Question

What is the best approximation of the projection of (5,-1) onto (2,6)?

Answer 1

Answer:

Hence, the scalar projection of $\vec a$ onto $\vec b$= $\frac{\sqrt{10} }{5}$, and  the vector projection of $\vec a$ onto $\vec b$ = $\frac{1}{5} \hat i+\frac{3}{5} \hat j$.

Step-by-step explanation:

We have given two points  $(5, -1)$ and $(2, 6)$.

Let,     $\vec a=5\hat {i}-\hat {j}$  and  $\vec b= 2\hat {i}+6\hat{j}$ .

and we have calculate the projection of $\vec a$ onto $\vec b$.

Now,

For the calculation of projection, first we need to calculate the dot product of  $\vec a$  and $\vec b$.

$\vec a.\vec b=(5\hat {i}-\hat{j}).(2\hat{i}+6\hat{j})$

     $=10-6$

     $=4$

then, we have to calculate the magnitude of $\vec b$.

   $\mid {\vec {b}}\mid$ = $\sqrt{2^{2}+6^{2} }$ = $\sqrt{40}$ = $2\sqrt{10}$.

Now, the scalar projection of $\vec a$ onto $\vec b$ = $\frac{\vec a.\vec b}{\mid b\mid}$

                                                                 = $\frac{4}{2\sqrt{10} }$$\frac{2}{\sqrt{10} } \times\frac{\sqrt{10} }{\sqrt{10} } =\frac{2\sqrt{10} }{10} = \frac{\sqrt{10} }{5}$

and the vector projection of $\vec a$ onto $\vec b$ = $\frac{\vec a. \vec b}{\mid\vec b \mid^{2} } . \vec b$

                                                               = $\frac{4}{40} . (2\hat i+ 6\hat j)$

                                                                = $\frac{1}{5} \hat i+\frac{3}{5} \hat j$

Hence, the scalar projection of $\vec a$ onto $\vec b$= $\frac{\sqrt{10} }{5}$, and  the vector projection of $\vec a$ onto $\vec b$ = $\frac{1}{5} \hat i+\frac{3}{5} \hat j$.