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

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 $\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{n}}_1=d_1$ and $\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{n}}_2=d_2$

Now, line of intersection of planes is perpendicular to $\overrightarrow{\mathbf{n}}_1$ and $\overrightarrow{\mathbf{n}}_2$.
Therefore, the line of intersection is parallel to vector $\overrightarrow{\mathbf{n}}_1 \times \overrightarrow{\mathbf{n}}_2$.
Hence the vector equation for the line of intersection is given by
$
\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{r}}_0+t \overrightarrow{\mathbf{v}}
$

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

To find the line of intersection of planes $a_1 x+b_1 y+c_1 z=d_1$ and $a_2 x+b_2 y+c_2 z=d_2$, 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 $a_1 x+b_1 y+c_1 z=d_1$ and $a_2 x+b_2 y+c_2 z=d_2$. Thus, by fixing the value of $z=\lambda$ we can find the corresponding value of $x$ and $y$ in terms of $\lambda$. After getting $x, y$, and $z$ in terms of $\lambda$ we can find the equation of line in symmetric form.|

Illustration

The line of intersection of two given plane $P_1:-3 x+2 y-3 z-1=0$ and $P_2: 2 x-y-4 z+2=0$ is

Let $z=\lambda$
Then $-3 x+2 y=1+3 \lambda$
and $2 x-y=-2+4 \lambda$
Solve these two equations, $x=-3+11 \lambda$ and $y=-4+18 \lambda$
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 $\mathrm{ax}+\mathrm{by}+\mathrm{cz}+\mathrm{d}=0$ and $\overrightarrow{\mathbf{n}}=a \hat{i}+b \hat{j}+c \hat{k}$
Then, $\overrightarrow{\mathbf{n}}_1=-3 \hat{i}+2 \hat{j}-3 \hat{k}$
and $\overrightarrow{\mathbf{n}}_2=2 \hat{i}-\hat{j}-4 \hat{k}$
$
\overrightarrow{\mathbf{s}}=\overrightarrow{\mathbf{n}}_1 \times \overrightarrow{\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}, \quad \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 points $\mathrm{P}\left(\mathrm{x}_0, \mathrm{y}_0, \mathrm{z}_0\right)$.

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 the equations of the given planes.

Therefore, by putting $z=0$ into $P_1$ and $P_2$ we get,

$
\begin{aligned}
& -3 x+2 y-3(0)-1=0 \\
& 2 x-y-4(0)+2=0 \\
& x=-3 \text { and } y=-4
\end{aligned}
$

So the line of intersection is $\frac{x+3}{-11}=\frac{y+4}{-18}=\frac{z}{-1}$ 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 $\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{r}}_0+\lambda \overrightarrow{\mathbf{b}}$ and the equation of the plane is $\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{n}}=d$. Then the angle $\theta$ between the line and the normal to the plane is
$
\cos \theta=\left|\frac{\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{n}}}{|\overrightarrow{\mathbf{b}}| \cdot|\overrightarrow{\mathbf{n}}|}\right|
$
and so the angle $\varphi$ between the line and the plane is given by $90^{\circ}-\theta$,
$
\begin{aligned}
\sin (90-\theta) & =\cos \theta \\
\sin \phi & =\left|\frac{\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{n}}}{|\overrightarrow{\mathbf{b}}| \cdot|\overrightarrow{\mathbf{n}}|}\right| \quad \text { or } \quad \phi=\sin ^{-1}\left|\frac{\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{n}}}{|\overrightarrow{\mathbf{b}}| \cdot|\overrightarrow{\mathbf{n}}|}\right|
\end{aligned}
$
i.e.

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

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

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)
$
be a point in the plane say $P$.
It must satisfy the equation of plane.
$
\begin{aligned}
& \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 \quad r=-\frac{\left(a x_1+b y_1+c z_1+d\right)}{a l+b m+c n}
\end{aligned}
$

Put the value of r in $\left(x=r l+x_1, y=m r+y_1, z=n r+z_1\right)$, you will get the coordinates of point P.

Condition for a Line to be Parallel to a Plane

Condition for a Line to be Parallel to a Plane
The line $\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$ is parallel to plane $\mathrm{ax}+\mathrm{by}+\mathrm{cz}+\mathrm{d}=0$ iff:
$
\theta=0 \text { or } \pi \text { or } \sin \theta=0 \Rightarrow a l+b m+c m=0
$

Condition for a Line to Lie in the Plane
Condition for the line $\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$ to lie in the plane $\mathrm{ax}+\mathrm{by}+\mathrm{cz}+\mathrm{d}=0$ are:
$
a l+b m+c n=0 \text { and } a x_1+b y_1+c z_1+d=0
$