Answer:
The answer is "[tex]\bold{\theta= 70.53^{\circ}}[/tex]"
Step-by-step explanation:
The given value:
[tex]8x+8y+8z = 1\ and \ 8x-8y+8z = 1[/tex]
When the normal vectors are in the plane:
[tex]n_1 = < 8,8,8 > \ and \ n_2 = < 8, -8, 8>:[/tex]
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.
[tex]n_1 \cdot n_2 =<8,8,8> < 8,—8,8>\\\\ = 64-64+ 64 \\\\ = 64 \neq 0[/tex]
Therefore, we won't find that angle between both the given planes as parallel.
[tex]\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}[/tex]
It is required to know the orientation of one plane relative to the other.
The angle between the planes is 70.5°.
The given planes are
8x + 8y + 8z = 1
8x − 8y + 8z = 1
Parallel
[tex]\dfrac{8}{8}=\dfrac{8}{-8}=\dfrac{8}{8}\\\Rightarrow 1\neq -1\neq 1[/tex]
The planes are not parallel.
Perpendicular
[tex]8\times 8+8\times -8+8\times8=64\neq 0[/tex]
The planes are not perpendicular as dot product is not 0.
Finding angle
[tex]\theta=\cos^{-1}\dfrac{8\times 8+8\times -8+8\times8}{\sqrt{8^2+8^2+8^2}\sqrt{8^2+(-8)^2+8^2}}\\\Rightarrow \theta=\cos^{-1}\dfrac{1}{3}\\\Rightarrow \theta=70.53^{\circ}[/tex]
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