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If ` vec a= hat i+ hat j+ hat k a n d`` vec b= hat j- hat k ,`find avector ` vec c`such that ` vec a "x" vec c= vec b`and ` vec adot vec c=3` |
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Answer» Here`vec(a)=hat(i) + hat(j) + hat(k) and vec(b) = hat(j)-hat(k)`. `Let vec(c)= c_(1) hat(i) + c_(2) hat(j) + c_(3)hat(k).` Then, `vec(a) * vec(c) = 3 and vec(a) xx vec(c) = vec(b)` ` rArr (hat(i) + hat(j) +hat(k))* (c_(1) hat(i) + c_(2) hat(j) + c_(3)hat(k))=3` `and (hat(i) + hat(j) + hat(k))xx ( c_(1) hat(i) + c_(2) hat(j) + c_(3) hat(k)) = ( hat(j) - hat(k))` `rArr c_(1) + c_(2) + C_(3) " "...(i) and |(hat(i),hat(j),hat(k)),(1,1,1),(c_(1), c_(2) , c_(3))|=(hat(j)-hat(k))`...(ii) Now, (ii) gives : `(c_(3) - c_(2) hat(i)- ( c_(3)-c_(1)) hat(j) + ( c_(2) - c_(1)) hat(k)= (hat(j) - hat(k))` `rArr c_(3)- c_(2) = 0, c_(1)-c_(3) = 1 and c_(2) - c_(1) =-1` Putting `C_(3) = c_(2) " in (i), we get " c_(1) + 2 c_(2) = 3.` On solving `c_(1)+ 2 c_(2) = 3 and c_(1) - c_(2) = 1," we get " c_(2) = 2/3 and c_(1)=5/3`. `:. c_(1)= 5/3, c_(2) = 2/3 and c_(3) = 2/3`. Hence, `vec(c) = 5/3 hat(i) + 2/3hat(j) + 2/3 hat(k) rArr vec(c) = 1/3 (5 hat(i) + 2 hat(j) +2hat(k)).` |
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