Question

Find the angle between the given planes.

$\mathbf{2}\mathit{x}\mathbf{}\mathbf{-}\mathbf{}\mathit{y}\mathbf{}\mathbf{-}\mathbf{}\mathit{z}\mathbf{}\mathbf{=}\mathbf{}\mathbf{4}$and$\mathbf{3}\mathit{x}\mathbf{}\mathbf{-}\mathbf{}\mathbf{2}\mathit{y}\mathbf{}\mathbf{-}\mathbf{}\mathbf{6}\mathit{z}\mathbf{}\mathbf{=}\mathbf{}\mathbf{7}$

Short answer

The angle between the planes $\theta =35.3$.

Step-by-step solution

Concept and formula used

The angle formed by the planes and the normal to the planes is same.

To determine the angle between the vectors by using the formula$\mathit{\theta }\mathbf{=}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{-}\mathbf{1}}\frac{\mathbf{A}\mathbf{\times }\mathbf{B}}{\mathbf{\left|}\mathbf{A}\mathbf{\right|}\mathbf{\left|}\mathbf{B}\mathbf{\right|}}$.

Find the angle

The given planes are,$2x-y-z=4$ and $3x-2y-6z=12$.

Calculating gives the values as below:

$$\begin{gathered} A=\left|A\right|=\sqrt{{\left(2\right)}^2+{(-1)}^2+{(-1)}^2}=\sqrt{6} \\ B=\left|B\right|=\sqrt{{\left(3\right)}^2+{(-2)}^2+{(-6)}^2}=7 \\ A\times B=\left(2\right)\left(3\right)+(-1)(-2)+(-1)(-6)=6+2+6=14 \end{gathered}$$

Now find the angle:

$$\begin{gathered} \theta ={\cos }^{-1}\frac{A\times B}{AB} \\ ={\cos }^{-1}\frac{14}{7\times \sqrt{6}} \\ ={\cos }^{-1}\frac{2}{\sqrt{6}} \\ =35.3 \end{gathered}$$

The angle between the planes $\theta =35.3$.