Respuesta :
Answer:
Hence, the scalar projection of [tex]\vec a[/tex] onto [tex]\vec b[/tex]= [tex]\frac{\sqrt{10} }{5}[/tex], and the vector projection of [tex]\vec a[/tex] onto [tex]\vec b[/tex] = [tex]\frac{1}{5} \hat i+\frac{3}{5} \hat j[/tex].
Step-by-step explanation:
We have given two points [tex](5, -1)[/tex] and [tex](2, 6)[/tex].
Let, [tex]\vec a=5\hat {i}-\hat {j}[/tex] and [tex]\vec b= 2\hat {i}+6\hat{j}[/tex] .
and we have calculate the projection of [tex]\vec a[/tex] onto [tex]\vec b[/tex].
Now,
For the calculation of projection, first we need to calculate the dot product of [tex]\vec a[/tex] and [tex]\vec b[/tex].
[tex]\vec a.\vec b=(5\hat {i}-\hat{j}).(2\hat{i}+6\hat{j})[/tex]
[tex]=10-6[/tex]
[tex]=4[/tex]
then, we have to calculate the magnitude of [tex]\vec b[/tex].
[tex]\mid {\vec {b}}\mid[/tex] = [tex]\sqrt{2^{2}+6^{2} }[/tex] = [tex]\sqrt{40}[/tex] = [tex]2\sqrt{10}[/tex].
Now, the scalar projection of [tex]\vec a[/tex] onto [tex]\vec b[/tex] = [tex]\frac{\vec a.\vec b}{\mid b\mid}[/tex]
= [tex]\frac{4}{2\sqrt{10} }[/tex][tex]\frac{2}{\sqrt{10} } \times\frac{\sqrt{10} }{\sqrt{10} } =\frac{2\sqrt{10} }{10} = \frac{\sqrt{10} }{5}[/tex]
and the vector projection of [tex]\vec a[/tex] onto [tex]\vec b[/tex] = [tex]\frac{\vec a. \vec b}{\mid\vec b \mid^{2} } . \vec b[/tex]
= [tex]\frac{4}{40} . (2\hat i+ 6\hat j)[/tex]
= [tex]\frac{1}{5} \hat i+\frac{3}{5} \hat j[/tex]
Hence, the scalar projection of [tex]\vec a[/tex] onto [tex]\vec b[/tex]= [tex]\frac{\sqrt{10} }{5}[/tex], and the vector projection of [tex]\vec a[/tex] onto [tex]\vec b[/tex] = [tex]\frac{1}{5} \hat i+\frac{3}{5} \hat j[/tex].
