Answer:
Neither planes are parallel nor perpendicular to each other. (NEITHER)
Step-by-step explanation:
Each expression can be represented as dot product, that is:
[tex](1, -1,-3)\bullet (x,y, z) = 1[/tex]
[tex](3, 1, -1)\bullet (x,y,z) = 2[/tex]
Where first vector is known as plane generator. It is also known that dot product equals to the following expression:
[tex]\vec u \bullet \vec v = \|\vec u\|\cdot \|\vec v \| \cdot \cos \theta[/tex]
The cosine is therefore cleared:
[tex]\cos \theta = \frac{\vec u \bullet \vec v}{\|\vec u\|\cdot \|\vec v\|}[/tex]
The norm of each vector is determined by these expressions:
[tex]\|\vec u\|=\sqrt{\vec u\bullet \vec u}[/tex] and [tex]\|\vec v\| =\sqrt{\vec v\bullet \vec v}[/tex]
Planes are parallel to each other when [tex]\cos \theta = \pm1[/tex] and perpendicular to each other if [tex]\cos \theta = 0[/tex].
If [tex]\vec u = (-1,-1,3)[/tex] and [tex]\vec v = (3,1,-1)[/tex], then:
[tex]\|\vec u\| = \sqrt{(-1,-1,3)\bullet (-1,-1,3)}[/tex]
[tex]\|\vec u\| =\sqrt{1+1+9}[/tex]
[tex]\|\vec u\| = \sqrt{10}[/tex]
[tex]\|\vec v\|=\sqrt{(3,1,-1)\bullet(3,1,-1)}[/tex]
[tex]\|\vec v\|=\sqrt{10}[/tex]
[tex]\vec u \bullet \vec v = (1)\cdot (3)+(-1)\cdot (1) +(-3)\cdot (-1)[/tex]
[tex]\vec u \bullet \vec v = 3-1+3[/tex]
[tex]\vec u\bullet \vec v = 5[/tex]
[tex]\cos \theta = \frac{5}{10}[/tex]
[tex]\cos \theta = \frac{1}{2}[/tex]
Neither planes are parallel nor perpendicular to each other.
