Question

Determine whether the planes are parallel, perpendicular, or neither.
x − y − 3z = 1, 3x + y − z = 2
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:

Neither planes are parallel nor perpendicular to each other. (NEITHER)

Step-by-step explanation:

Each expression can be represented as dot product, that is:

$(1, -1,-3)\bullet (x,y, z) = 1$

$(3, 1, -1)\bullet (x,y,z) = 2$

Where first vector is known as plane generator. It is also known that dot product equals to the following expression:

$\vec u \bullet \vec v = \|\vec u\|\cdot \|\vec v \| \cdot \cos \theta$

The cosine is therefore cleared:

$\cos \theta = \frac{\vec u \bullet \vec v}{\|\vec u\|\cdot \|\vec v\|}$

The norm of each vector is determined by these expressions:

$\|\vec u\|=\sqrt{\vec u\bullet \vec u}$ and $\|\vec v\| =\sqrt{\vec v\bullet \vec v}$

Planes are parallel to each other when $\cos \theta = \pm1$ and perpendicular to each other if $\cos \theta = 0$.

If $\vec u = (-1,-1,3)$ and $\vec v = (3,1,-1)$, then:

$\|\vec u\| = \sqrt{(-1,-1,3)\bullet (-1,-1,3)}$

$\|\vec u\| =\sqrt{1+1+9}$

$\|\vec u\| = \sqrt{10}$

$\|\vec v\|=\sqrt{(3,1,-1)\bullet(3,1,-1)}$

$\|\vec v\|=\sqrt{10}$

$\vec u \bullet \vec v = (1)\cdot (3)+(-1)\cdot (1) +(-3)\cdot (-1)$

$\vec u \bullet \vec v = 3-1+3$

$\vec u\bullet \vec v = 5$

$\cos \theta = \frac{5}{10}$

$\cos \theta = \frac{1}{2}$

Neither planes are parallel nor perpendicular to each other.