Question

Find the cosine of the angle between the planes −1x+3y+1z=0 and the plane 5x+5y+4z=−4

Answer 1

Answer:

The he cosine of the angle between the planes is $\frac{14}{11\sqrt{6}}$.

Step-by-step explanation:

Using the definition of the dot product:

$\cos\theta =\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}$

The given planes are

$-1x+3y+1z=0$

$5x+5y+4z=-4$

The angle between two normal vectors of the planes is the same as one of

the angles between the planes. We can find a normal vector to each of the

planes by looking at the coefficients of x, y, z.

$\overrightarrow{n_1}=$

$\overrightarrow{n_2}=$

$\overrightarrow{n_1}\cdot \overrightarrow{n_2}=(-1)(5)+(3)(5)+(1)(4)=14$

$|n_1|=\sqrt{(-1)^2+(3)^2+(1)^2}=\sqrt{11}$

$|n_2|=\sqrt{(5)^2+(5)^2+(4)^2}=\sqrt{66}$

The cosine of the angle between the planes

$\cos\theta =\frac{\overrightarrow{n_1}\cdot \overrightarrow{n_2}}{|\overrightarrow{n_1}||\overrightarrow{n_2}|}$

$\cos\theta =\frac{14}{\sqrt{11}\sqrt{66}}$

$\cos\theta =\frac{14}{11\sqrt{6}}$

Therefore the cosine of the angle between the planes is $\frac{14}{11\sqrt{6}}$.