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Sagot :
Sure, let's go step-by-step through the solution.
### Given Vectors:
[tex]\[ v = \sqrt{3}i + 3j \][/tex]
[tex]\[ w = -i + j \][/tex]
#### Part (a): Finding the Dot Product [tex]\( v \cdot w \)[/tex]
The dot product of two vectors [tex]\( v = [v_1, v_2] \)[/tex] and [tex]\( w = [w_1, w_2] \)[/tex] is calculated as:
[tex]\[ v \cdot w = v_1w_1 + v_2w_2 \][/tex]
For our vectors [tex]\( v \)[/tex] and [tex]\( w \)[/tex]:
[tex]\[ v = [\sqrt{3}, 3] \][/tex]
[tex]\[ w = [-1, 1] \][/tex]
So,
[tex]\[ v \cdot w = (\sqrt{3} \cdot -1) + (3 \cdot 1) \][/tex]
[tex]\[ v \cdot w = -\sqrt{3} + 3 \][/tex]
To simplify:
[tex]\[ v \cdot w = 3 - \sqrt{3} \][/tex]
Therefore, the exact dot product is:
[tex]\[ v \cdot w = 3 - \sqrt{3} \][/tex]
### Given Result Verification
The calculated numerical result for the dot product using the necessary computations simplifies to approximately:
[tex]\[ 1.2679491924311228 \][/tex]
matching the provided answer in the numerical format.
### Continuing to Parts (b) and (c)
### Part (b): Finding the Angle Between [tex]\( v \)[/tex] and [tex]\( w \)[/tex]
First, we calculate the magnitudes (or norms) of [tex]\( v \)[/tex] and [tex]\( w \)[/tex].
The magnitude of [tex]\( v \)[/tex] is:
[tex]\[ \| v \| = \sqrt{(\sqrt{3})^2 + 3^2} = \sqrt{3 + 9} = \sqrt{12} = 2\sqrt{3} \][/tex]
The magnitude of [tex]\( w \)[/tex] is:
[tex]\[ \| w \| = \sqrt{(-1)^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \][/tex]
The cosine of the angle [tex]\( \theta \)[/tex] between the vectors is given by:
[tex]\[ \cos\theta = \frac{v \cdot w}{\| v \| \| w \|} \][/tex]
Substituting the values:
[tex]\[ \cos\theta = \frac{3 - \sqrt{3}}{2\sqrt{3} \cdot \sqrt{2}} \][/tex]
This simplifies to:
[tex]\[ \cos\theta = \frac{3 - \sqrt{3}}{2\sqrt{6}} \][/tex]
To find the angle [tex]\( \theta \)[/tex], we use the arccosine function:
[tex]\[ \theta = \arccos\left(\frac{3 - \sqrt{3}}{2\sqrt{6}}\right) \][/tex]
The result for this angle in degrees is:
[tex]\[ \theta \approx 75^\circ \][/tex]
### Part (c): Determining the Relationship Between the Vectors
Two vectors are orthogonal if their dot product is zero. In our case:
[tex]\[ v \cdot w = 1.2679491924311228 \][/tex]
which is not zero, so they are not orthogonal.
Two vectors are parallel if the cosine of the angle between them is either 1 or -1. In our case:
[tex]\[ \cos\theta \approx 0.258819 \][/tex]
which is not 1 or -1, so they are not parallel.
Therefore, the vectors [tex]\( v \)[/tex] and [tex]\( w \)[/tex] are neither parallel nor orthogonal.
### Summary:
(a) [tex]\( v \cdot w = 3 - \sqrt{3} \)[/tex]
(b) The angle between [tex]\( v \)[/tex] and [tex]\( w \)[/tex] is [tex]\( 75^\circ \)[/tex]
(c) The vectors are neither parallel nor orthogonal.
### Given Vectors:
[tex]\[ v = \sqrt{3}i + 3j \][/tex]
[tex]\[ w = -i + j \][/tex]
#### Part (a): Finding the Dot Product [tex]\( v \cdot w \)[/tex]
The dot product of two vectors [tex]\( v = [v_1, v_2] \)[/tex] and [tex]\( w = [w_1, w_2] \)[/tex] is calculated as:
[tex]\[ v \cdot w = v_1w_1 + v_2w_2 \][/tex]
For our vectors [tex]\( v \)[/tex] and [tex]\( w \)[/tex]:
[tex]\[ v = [\sqrt{3}, 3] \][/tex]
[tex]\[ w = [-1, 1] \][/tex]
So,
[tex]\[ v \cdot w = (\sqrt{3} \cdot -1) + (3 \cdot 1) \][/tex]
[tex]\[ v \cdot w = -\sqrt{3} + 3 \][/tex]
To simplify:
[tex]\[ v \cdot w = 3 - \sqrt{3} \][/tex]
Therefore, the exact dot product is:
[tex]\[ v \cdot w = 3 - \sqrt{3} \][/tex]
### Given Result Verification
The calculated numerical result for the dot product using the necessary computations simplifies to approximately:
[tex]\[ 1.2679491924311228 \][/tex]
matching the provided answer in the numerical format.
### Continuing to Parts (b) and (c)
### Part (b): Finding the Angle Between [tex]\( v \)[/tex] and [tex]\( w \)[/tex]
First, we calculate the magnitudes (or norms) of [tex]\( v \)[/tex] and [tex]\( w \)[/tex].
The magnitude of [tex]\( v \)[/tex] is:
[tex]\[ \| v \| = \sqrt{(\sqrt{3})^2 + 3^2} = \sqrt{3 + 9} = \sqrt{12} = 2\sqrt{3} \][/tex]
The magnitude of [tex]\( w \)[/tex] is:
[tex]\[ \| w \| = \sqrt{(-1)^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \][/tex]
The cosine of the angle [tex]\( \theta \)[/tex] between the vectors is given by:
[tex]\[ \cos\theta = \frac{v \cdot w}{\| v \| \| w \|} \][/tex]
Substituting the values:
[tex]\[ \cos\theta = \frac{3 - \sqrt{3}}{2\sqrt{3} \cdot \sqrt{2}} \][/tex]
This simplifies to:
[tex]\[ \cos\theta = \frac{3 - \sqrt{3}}{2\sqrt{6}} \][/tex]
To find the angle [tex]\( \theta \)[/tex], we use the arccosine function:
[tex]\[ \theta = \arccos\left(\frac{3 - \sqrt{3}}{2\sqrt{6}}\right) \][/tex]
The result for this angle in degrees is:
[tex]\[ \theta \approx 75^\circ \][/tex]
### Part (c): Determining the Relationship Between the Vectors
Two vectors are orthogonal if their dot product is zero. In our case:
[tex]\[ v \cdot w = 1.2679491924311228 \][/tex]
which is not zero, so they are not orthogonal.
Two vectors are parallel if the cosine of the angle between them is either 1 or -1. In our case:
[tex]\[ \cos\theta \approx 0.258819 \][/tex]
which is not 1 or -1, so they are not parallel.
Therefore, the vectors [tex]\( v \)[/tex] and [tex]\( w \)[/tex] are neither parallel nor orthogonal.
### Summary:
(a) [tex]\( v \cdot w = 3 - \sqrt{3} \)[/tex]
(b) The angle between [tex]\( v \)[/tex] and [tex]\( w \)[/tex] is [tex]\( 75^\circ \)[/tex]
(c) The vectors are neither parallel nor orthogonal.
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