The angle between two planes is the angle between their normals.

Let the equations of two planes be \[A_1x+B_1y+C_1z+D_1=0\] \[A_2x+B_2y+C_2z+D_2=0\] The normals of the above two planes have their direction cosines proportional to $A_1,B_1,C_1$ and $A_2,B_2,C_2$ respectively.

Let $\theta$ be the angle between the planes i.e. angle between their normals. Hence, \[\cos\theta=\frac{A_1A_2+B_1B_2+C_1C_2}{\sqrt{\sum A_1^2}\sqrt{\sum A_2^2}}\]

## Perpendicularity

If the two planes be perpendicular, then their normals with direction cosines proportional to $A_1,B_1,C_1$ and $A_2,B_2,C_2$ are also perpendicular. The condition for two lines to be perpendicular is \[A_1A_2+B_1B_2+C_1C_2=0\] which is the required condition for perpendicularity of two planes. [Condition for perpendicularity of two lines with direction ratios]

## Parallelism

If the two planes be parallel, then their normals with direction cosines proportional to $A_1,B_1,C_1$ and $A_2,B_2,C_2$ are also parallel. The condition for two lines to be parallel is \[\frac{A_1}{A_2}=\frac{B_1}{B_2}=\frac{C_1}{C_2}\] which is the required condition for parallelism of two planes. [Condition for parallelism of two lines with direction ratios]

**Cor.** We have provided that the two planes \[A_1x+B_1y+C_1z+D_1=0\text{ __(1)}\] \[A_1x+B_2y+C_2z+D_2=0\text{ __(2)}\] are parallel if \[\frac{A_1}{A_2}=\frac{B_1}{B_2}=\frac{C_1}{C_2}=k\;\text{(say) and}\;k≠0. \] Then, \[A_1=A_2k,\;B_1=B_2k\:\;\text{and}\:\;C_1=C_2k\]

Substituting the values of $A_1,B_1,C_1$ in equation $\text{(1)}$, we get \[kA_2x+kB_2y+kC_2z+D_1=0\] \[A_2x+B_2y+C_2z+\frac{D_1}{k}=0\text{ __(3)}\]

Thus if plane $\text{(1)}$ is parallel to the plane $\text{(2)}$ then plane $\text{(1)}$ can be expressed in the form of $\text{(3)}$ which shows that we can write the equations of two parallel planes in such a way that the left hand sides of equations differ only by a constant.

### Find the angle between two planes $x+2y+z+7=0$ and $2x+y-z+13=0$.

Given equations of planes are \[x+2y+z+7=0\] \[\therefore A_1=1,\;B_1=2,\;C_1=1\:\;\text{and}\:\;D_1=7\] \[2x+y-z+13=0\] \[\therefore A_2=2,\;B_2=1,\;C_2=-1\:\;\text{and}\:\;D_2=13\]

Let $\theta$ be the angle between the two given planes. Then, we have \[\cos\theta=\frac{A_1A_2+B_1B_2+C_1C_2}{\sqrt{A_1^2+B_1^2+C_1^2}\sqrt{A_2^2+B_2^2+C_2^2}}\] \[\cos\theta=\frac{2+2-1}{\sqrt{1+4+1}\sqrt{4+1+1}}\] \[\cos\theta=\frac{3}{6}=\frac{1}{2}=\cos\frac{π}{3}\] \[\therefore\theta=\frac{π}{3}\]

### Show that the equation of the plane through $(\alpha,\beta,\gamma)$ and parallel to the plane $ax+by+cz=0$ is $ax+by+cz=a\alpha+b\beta+c\gamma$.

Equation of any plane parallel to the plane $ax+by+cz=0$ is \[ax+by+cz+k=0\text{ __(1)}\] Since this plane passes through the point $(\alpha,\beta,\gamma)$, we have \[\alpha a+\beta b+\gamma c+k=0\] \[\therefore k=-(\alpha a+\beta b+\gamma c)\]

Putting the value of $k$ in equation $\text{(1)}$, we have, \[ax+by+cz-(\alpha a+\beta b+\gamma c)=0\] \[\therefore ax+by+cz=\alpha a +\beta b+\gamma c\]

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