Question

Determine whether the planes are parallel, perpendicular, or neither.8x + 8y + 8z = 1, 8x − 8y + 8z = 1 parallel perpendicular neither Correct: Your answer is correct.If neither, find the angle between them. (Round your answer to one decimal place. If the planes are parallel or perpendicular, enter PARALLEL or PERPENDICULAR, respectively.)

Answer 1

Answer:

The answer is "$\bold{\theta= 70.53^{\circ}}$"

Step-by-step explanation:

The given value:

$8x+8y+8z = 1\ and \ 8x-8y+8z = 1$  

When the normal vectors are in the plane:  

$n_1 = < 8,8,8 > \ and \ n_2 = < 8, -8, 8>:$

Because the two variables really aren't proportional to each other and the wings are not parallel.  

Let us just find that two vectors' dot product.

$n_1 \cdot n_2 = < 8,—8,8>\\\\ = 64-64+ 64 \\\\ = 64 \neq 0$

Therefore, we won't find that angle between both the given planes as parallel.

$\cos \theta = \frac{n_1 \cdot n_2}{|n_1||n_2|}\\\\\cos \theta =\frac{64}{\sqrt{192} \sqrt{192}}\\\\\cos \theta =\frac{64}{8\sqrt{3} 8\sqrt{3}}\\\\\cos \theta = \frac{1}{3}\\\\ \theta = 70.53^{\circ}$