Answer :
Let's tackle each part of the question step-by-step, providing a detailed solution to each.
### Part (a): Find the dot product [tex]\( v \cdot w \)[/tex]
Given vectors:
[tex]\[ v = \sqrt{3}i + 3j \][/tex]
[tex]\[ w = i + j \][/tex]
The dot product of two vectors [tex]\( v = [v_1, v_2] \)[/tex] and [tex]\( w = [w_1, w_2] \)[/tex] is given by:
[tex]\[ v \cdot w = v_1 \cdot w_1 + v_2 \cdot w_2 \][/tex]
In our case:
[tex]\[ v_1 = \sqrt{3}, \quad v_2 = 3 \][/tex]
[tex]\[ w_1 = 1, \quad w_2 = 1 \][/tex]
Now calculate the dot product:
[tex]\[ v \cdot w = (\sqrt{3} \cdot 1) + (3 \cdot 1) \][/tex]
[tex]\[ v \cdot w = \sqrt{3} + 3 \][/tex]
Hence:
[tex]\[ v \cdot w = \sqrt{3} + 3 \][/tex]
To give you the finalized value of dot product:
[tex]\[ v \cdot w = 4.732050807568877 \][/tex]
### Part (b): Find the angle [tex]\(\theta\)[/tex] 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]
First, calculate the magnitudes of [tex]\(v\)[/tex] and [tex]\(w\)[/tex]:
[tex]\[ \|v\| = \sqrt{(\sqrt{3})^2 + 3^2} = \sqrt{3 + 9} = \sqrt{12} = 2\sqrt{3} \][/tex]
[tex]\[ \|w\| = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \][/tex]
Now substitute the values:
[tex]\[ \cos \theta = \frac{\sqrt{3} + 3}{2\sqrt{3} \cdot \sqrt{2}} = \frac{4.732050807568877}{2\sqrt{6}} \][/tex]
[tex]\[ \cos \theta = \frac{4.732050807568877}{4.898979485566356} \approx 0.965925826289068 \][/tex]
Therefore:
[tex]\[ \theta = \arccos(0.965925826289068) \approx 0.2617993877991502 \text{ radians} \][/tex]
To convert to degrees:
[tex]\[ \theta \approx 15.000000000000043^\circ \][/tex]
### Part (c): State whether the vectors are parallel, orthogonal, or neither
Two vectors are:
- Parallel if the cosine of the angle between them is ±1.
- Orthogonal if the dot product is 0.
Given the calculations:
- [tex]\(\cos(\theta) = 0.965925826289068\)[/tex], which is neither 1 nor -1.
- The dot product [tex]\(\sqrt{3} + 3\)[/tex] is not 0.
Hence:
The vectors [tex]\(v\)[/tex] and [tex]\(w\)[/tex] are neither parallel nor orthogonal.
### Final Answer:
(a) The dot product [tex]\( v \cdot w = 4.732050807568877 \)[/tex]
(b) The angle between [tex]\( v \)[/tex] and [tex]\( w \)[/tex] is approximately [tex]\( 0.2617993877991502 \)[/tex] radians or [tex]\( 15.000000000000043^\circ \)[/tex] degrees.
(c) The vectors are neither parallel nor orthogonal.
### Part (a): Find the dot product [tex]\( v \cdot w \)[/tex]
Given vectors:
[tex]\[ v = \sqrt{3}i + 3j \][/tex]
[tex]\[ w = i + j \][/tex]
The dot product of two vectors [tex]\( v = [v_1, v_2] \)[/tex] and [tex]\( w = [w_1, w_2] \)[/tex] is given by:
[tex]\[ v \cdot w = v_1 \cdot w_1 + v_2 \cdot w_2 \][/tex]
In our case:
[tex]\[ v_1 = \sqrt{3}, \quad v_2 = 3 \][/tex]
[tex]\[ w_1 = 1, \quad w_2 = 1 \][/tex]
Now calculate the dot product:
[tex]\[ v \cdot w = (\sqrt{3} \cdot 1) + (3 \cdot 1) \][/tex]
[tex]\[ v \cdot w = \sqrt{3} + 3 \][/tex]
Hence:
[tex]\[ v \cdot w = \sqrt{3} + 3 \][/tex]
To give you the finalized value of dot product:
[tex]\[ v \cdot w = 4.732050807568877 \][/tex]
### Part (b): Find the angle [tex]\(\theta\)[/tex] 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]
First, calculate the magnitudes of [tex]\(v\)[/tex] and [tex]\(w\)[/tex]:
[tex]\[ \|v\| = \sqrt{(\sqrt{3})^2 + 3^2} = \sqrt{3 + 9} = \sqrt{12} = 2\sqrt{3} \][/tex]
[tex]\[ \|w\| = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \][/tex]
Now substitute the values:
[tex]\[ \cos \theta = \frac{\sqrt{3} + 3}{2\sqrt{3} \cdot \sqrt{2}} = \frac{4.732050807568877}{2\sqrt{6}} \][/tex]
[tex]\[ \cos \theta = \frac{4.732050807568877}{4.898979485566356} \approx 0.965925826289068 \][/tex]
Therefore:
[tex]\[ \theta = \arccos(0.965925826289068) \approx 0.2617993877991502 \text{ radians} \][/tex]
To convert to degrees:
[tex]\[ \theta \approx 15.000000000000043^\circ \][/tex]
### Part (c): State whether the vectors are parallel, orthogonal, or neither
Two vectors are:
- Parallel if the cosine of the angle between them is ±1.
- Orthogonal if the dot product is 0.
Given the calculations:
- [tex]\(\cos(\theta) = 0.965925826289068\)[/tex], which is neither 1 nor -1.
- The dot product [tex]\(\sqrt{3} + 3\)[/tex] is not 0.
Hence:
The vectors [tex]\(v\)[/tex] and [tex]\(w\)[/tex] are neither parallel nor orthogonal.
### Final Answer:
(a) The dot product [tex]\( v \cdot w = 4.732050807568877 \)[/tex]
(b) The angle between [tex]\( v \)[/tex] and [tex]\( w \)[/tex] is approximately [tex]\( 0.2617993877991502 \)[/tex] radians or [tex]\( 15.000000000000043^\circ \)[/tex] degrees.
(c) The vectors are neither parallel nor orthogonal.