Respuesta :

Take into account that the scalar projection is given by:

[tex]comp_a\vec{b}=\frac{\vec{a}\cdot\vec{b}}{a}[/tex]

the magnitude of a vector is:

[tex]\begin{gathered} a=\sqrt[]{(-3^{})^2+4^2} \\ a=\sqrt[]{25}=5 \end{gathered}[/tex]

The dot product between a and b is:

[tex]\vec{a}\cdot\vec{b}=(-3,4)\cdot(7,9)=-3\cdot7+4\cdot9=-21+36=15[/tex]

Then, the scalar projection is:

[tex]\text{comp}_a\vec{b}=\frac{15}{5}=3[/tex]

Now, consider that the vector projections is given by:

[tex]\begin{gathered} \text{proj}_a\vec{b}=(\frac{\vec{a}\cdot\vec{b}}{a})\frac{a}{\lvert a\rvert} \\ \text{proj}_a\vec{b}=\frac{3}{5}\cdot(-3,4)=(-\frac{9}{5},\frac{12}{5}) \end{gathered}[/tex]

Hence, the answer is:

scalar projection of b onto a = 3

vector projection of b onto a = (-9/5 . 12/5)

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