Let `vecb= 4 hati + 3hatj and vecc` be two vectors perpendicular to each other in the xy- plane. All vectors in the sme plane having projections 1 and 2 along `vecb and vecc`., respectively, are given by ________

Answer

`2hati-hatj`

Explanation

Let `vecc = alpha hati + beta hatj`
Give n that ` vecb bot vecc`
` vecb.vecc=0`
`Rightarrow (4hati+3hatj) .(alphahati+betahatj)=0`
`or 4alpha + 3beta=0`
`or alpha/3 = beta/(-4) =0`
`or alpha=3 lambda, beta= -4 lambda`
Now let `veca = xhati + yhatj` be the required vectors,porjection of `veca " along " vecb` given
`(veca.vecb)/(|vecb|) = (4x + 3y)/(sqrt(4^(2)+3^(2)))=1`
`or 4x + 3y =5 `
Also projection of `veca ` along `vecc` given
`(veca.vecc)/(|vecc|)=2`
` Rightarrow (alphax +betay)/(sqrt(alpha^(2) +beta^(2)))=2`
`or 3lambda xx - 4 lambday = 10 lambda`
`or 3x - 4y =10`
solving (ii) and (iii) we get x=2,y=-1
therefore, the required vector is `2hati - hatj`