Respuesta :
Answer:
[tex]comp_ab=\left(\dfrac{-5}{7}\right)[/tex]
[tex]Proj_ab=\left<\dfrac{15}{49},\dfrac{30}{49},\dfrac{-10}{49}\right>[/tex].
Step-by-step explanation:
The given vectors are
[tex]\vec{a}=<-3,-6,2>,\vec{b}=<1,1,2>[/tex]
Now,
[tex]|\vec{a}|=\sqrt{(-3)^2+(-6)^2+(2)^2}[/tex]
[tex]|\vec{a}|=\sqrt{9+36+4}[/tex]
[tex]|\vec{a}|=\sqrt{49}[/tex]
[tex]|\vec{a}|=7[/tex]
[tex]\vec{a}\cdot \vec{b}=(-3)(1)+(-6)(1)+(2)(2)[/tex]
[tex]\vec{a}\cdot \vec{b}=-3-6+4[/tex]
[tex]\vec{a}\cdot \vec{b}=-5[/tex]
The scalar projection of b onto a is
[tex]comp_ab=\left(\dfrac{\vec{a}\cdot \vec{b}}{|\vec{a}|}\right)[/tex]
[tex]comp_ab=\left(\dfrac{-5}{7}\right)[/tex]
Vector projection of b onto a is
[tex]Proj_ab=\left(\dfrac{\vec{a}\cdot \vec{b}}{|\vec{a}|}\right)\left(\dfrac{\vec{a}}{|\vec {a}|}\right)[/tex]
[tex]Proj_ab=\left(\dfrac{-5}{7}\right)\left(\dfrac{<-3,-6,2>}{7}\right)[/tex]
[tex]Proj_ab=\left<\dfrac{15}{49},\dfrac{30}{49},\dfrac{-10}{49}\right>[/tex]
Therefore, scale and vector projections of b onto a are [tex]comp_ab=\left(\dfrac{-5}{7}\right)[/tex] and [tex]Proj_ab=\left<\dfrac{15}{49},\dfrac{30}{49},\dfrac{-10}{49}\right>[/tex] receptively.
