Respuesta :
The formula of the dot product of two vectors:
[tex] \overrightarrow{a}\circ\overrightarrow{b}=|\overrightarrow{a}|\cdot|\overrightarrow{b}|\cos\alpha\to\cos\alpha=\dfrac{\overrightarrow{a}\circ\overrightarrow{b}}{|\overrightarrow{a}|\cdot|\overrightarrow{b}|} [/tex]
We have:
[tex] \overrightarrow{a}=[1,\ -4,\ 1]\\\\\overrightarrow{b}=[0,\ 3,\ -3] [/tex]
Calculate the dot product:
[tex] \overrightarrow{a}\circ\overrightarrow{b}=1\cdot0+(-4)\cdot3+1\cdot(-3)=-15 [/tex]
Calculate the lengths of vectors:
[tex] |\overrightarrow{a}|=\sqrt{1^2+(-4)^2+1^2}=\sqrt{18}=\sqrt{9\cdot2}=\sqrt9\cdot\sqrt2=3\sqrt2\\\\|\overrightarrow{b}|=\sqrt{0^2+3^2+(-3)^2}=\sqrt{18}=\sqrt{9\cdot2}=\sqrt9\cdot\sqrt2=3\sqrt2 [/tex]
Substitute:
[tex] \cos\alpha=\dfrac{-15}{3\sqrt2\cdot3\sqrt2}=\dfrac{-15}{9\sqrt4}=-\dfrac{15}{18}\approx-0.8333\to\alpha\approx146^o [/tex]
Answer: Measure of the angle between the vectors a and b is equal 146°.
The angle between the two vectors given is of 146 degrees.
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- Suppose there are two vectors, a and b.
- The angle between them is [tex]\theta[/tex], given by the following expression:
[tex]\cos{\theta} = \frac{u.v}{|u||v|}[/tex]
In which
- u.v is the dot product between them.
- |u| and |v| are their norms.
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- The vectors are: [tex]a = (1,-4,1), b = (0, 3, -3)[/tex]
- The dot product is: [tex]ab = (1, -4, 1).(0, 3, -3) = 1\times0 -4\times3 + 1\times(-3) = -15[/tex]
- The norms are:
[tex]|u| = \sqrt{1^2 + (-4)^2 + 1^2} = \sqrt{18} = \sqrt{2 \times 9} = \sqrt{2}\sqrt{9} = 3\sqrt{2}[/tex]
[tex]|v| = \sqrt{0^2 + 3^2 + (-3)^2} = \sqrt{18} = 3\sqrt{2}[/tex]
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Finding the angle:
[tex]\cos{\theta} = \frac{u.v}{|u||v|}[/tex]
[tex]\cos{\theta} = \frac{-15}{(3\sqrt{2})^2}[/tex]
[tex]\cos{\theta} = -\frac{15}{18}[/tex]
[tex]\cos{\theta} = -\frac{5}{6}[/tex]
[tex]\theta = \cos^{-1}{(-\frac{5}{6})}[/tex]}
Using a calculator:
[tex]\theta = 146[/tex]
The angle between the two vectors is of 146 degrees.
A similar problem is given at https://brainly.com/question/15006306
