Let's address each of the parts of the question step-by-step:
### b. Can the sum of two equal vectors be equal to either of the vectors?
Given two equal vectors [tex]\(\vec{A}\)[/tex] and [tex]\(\vec{A}\)[/tex], their sum is:
[tex]\[
\vec{A} + \vec{A} = 2\vec{A}
\][/tex]
If we compare [tex]\(2\vec{A}\)[/tex] with [tex]\(\vec{A}\)[/tex], we can see that [tex]\(2\vec{A}\)[/tex] is not equal to [tex]\(\vec{A}\)[/tex] because it is twice the magnitude of [tex]\(\vec{A}\)[/tex].
Hence, the sum of two equal vectors [tex]\(\vec{A} + \vec{A}\)[/tex] cannot be equal to either one of the original vectors. Thus, the answer to this question is False.
### c. If [tex]\(\vec{A} \cdot \vec{B} = 0\)[/tex], what is the angle between [tex]\(\vec{A}\)[/tex] and [tex]\(\vec{B}\)[/tex]?
When the dot product of two vectors [tex]\(\vec{A}\)[/tex] and [tex]\(\vec{B}\)[/tex] is zero, it indicates that the vectors are orthogonal (perpendicular) to each other. The dot product formula is given by:
[tex]\[
\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta
\][/tex]
Where [tex]\(\theta\)[/tex] is the angle between the vectors [tex]\(\vec{A}\)[/tex] and [tex]\(\vec{B}\)[/tex]. Given [tex]\(\vec{A} \cdot \vec{B} = 0\)[/tex], we can deduce:
[tex]\[
0 = |\vec{A}| |\vec{B}| \cos \theta
\][/tex]
Since the magnitudes [tex]\(|\vec{A}|\)[/tex] and [tex]\(|\vec{B}|\)[/tex] are not zero, the equation reduces to:
[tex]\[
\cos \theta = 0
\][/tex]
The angle [tex]\(\theta\)[/tex] for which [tex]\(\cos \theta = 0\)[/tex] is:
[tex]\[
\theta = 90^\circ
\][/tex]
Therefore, the angle between [tex]\(\vec{A}\)[/tex] and [tex]\(\vec{B}\)[/tex] is [tex]\(90\)[/tex] degrees.
### a. If [tex]\(\hat{i}, \hat{j}\)[/tex], and [tex]\(\hat{k}\)[/tex] are unit vectors along the [tex]\(x\)[/tex], [tex]\(y\)[/tex], and [tex]\(z\)[/tex]-axis respectively, find [tex]\(\hat{i} \cdot (\hat{j} \times \hat{k})\)[/tex].
The cross product [tex]\(\hat{j} \times \hat{k}\)[/tex] yields a vector. The unit vectors [tex]\(\hat{j}\)[/tex] and [tex]\(\hat{k}\)[/tex] follow the right-hand rule in a Cartesian coordinate system. According to the right-hand rule, the cross product [tex]\(\hat{j} \times \hat{k}\)[/tex] results in the unit vector along the [tex]\(x\)[/tex]-axis:
[tex]\[
\hat{j} \times \hat{k} = \hat{i}
\][/tex]
Then, the dot product of [tex]\(\hat{i}\)[/tex] with itself is:
[tex]\[
\hat{i} \cdot \hat{i} = |\hat{i}| |\hat{i}| \cos 0^\circ = 1 \cdot 1 \cdot 1 = 1
\][/tex]
Therefore:
[tex]\[
\hat{i} \cdot (\hat{j} \times \hat{k}) = 1
\][/tex]
So, the result for [tex]\(\hat{i} \cdot (\hat{j} \times \hat{k})\)[/tex] is [tex]\(1\)[/tex].
