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Let `vecb=4hati+3hatj and vecc` be two vectors perpendicular to each other in the xy-plane. Find all vetors in te same plane having projection 1 and 2 along `vecb and vecc` respectively. |
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Answer» Correct Answer - `2hati-hatj` 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` |
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