Respuesta :

frika

If you have to find the angle between two vectors, first of all, you should draw them with their beginnings at the common point. Then the smallest of two possible angles is the angle between these vectors (see attached diagram for details).

Now, use the definition of dot product:

[tex]\vec{a}\cdot \vec{b}=|\vec{a}|\cdot |\vec{b}|\cdot \cos \theta_{\vec{a},\vec{b}}.[/tex]

Express the cosine from the previous expression:

[tex]\cos \theta_{\vec{a},\vec{b}}=\dfrac{\vec{a}\cdot \vec{b}}{|\vec{a}|\cdot |\vec{b}|}.[/tex]

Moreover, if  [tex]\vec{a}=(a_1,a_2,a_3)[/tex] and  [tex]\vec{b}=(b_1,b_2,b_3),[/tex]  then

[tex]\cos \theta_{\vec{a},\vec{b}}=\dfrac{a_1\cdot b_1+a_2\cdot b_2+a_3\cdot b_3}{\sqrt{a_1^2+a_2^2+a_3^2}\cdot \sqrt{b_1^2+b_2^2+b_3^2}}.[/tex]

Ver imagen frika

Let vectors A⃗ =(2,1,−4), B⃗ =(−3,0,1), What is the angle θab between a⃗ and b⃗ ? express your answer using one significant figure?

Answer:

134°

Step-by-step explanation:

Magnitude of A = [tex]\sqrt{A_x^2 + A_y^2+A_z^2} =\sqrt{2^2+1^2+(-4)^2} \\[/tex]

=4.58

Magnitude of B = [tex]=\sqrt{B_x^2+B_y^2+B_z^2} =\sqrt{(-3)^2+0^2+1^2}[/tex]

=3.16

Dot Product

[tex]\hat{A} \cdot \hat{B}= A_x \cdot B_x +A_y \cdot B_y +A_z \cdot B_z\\\\=2*-3+ 1*0+ (-4)*1[/tex]

=-10

the angle θab between a⃗ and b⃗ ⇒

θab = [tex]cos^{-1}(\frac{-10}{4.58 /cdot 3.16} )[/tex]

=134°