Let $\theta$ be the angle between a line and a plane. Then, the angle between the line and the normal to the plane is $\frac{π}{2}-\theta$. Let the equation of the plane be \[A_1x+B_1y+C_1z+D=0\]

Hence, the direction cosines of the normal to the plane are proportional to $A_1,B_1,C_1$. Let the direction cosines of the line be proportional to $A_2,B_2,C_2$. Then, we have \[\cos\left(\frac{π}{2}-\theta\right)=\sin\theta=\frac{A_1A_2+B_1B_2+C_1C_2}{\sqrt{\sum A_1^2}\sqrt{\sum A_2^2}}\]

**Cor.** If the line lies on the plane (or parallel to the plane) then clearly \[A_1A_2+B_1B_2+C_1C_2=0\]

Now, let us find the equation of a plane under given conditions.

Let the equations of two planes be \[a_1x+b_1y+c_1z+d_1=0\] \[a_2x+b_2y+c_2z+d_2=0\] Let $l,m,n$ be the direction cosines of a line.

If two planes are parallel, then the lines normal to the planes are also parallel. \[\therefore\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\] [Condition for the parallelism of two lines]

If two planes are perpendicular to each other, then the lines normal to the planes are also perpendicular. \[\therefore a_1a_2+b_1b_2+c_1c_2=0\] [Condition for the perpendicularity of two lines]

If a line is parallel to the plane, then the line and the line normal to the plane are perpendicular to each other. \[\therefore a_1l+b_1m+c_1n=0\]

If a line is perpendicular to the plane, then the line and the line normal to the plane are parallel. \[\therefore\frac{a_1}{l}=\frac{b_1}{m}=\frac{c_1}{n}\]

### Find the equation of the plane through the point $(2,-3,1)$ and perpendicular to the line joining the two points $(3,4,-1)$ and $(2,-1,5)$.

Equation of a plane through $(2,-3,1)$ is \[A(x-2)+B(y+3)+C(z-1)=0\text{ __(1)}\]

The direction cosines of the line joining the points $(3,4,-1)$ and $(2,-1,5)$ are proportional to $-1,-5,6$.

The plane $\text{(1)}$ is perpendicular to this line, if any normal to the plane is parallel to this line. But direction cosines of the normal to the plane are proportional to $A,B,C$ and therefore the condition for parallelism gives \[\frac{A}{-1}=\frac{B}{-5}=\frac{C}{6}=k\]

\[\therefore A=-k,\;B=-5k\:\;\text{and}\:\;C=6k\]

Putting the values of $A$, $B$ and $C$ in equation $\text{(1)}$, we get \[-k(x-2)-5k(y+3)+6k(z-1)=0\] \[-x+2-5y-15+6z-6=0\] \[-x-5y+6z-19=0\] \[\therefore x+5y-6z+19=0\] which is the required equation of the plane.

### Find the equation of the plane through $(-1,1,-1)$ and $(6,2,1)$ and perpendicular to the plane $2x+y+z=5$.

Equation of a plane through $(-1,1,-1)$ is \[a(x+1)+b(y-1)+c(z+1)=0\text{ __(1)}\]

Given points are $(-1,1,-1)$ and $(6,2,1)$. The direction cosines of the line joining the given points are proportional to $7,1,2$.

If two planes are perpendicular to each other, then the lines normal to the planes are also perpendicular to each other. \[\therefore 7a+b+2c=0\text{ __(2)}\]

Also, the plane is perpendicular to $2x+y+z=5$. \[\therefore 2a+b+c=0\text{ __(3)}\]

Solving $\text{(2)}$ and $\text{(3)}$, \[\frac{a}{1-2}=\frac{b}{4-7}=\frac{c}{7-2}=k\] \[\frac{a}{-1}=\frac{b}{-3}=\frac{c}{5}=k\] \[\therefore a=-k,\;b=-3k\:\;\text{and}\:\;c=5k\]

Putting the values of $a$, $b$ and $c$ in equation $\text{(1)}$, we get \[-k(x+1)-3k(y-1)+5k(z+1)=0\] \[-x-1-3y+3+5z+5=0\] \[\therefore x+3y-5z-7=0\]