Find the components of a vector `vecA=2hati+3hatj` along the directions of `hati+hatj` and `hati-hatj`

Explanation

Magnitude of the vector ` hati+ hatj`
`|vec hati + hatj| =sqrt((1)^(2)+(1)^(2))=sqrt2`
`tan theta =(|hatj|)/(|hati|)=1`
`hati+hatj` makes an angle of `45^(@)` with direction of the X-axis .
Magnitude of the `(hati-hatj)`
`|hati-hatj|=sqrt(1^(2)+(-1)^(2))=sqrt2`

`tan theta =-(|hatj|)/(|hati|)=-1`
`hati-hatj` makes an angle of `-45^(@)` with direction of the X-axis.
To find the Component of vector `vec A=2hati+3hatj` alomg the direction `(hati+hatj)`
Let `vec B =hati +hatj`
`vec A.vecB=AB cos theta =(a cos theta )B`
`therefore A cos theta (vecA.vecB)/(B)`
`=((2hati+3hatj).(hati+hatj))/(sqrt((1)^(2)+(1)^(2))`
`=((2+3))/(sqrt2)=(5)/(sqrt2) [ :. A cos theta =(|vecA|)/(|vecB|)]`
To find the component of vector `vec A=2 hati +3 hatj ` along the direction `(hati-hatj)`
Let `vec B=hati-hatj`
Similarly `A cos theta=(vecA.vecB)/(b)`
`=((2hati+3hatj).(hati-hatj))/(sqrt((1)^(2)+(-1))^(2))`
`((2-3))/(sqrt2)=(-1)/(sqrt2)`