Answer: Projection of (2,6) onto (-1,5) is given by
[tex]\dfrac{28}{\sqrt{26}}[/tex]
Step-by-step explanation:
Let [tex]\vec{a}=2\hat{i}+6\hat{j}[/tex]
and [tex]\vec{b}=-1\jhat{i}+5\hat{j}[/tex]
we need to find the projection of [tex]\vec{a}[/tex] onto [tex]\vec{b}[/tex]
As we know the formula for projection:
[tex]Projection=\frac{\vec{b}}{\mid b\mid}. \vec{a}[/tex]
Since we know the value of [tex]\vec{b}=-1\jhat{i}+5\hat{j}[/tex]
so,
[tex]\mid b\mid=\sqrt{(-1)^2+(5)^2}\\\\\mid b\mid=\sqrt{1+25}\\\\\mid b \mid=\sqrt{26}[/tex]
so, [tex]\dfrac{\vec{b}}{\mid b\mid}=\dfrac{-1\hat{i}+5\hat{j}}{\sqrt{26}}[/tex]
Now, it becomes,
[tex]Projection=\dfrac{(2\ht{i}+6\hat{j}).(-1\hat{i}+5\hat{j})}{\sqrt{26}}\\\\Projection=\dfrac{-2+30}{\sqrt{26}}\\\\Projection=\dfrac{28}{\sqrt{26}}[/tex]
Hence, projection of (2,6) onto (-1,5) is given by
[tex]\dfrac{28}{\sqrt{26}}[/tex]
