Answer :
Let's go step by step to solve the given problem with vectors:
Given:
[tex]\[ \mathbf{v} = \mathbf{i} - \mathbf{j} \][/tex]
[tex]\[ \mathbf{w} = -\mathbf{i} - \mathbf{j} \][/tex]
### (a) Find the dot product [tex]\(\mathbf{v} \cdot \mathbf{w}\)[/tex]
The dot product of two vectors [tex]\(\mathbf{v}\)[/tex] and [tex]\(\mathbf{w}\)[/tex] is given by:
[tex]\[ \mathbf{v} \cdot \mathbf{w} = v_1 w_1 + v_2 w_2 \][/tex]
For [tex]\(\mathbf{v}\)[/tex], the components are:
[tex]\[ \mathbf{v} = 1\mathbf{i} - 1\mathbf{j} \][/tex]
So, [tex]\( v_1 = 1 \)[/tex] and [tex]\( v_2 = -1 \)[/tex].
For [tex]\(\mathbf{w}\)[/tex], the components are:
[tex]\[ \mathbf{w} = -1\mathbf{i} - 1\mathbf{j} \][/tex]
So, [tex]\( w_1 = -1 \)[/tex] and [tex]\( w_2 = -1 \)[/tex].
Now, calculate the dot product:
[tex]\[ \mathbf{v} \cdot \mathbf{w} = (1)(-1) + (-1)(-1) = -1 + 1 = 0 \][/tex]
So, the dot product is:
[tex]\[ \mathbf{v} \cdot \mathbf{w} = 0 \][/tex]
### (b) Find the angle between [tex]\(\mathbf{v}\)[/tex] and [tex]\(\mathbf{w}\)[/tex]
The angle [tex]\(\theta\)[/tex] between two vectors can be found using the dot product formula:
[tex]\[ \cos(\theta) = \frac{\mathbf{v} \cdot \mathbf{w}}{\|\mathbf{v}\| \|\mathbf{w}\|} \][/tex]
We've already calculated that [tex]\(\mathbf{v} \cdot \mathbf{w} = 0\)[/tex].
Next, we need the magnitudes (norms) of the vectors [tex]\(\|\mathbf{v}\|\)[/tex] and [tex]\(\|\mathbf{w}\|\)[/tex]:
[tex]\[ \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2} = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \][/tex]
[tex]\[ \|\mathbf{w}\| = \sqrt{w_1^2 + w_2^2} = \sqrt{(-1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \][/tex]
Now, using the dot product formula:
[tex]\[ \cos(\theta) = \frac{0}{\sqrt{2} \cdot \sqrt{2}} = \frac{0}{2} = 0 \][/tex]
To find the angle:
[tex]\[ \theta = \arccos(0) = \frac{\pi}{2} \text{ radians} \][/tex]
Converting radians to degrees:
[tex]\[ \theta = \frac{\pi}{2} \times \frac{180}{\pi} = 90^\circ \][/tex]
The angle between [tex]\(\mathbf{v}\)[/tex] and [tex]\(\mathbf{w}\)[/tex] is:
[tex]\[ \theta = 90^\circ \][/tex]
### (c) State whether the vectors are parallel, orthogonal, or neither
Two vectors are orthogonal if their dot product is 0. In this case:
[tex]\[ \mathbf{v} \cdot \mathbf{w} = 0 \][/tex]
Since the dot product is zero, the vectors [tex]\(\mathbf{v}\)[/tex] and [tex]\(\mathbf{w}\)[/tex] are orthogonal.
### Summary:
(a) The dot product [tex]\(\mathbf{v} \cdot \mathbf{w} = 0\)[/tex].
(b) The angle between [tex]\(\mathbf{v}\)[/tex] and [tex]\(\mathbf{w}\)[/tex] is [tex]\(90^\circ\)[/tex].
(c) The vectors are orthogonal.
Given:
[tex]\[ \mathbf{v} = \mathbf{i} - \mathbf{j} \][/tex]
[tex]\[ \mathbf{w} = -\mathbf{i} - \mathbf{j} \][/tex]
### (a) Find the dot product [tex]\(\mathbf{v} \cdot \mathbf{w}\)[/tex]
The dot product of two vectors [tex]\(\mathbf{v}\)[/tex] and [tex]\(\mathbf{w}\)[/tex] is given by:
[tex]\[ \mathbf{v} \cdot \mathbf{w} = v_1 w_1 + v_2 w_2 \][/tex]
For [tex]\(\mathbf{v}\)[/tex], the components are:
[tex]\[ \mathbf{v} = 1\mathbf{i} - 1\mathbf{j} \][/tex]
So, [tex]\( v_1 = 1 \)[/tex] and [tex]\( v_2 = -1 \)[/tex].
For [tex]\(\mathbf{w}\)[/tex], the components are:
[tex]\[ \mathbf{w} = -1\mathbf{i} - 1\mathbf{j} \][/tex]
So, [tex]\( w_1 = -1 \)[/tex] and [tex]\( w_2 = -1 \)[/tex].
Now, calculate the dot product:
[tex]\[ \mathbf{v} \cdot \mathbf{w} = (1)(-1) + (-1)(-1) = -1 + 1 = 0 \][/tex]
So, the dot product is:
[tex]\[ \mathbf{v} \cdot \mathbf{w} = 0 \][/tex]
### (b) Find the angle between [tex]\(\mathbf{v}\)[/tex] and [tex]\(\mathbf{w}\)[/tex]
The angle [tex]\(\theta\)[/tex] between two vectors can be found using the dot product formula:
[tex]\[ \cos(\theta) = \frac{\mathbf{v} \cdot \mathbf{w}}{\|\mathbf{v}\| \|\mathbf{w}\|} \][/tex]
We've already calculated that [tex]\(\mathbf{v} \cdot \mathbf{w} = 0\)[/tex].
Next, we need the magnitudes (norms) of the vectors [tex]\(\|\mathbf{v}\|\)[/tex] and [tex]\(\|\mathbf{w}\|\)[/tex]:
[tex]\[ \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2} = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \][/tex]
[tex]\[ \|\mathbf{w}\| = \sqrt{w_1^2 + w_2^2} = \sqrt{(-1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \][/tex]
Now, using the dot product formula:
[tex]\[ \cos(\theta) = \frac{0}{\sqrt{2} \cdot \sqrt{2}} = \frac{0}{2} = 0 \][/tex]
To find the angle:
[tex]\[ \theta = \arccos(0) = \frac{\pi}{2} \text{ radians} \][/tex]
Converting radians to degrees:
[tex]\[ \theta = \frac{\pi}{2} \times \frac{180}{\pi} = 90^\circ \][/tex]
The angle between [tex]\(\mathbf{v}\)[/tex] and [tex]\(\mathbf{w}\)[/tex] is:
[tex]\[ \theta = 90^\circ \][/tex]
### (c) State whether the vectors are parallel, orthogonal, or neither
Two vectors are orthogonal if their dot product is 0. In this case:
[tex]\[ \mathbf{v} \cdot \mathbf{w} = 0 \][/tex]
Since the dot product is zero, the vectors [tex]\(\mathbf{v}\)[/tex] and [tex]\(\mathbf{w}\)[/tex] are orthogonal.
### Summary:
(a) The dot product [tex]\(\mathbf{v} \cdot \mathbf{w} = 0\)[/tex].
(b) The angle between [tex]\(\mathbf{v}\)[/tex] and [tex]\(\mathbf{w}\)[/tex] is [tex]\(90^\circ\)[/tex].
(c) The vectors are orthogonal.