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Line of Intersection of Two Plane and Angle Between a Line and a Plane - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Syllabus

Quick Facts

Line of Intersection of Two Plane and Angle Between a Line and a Plane, Intersection of Line and Plane is considered one of the most asked concept.
32 Questions around this concept.

Solve by difficulty

EASY

The distance of the point (1, 0, 2) from the line point of intersection of the line

$\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ and the plane $x-y+z=16$ , is:

$2\sqrt{14}$

$8$

$3\sqrt{21}$

$13$

View Solution

If the line, $\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z+4}{3}$ lies in

the plane, $lx+my-z=9,$ then $l^{2}+m^{2}$ is equal to:

View Solution

Concepts Covered - 1

Line of Intersection of Two Plane and Angle Between a Line and a Plane

The intersection of two non-parallel planes always form a line, for example in three dimensional coordinate system, intersection of the XY plane with XZ plane forms X-axis.

What is the equation of line when YZ plane and XZ plane intersects?

Let the equation of two non-parallel planes be $\vec{\mathbf r}\cdot \vec{\mathbf n}_1=d_1$ and $\vec{\mathbf r}\cdot \vec{\mathbf n}_2=d_2$ .

Now, line of intersection of planes is perpendicular to $\vec{\mathbf n}_1$ and $\vec{\mathbf n}_2$ .

Therefore, line of intersection is parallel to vector $\vec{\mathbf n}_1\times \vec{\mathbf n}_2$ .

Hence the vector equation for the line of intersection is given by

$\vec{\mathbf r}=\vec{\mathbf r}_0+t\vec{\mathbf v}$

Where, $\vec {\mathbf v}$ is the vector result of the normal vector of the two planes.

To find the line of intersection of planes a₁x+b₁y+c₁z=d₁ and a₂x+b₂y+c₂z=d₂, then first find any point on the line by putting z = 0 (say), then we can find corresponding values of x and y be solving equations a₁x+b₁y+c₁z=d₁ and a₂x+b₂y+c₂z=d₂. Thus, by fixing the value of z = λ we can find the corresponding value of x and y in terms of λ. After getting x, y and z in terms of λ we can find the equation of line in symmetric form.

Illustration

The line of intersection of two given plane P₁: -3x + 2y - 3z - 1 = 0 and P₂: 2x - y - 4z + 2 = 0 is

Let z = λ

Then, -3x + 2y = 1 + 3λ

and 2x - y = -2 + 4λ

Solve these two equations, x = -3 + 11λ and y = -4 + 18λ

The equation of the line is

$\frac{x+3}{11}=\frac{y+4}{18}=\frac{z-0}{1}=\lambda$

Aliter

The general equation of plane and its normal is ax + by + cz + d = 0 and $\vec{\mathbf n}=a\hat i+b\hat j+c\hat k$

Then, $\vec{\mathbf n}_1=-3\hat i+2\hat j-3\hat k$

and $\vec{\mathbf n}_2=2\hat i-\hat j-4\hat k$

$\vec{\mathbf s}=\vec{\mathbf n}_1\times \vec{\mathbf n}_2=\left|\begin{array}{ccc}{\vec{i}} & {\vec{j}} & {\vec{k}} \\ {-3} & {2} & {-3} \\ {2} & {-1} & {-4}\end{array}\right|=-11 \vec{i}-18 \vec{j}-\vec{k},\;\;\vec{s}=a \vec{i}+b \vec{j}+c \vec{k}$

To write the equation of the line of intersection, i.e., $\frac{x-x_{0}}{a}=\frac{y-y_{0}}{b}=\frac{z-z_{0}}{c}$ , we still need the coordinates of any of its point P(x₀, y₀, z₀).

Let this point be the intersection of the intersection line and the XY coordinate plane.

Then, the coordinates of the point of intersection (x, y, 0) must satisfy equations of the given planes.

Therefore, by putting z = 0 into P₁ and P₂ we get,

-3x + 2y - 3(0) - 1 = 0

2x - y - 4(0) + 2 = 0

x = -3 and y = -4

So the line of intersection is $\frac{x+3}{-11}=\frac{y+4}{-18}=\frac{z}{-1} \text { or } \frac{x+3}{11}=\frac{y+4}{18}=\frac{z}{1}$ .

Angle Between a Line and a Plane

The angle between a line and a plane is the complement of the angle between the line and normal to the plane.

Vector Form

If the equation of the line is $\vec{\mathbf r}=\vec{\mathbf r}_0+\lambda \vec{\mathbf b}$ and the equation of the plane is $\vec{\mathbf r}\cdot \vec{\mathbf n}=d$ . Then the angle θ between the line and the normal to the plane is

$\cos \theta=\left|\frac{\vec{\mathbf b} \cdot \vec{\mathbf n}}{|\vec{\mathbf b}| \cdot|\vec{\mathbf n}|}\right|$

and so the angle φ between the line and the plane is given by 90⁰ – θ,

$\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\sin (90-\theta)=\cos \theta\\\mathrm{i.e.\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\sin \phi=\left|\frac{\vec{\mathbf b} \cdot \vec{\mathbf n}}{|\vec{\mathbf b}| \cdot|\vec{\mathbf n}|}\right|\;\;\;\text{or}\;\;\;\phi=\sin^{-1}\left|\frac{\vec{\mathbf b} \cdot \vec{\mathbf n}}{|\vec{\mathbf b}| \cdot|\vec{\mathbf n}|}\right|$

Cartesian Form

The angle between a line and a plane

If the line is $\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}$

and the plane is $a_{1}x+b_{1}y+c_{1}z+d=0$ is given by

$\sin \Theta = \frac{a_{1}a+b_{1}b+c_{1}c}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a^{2}+b^{2}+c^{2}}}$

NOTE:
Line $\vec{\mathbf r}=\vec{\mathbf r}_0+\lambda \vec{\mathbf b}$ and plane $\vec{\mathbf r}\cdot \vec{\mathbf n}=d$ are perpendicular if $\vec{\mathbf b}=\lambda \vec{\mathbf n}$ or $\vec{\mathbf b}\times \vec{\mathbf n} = 0$ and parallel if $\vec{\mathbf b}\perp \vec{\mathbf n}$ or $\vec{\mathbf b}\cdot \vec{\mathbf n}=0$

Intersection of Line and Plane

Given equation of the line is $\frac{x-x_{1}}{l}=\frac{y-y_{1}}{m}=\frac{z-z_{1}}{n}$ and the equation of the plane is ax + by + cz + d = 0

$\\\text {Let } \quad \frac{x-x_{1}}{l}=\frac{y-y_{1}}{m}=\frac{z-z_{1}}{n}=r\\\\\therefore \quad\left(x=r l+x_{1}, y=m r+y_{1}, z=n r+z_{1}\right)\\\text {be a point in the plane say } P .\\\text {It must satisfy the equation of plane. }\\\therefore\; \quad a\left(x_{1}+l r\right)+b\left(y_{1}+m r\right)+c\left(z_{1}+n r\right)+d=0\\\Rightarrow \quad\left(a x_{1}+b y_{1}+c z_{1}+d\right)+r(a l+b m+c n)=0\\\\\Rightarrow \;\;\;\;\;\;\;\;\;\;r=-\frac{\left(a x_{1}+b y_{1}+c z_{1}+d\right)}{a l+b m+c n}$