Question

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

Answer 1

Is this vector calculus? If so, please re-write your two numerical expressions as vectors. Otherwise other people will think you're discussing points (x,y).

Answer 2

This is what is given in the question, I am not sure how to write them otherwise

Answer 3

try using notation <2,6> for vectors

Answer 4

reference:
https://en.wikipedia.org/wiki/Vector_projection#Vector_projection

Answer 5

first determine magnitude of both vectors
\[|<2,6>| = \sqrt{2^2 + 6^2} = 2 \sqrt{10}\]
\[|<5,-1>| = \sqrt{5^2 + 1^2} = \sqrt{26}\]
Next find dot product:
\[\rightarrow (2*5) + (6*-1) = 4\]
Determine unit vector of 2nd vector
\[\rightarrow <\frac{5}{\sqrt{26}}, -\frac{1}{\sqrt{26}}>\]
Now plug into projection equation:
\[(2 \sqrt{10})(\frac{4}{2 \sqrt{10} \sqrt{26}}) <\frac{5}{\sqrt{26}},-\frac{1}{\sqrt{26}}>\]
Finally the projection vector is:
\[<\frac{10}{13}, -\frac{2}{13}>\]

Answer 6

"best approximation of the projection "

What does that mean?

It IS the projection. It's not an approximation.