$6 \lambda+2-4 \lambda+2-6 \lambda+9-1=0 $
$4 \lambda=12$
$\lambda=3$
Therefore,
$P =(10,10,-3))$
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Concepts Used:
Plane
A surface comprising all the straight lines that join any two points lying on it is called a plane in geometry. A plane is defined through any of the following uniquely:
- Using three non-collinear points
- Using a point and a line not on that line
- Using two distinct intersecting lines
- Using two separate parallel lines
Properties of a Plane:
- In a three-dimensional space, if there are two different planes than they are either parallel to each other or intersecting in a line.
- A line could be parallel to a plane, intersects the plane at a single point or is existing in the plane.
- If there are two different lines that are perpendicular to the same plane then they must be parallel to each other.
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