Let's go through the steps to solve the given question step-by-step.
### (a) Calculate the dot product [tex]\(v \cdot w\)[/tex]
Given vectors:
[tex]\[ v = i - j \][/tex]
[tex]\[ w = -i - j \][/tex]
The dot product of two vectors [tex]\( \mathbf{a} = a_1 i + a_2 j \)[/tex] and [tex]\( \mathbf{b} = b_1 i + b_2 j \)[/tex] is given by:
[tex]\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \][/tex]
For [tex]\(v\)[/tex] and [tex]\(w\)[/tex]:
[tex]\[
v \cdot w = (1)(-1) + (-1)(-1)
= -1 + 1
= 0
\][/tex]
So, the dot product [tex]\( v \cdot w = 0 \)[/tex].
### (b) Calculate the angle between [tex]\(v\)[/tex] and [tex]\(w\)[/tex]
The angle [tex]\( \theta \)[/tex] between two vectors can be found using the formula:
[tex]\[ \cos \theta = \frac{v \cdot w}{\|v\| \|w\|} \][/tex]
Firstly, we need the magnitudes of [tex]\(v\)[/tex] and [tex]\(w\)[/tex].
The magnitude of vector [tex]\( v = i - j \)[/tex]:
[tex]\[ \|v\| = \sqrt{(1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \][/tex]
The magnitude of vector [tex]\( w = -i - j \)[/tex]:
[tex]\[ \|w\| = \sqrt{(-1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \][/tex]
Given that the dot product [tex]\( v \cdot w = 0 \)[/tex], we substitute into the formula:
[tex]\[ \cos \theta = \frac{0}{\sqrt{2} \cdot \sqrt{2}} = \frac{0}{2} = 0 \][/tex]
Solving for [tex]\( \theta \)[/tex]:
[tex]\[ \theta = \arccos(0) = \frac{\pi}{2} \ \text{radians} = 90 \ \text{degrees} \][/tex]
So, the angle between [tex]\( v \)[/tex] and [tex]\( w \)[/tex] is [tex]\( \frac{\pi}{2} \)[/tex] radians or 90 degrees.
### (c) Determine if the vectors are parallel, orthogonal, or neither
Vectors are:
- Parallel if they have the same or exact opposite direction, i.e., their scalar multiples are identical.
- Orthogonal if their dot product is zero.
- Neither if they are neither parallel nor orthogonal.
Given [tex]\( v \cdot w = 0 \)[/tex], the vectors are orthogonal.
So, the vectors [tex]\( v \)[/tex] and [tex]\( w \)[/tex] are orthogonal.
### Summary
(a) The dot product [tex]\( v \cdot w = 0 \)[/tex] \\
(b) The angle between [tex]\( v \)[/tex] and [tex]\( w \)[/tex] is 90 degrees ([tex]\( \frac{\pi}{2} \)[/tex] radians) \\
(c) The vectors are orthogonal.
