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: Jun 4, 2023
- (10, -10, 3)
- (10, 10, -3)
- (-10, 10, 3)
- none of these
Hide Solution
Verified By Collegedunia
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,
$P =(10,10,-3))$
Was this answer helpful?
1
0
Top Questions on Plane
- Let the plane $P : 8 x+\alpha_1 y+\alpha_2 z+12=0$ be parallel to the line $L : \frac{x+2}{2}=\frac{y-3}{3}=\frac{z+4}{5}$. If the intercept of $P$ on the $y$-axis is 1 , then the distance between $P$ and $L$ is :
- The locus of points(x,y) in the plane satisfying sin2x + sin2y =1 consists of
- A vector $\vec{v}$ in the first octant is inclined to the $x$-axis at $60^{\prime}$, to the $y$-axis at $45$ and to the $z$-axis at an acute angle If a plane passing through the points $(\sqrt{2},-1,1)$ and $(a, b, c)$, is normal to $\vec{v}$, then
- A vector $\vec{v}$ in the first octant is inclined to the $x$-axis at $60^{\prime}$, to the $y$-axis at $45$ and to the $z$-axis at an acute angle If a plane passing through the points $(\sqrt{2},-1,1)$ and $(a, b, c)$, is normal to $\vec{v}$, then
- Let $P$ be the plane, passing through the point $(1,-1,-5)$ and perpendicular to the line joining the points $(4,1,-3)$ and $(2,4,3)$. Then the distance of $P$ from the point $(3,-2,2)$ is
View More Questions
Questions Asked in BITSAT exam
- Minimum value of 5cos(2x) + 5sin(2x)
- BITSAT - 2023
- Trigonometric Functions
- Two capacitors, of capacitance C, are connected in series. If one of them is filled with a dielectric substance K, what is the effective capacitance?
- BITSAT - 2023
- electrostatic potential and capacitance
- What is the dimension of the Gravitational constant?
- BITSAT - 2023
- Gravitation
NaOH is deliquescent
- BITSAT - 2023
- States of matter
- The freezing point of the 0.05 molal solution of non - electrolyte in water is? (kf = 1.86)
- BITSAT - 2023
- Solutions
View More Questions
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.
- If there are two separate planes which are perpendicular to the same line then they must be parallel to each other.