Question

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

Answer 1

Hence, the scalar projection of
`\vec a` onto
`\vec b`=
`(√(10) )/(5)`, and the vector projection of
`\vec a` onto
`\vec b` =
`(1)/(5) \hat i+(3)/(5) \hat j`.

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) }` =
`√(40)` =
`2√(10)`.

Now, the scalar projection of
`\vec a` onto
`\vec b` =
`(\vec a.\vec b)/(\mid b\mid)`

=
`(4)/(2√(10) )`
`(2)/(√(10) ) *(√(10) )/(√(10) ) =(2√(10) )/(10) = (√(10) )/(5)`

and the vector projection of
`\vec a` onto
`\vec b` =
`(\vec a. \vec b)/(\mid\vec b \mid^(2) ) . \vec b`

=
`(4)/(40) . (2\hat i+ 6\hat j)`

=
`(1)/(5) \hat i+(3)/(5) \hat j`

Hence, the scalar projection of
`\vec a` onto
`\vec b`=
`(√(10) )/(5)`, and the vector projection of
`\vec a` onto
`\vec b` =
`(1)/(5) \hat i+(3)/(5) \hat j`.