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Sagot :
To find the angle between vectors [tex]\(\vec{a}\)[/tex] and [tex]\(\vec{b}\)[/tex], we need to use the information given and apply some vector algebra and properties of the dot product.
Given:
1. [tex]\(\vec{a} + \vec{b} + \vec{c} = 0 \)[/tex]
2. [tex]\(|\vec{a}| = 6 \)[/tex]
3. [tex]\(|\vec{b}| = 7 \)[/tex]
4. [tex]\(|\vec{c}| = \sqrt{127} \)[/tex]
Since [tex]\(\vec{a} + \vec{b} + \vec{c} = 0 \)[/tex], it means:
[tex]\[ \vec{c} = -(\vec{a} + \vec{b}) \][/tex]
To find the magnitude of [tex]\(\vec{c}\)[/tex], we notice:
[tex]\[ |\vec{c}| = |- (\vec{a} + \vec{b})| = |\vec{a} + \vec{b}| \][/tex]
Given [tex]\(|\vec{c}| = \sqrt{127} \)[/tex], we use the property of magnitudes:
[tex]\[ |\vec{a} + \vec{b}| = \sqrt{127} \][/tex]
We know that the square of the magnitude of the sum of two vectors can be expressed as:
[tex]\[ |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2 \vec{a} \cdot \vec{b} \][/tex]
Substituting the values:
[tex]\[ 127 = 6^2 + 7^2 + 2 \vec{a} \cdot \vec{b} \][/tex]
[tex]\[ 127 = 36 + 49 + 2 \vec{a} \cdot \vec{b} \][/tex]
[tex]\[ 127 = 85 + 2 \vec{a} \cdot \vec{b} \][/tex]
Solving for [tex]\( \vec{a} \cdot \vec{b} \)[/tex]:
[tex]\[ 127 - 85 = 2 \vec{a} \cdot \vec{b} \][/tex]
[tex]\[ 42 = 2 \vec{a} \cdot \vec{b} \][/tex]
[tex]\[ \vec{a} \cdot \vec{b} = 21 \][/tex]
Now, the dot product [tex]\( \vec{a} \cdot \vec{b} \)[/tex] can also be written as:
[tex]\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(\theta) \][/tex]
where [tex]\( \theta \)[/tex] is the angle between [tex]\(\vec{a}\)[/tex] and [tex]\(\vec{b}\)[/tex].
We substitute the known values and solve for [tex]\(\cos(\theta)\)[/tex]:
[tex]\[ 21 = 6 \times 7 \times \cos(\theta) \][/tex]
[tex]\[ 21 = 42 \cos(\theta) \][/tex]
[tex]\[ \cos(\theta) = \frac{21}{42} \][/tex]
[tex]\[ \cos(\theta) = 0.5 \][/tex]
The angle [tex]\(\theta\)[/tex] is found by taking the inverse cosine (arccos) of [tex]\(0.5\)[/tex]:
[tex]\[ \theta = \arccos(0.5) \][/tex]
In degrees, [tex]\(\arccos(0.5)\)[/tex] is:
[tex]\[ \theta = 60^\circ \][/tex]
So, the angle between [tex]\(\vec{a}\)[/tex] and [tex]\(\vec{b}\)[/tex] is [tex]\(60^\circ\)[/tex].
Given:
1. [tex]\(\vec{a} + \vec{b} + \vec{c} = 0 \)[/tex]
2. [tex]\(|\vec{a}| = 6 \)[/tex]
3. [tex]\(|\vec{b}| = 7 \)[/tex]
4. [tex]\(|\vec{c}| = \sqrt{127} \)[/tex]
Since [tex]\(\vec{a} + \vec{b} + \vec{c} = 0 \)[/tex], it means:
[tex]\[ \vec{c} = -(\vec{a} + \vec{b}) \][/tex]
To find the magnitude of [tex]\(\vec{c}\)[/tex], we notice:
[tex]\[ |\vec{c}| = |- (\vec{a} + \vec{b})| = |\vec{a} + \vec{b}| \][/tex]
Given [tex]\(|\vec{c}| = \sqrt{127} \)[/tex], we use the property of magnitudes:
[tex]\[ |\vec{a} + \vec{b}| = \sqrt{127} \][/tex]
We know that the square of the magnitude of the sum of two vectors can be expressed as:
[tex]\[ |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2 \vec{a} \cdot \vec{b} \][/tex]
Substituting the values:
[tex]\[ 127 = 6^2 + 7^2 + 2 \vec{a} \cdot \vec{b} \][/tex]
[tex]\[ 127 = 36 + 49 + 2 \vec{a} \cdot \vec{b} \][/tex]
[tex]\[ 127 = 85 + 2 \vec{a} \cdot \vec{b} \][/tex]
Solving for [tex]\( \vec{a} \cdot \vec{b} \)[/tex]:
[tex]\[ 127 - 85 = 2 \vec{a} \cdot \vec{b} \][/tex]
[tex]\[ 42 = 2 \vec{a} \cdot \vec{b} \][/tex]
[tex]\[ \vec{a} \cdot \vec{b} = 21 \][/tex]
Now, the dot product [tex]\( \vec{a} \cdot \vec{b} \)[/tex] can also be written as:
[tex]\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(\theta) \][/tex]
where [tex]\( \theta \)[/tex] is the angle between [tex]\(\vec{a}\)[/tex] and [tex]\(\vec{b}\)[/tex].
We substitute the known values and solve for [tex]\(\cos(\theta)\)[/tex]:
[tex]\[ 21 = 6 \times 7 \times \cos(\theta) \][/tex]
[tex]\[ 21 = 42 \cos(\theta) \][/tex]
[tex]\[ \cos(\theta) = \frac{21}{42} \][/tex]
[tex]\[ \cos(\theta) = 0.5 \][/tex]
The angle [tex]\(\theta\)[/tex] is found by taking the inverse cosine (arccos) of [tex]\(0.5\)[/tex]:
[tex]\[ \theta = \arccos(0.5) \][/tex]
In degrees, [tex]\(\arccos(0.5)\)[/tex] is:
[tex]\[ \theta = 60^\circ \][/tex]
So, the angle between [tex]\(\vec{a}\)[/tex] and [tex]\(\vec{b}\)[/tex] is [tex]\(60^\circ\)[/tex].
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