Find the angle between force `vecF=(3hati+4hatj-5hatk)` and displacement `vecd=(5hati+4hatj-3hatk)`. Also find the projection of `vecF " ""on" " " vecd`.

Explanation

Let `theta` be the angle between the vectors `vecF and vecd`. Then by the definition of scalar product of two vector, we have
`vecF*vecd=Fd cos theta or, cos theta =(vecF*vecd)/(Fd)`
Now, `|vecF|=sqrt(3^2+4^2+(-5)^2)=sqrt(50)`
`and |vecd|=sqrt(5^2+4^2+(-3)^2)=sqrt(50)`
`|vecF|=sqrt(3^2+4^2+(-5)^2)=sqrt(50)`
`and |vecd|=sqrt(5^2+4^2+(-3)^2)=sqrt(50)`
`therefore cos theta =(46)/(sqrt(50)xxsqrt(50))=(46)/(50)=0.92 `
`or, theta = cos ^(-1) 0.92 ~~23^@`
Hence, the projection of `vecF ` on `vecd` is
`F cos theta =sqrt(50)xx(0.92)=7.07xx0.92=6.50` units.