Let bar{a} = 4bar{i} + 3bar{j} and bar{b} be | vedclass

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(D) Given $\bar{a} = 4\bar{i} + 3\bar{j}$. Since $\bar{b}$ is perpendicular to $\bar{a}$ in the $XOY$-plane,$\bar{b}$ must be in the direction of $3\b

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$2\bar{i} - \bar{j}$

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(D) Given $\bar{a} = 4\bar{i} + 3\bar{j}$. Since $\bar{b}$ is perpendicular to $\bar{a}$ in the $XOY$-plane,$\bar{b}$ must be in the direction of $3\bar{i} - 4\bar{j}$ or $-3\bar{i} + 4\bar{j}$. Let $\bar{b} = 3\bar{i} - 4\bar{j}$.Let $\bar{c} = x\bar{i} + y\bar{j}$.The projection of $\bar{c}$ on $\bar{a}$ is $\frac{\bar{c} \cdot \bar{a}}{|\bar{a}|} = 1 \implies \frac{4x + 3y}{5} = 1 \implies 4x + 3y = 5$.The projection of $\bar{c}$ on $\bar{b}$ is $\frac{\bar{c} \cdot \bar{b}}{|\bar{b}|} = 2 \implies \frac{3x - 4y}{5} = 2 \implies 3x - 4y = 10$.Solving the system of equations:Multiply the first by $4$ and the second by $3$: $16x + 12y = 20$ and $9x - 12y = 30$.Adding them: $25x = 50 \implies x = 2$.Substituting $x = 2$ into $4x + 3y = 5$: $8 + 3y = 5 \implies 3y = -3 \implies y = -1$.Thus,$\bar{c} = 2\bar{i} - \bar{j}$.

Let $\bar{a} = 4\bar{i} + 3\bar{j}$ and $\bar{b}$ be two perpendicular vectors in the $XOY$-plane. $A$ vector $\bar{c}$ in the same plane and having projections $1$ and $2$ respectively on $\bar{a}$ and $\bar{b}$ is

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