Angle between a Line and a Plane

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 parallel, then the lines normal to the planes are also parallel.

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 two planes are perpendicular to each other, then the lines normal to the planes are also perpendicular.

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 parallel to the plane, then the line and the line normal to the plane are perpendicular to each other.

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

If a line is perpendicular to the plane, then the line and the line normal to the plane are parallel.

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 \[-k(x+1)-3k(y-1)+5k(z+1)=0\] \[-x-1-3y+3+5z+5=0\] \[\therefore x+3y-5z-7=0\]


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