a = 4 hat{i} + 3 hat{j} and b are two vectors | vedclass

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(A) Given $a = 4 \hat{i} + 3 \hat{j}$. Since $a \cdot b = 0$ and $b$ is in the $XOY$ plane,$b$ must be of the form $k(3 \hat{i} - 4 \hat{j})$. Let $b

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

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(A) Given $a = 4 \hat{i} + 3 \hat{j}$. Since $a \cdot b = 0$ and $b$ is in the $XOY$ plane,$b$ must be of the form $k(3 \hat{i} - 4 \hat{j})$. Let $b = 3 \hat{i} - 4 \hat{j}$.Let $c = x \hat{i} + y \hat{j}$.The projection of $c$ on $a$ is $\frac{a \cdot c}{|a|} = 1 \implies \frac{4x + 3y}{5} = 1 \implies 4x + 3y = 5$ $(i)$.The projection of $c$ on $b$ is $\frac{b \cdot c}{|b|} = 2 \implies \frac{3x - 4y}{5} = 2 \implies 3x - 4y = 10$ (ii).Multiplying $(i)$ by $4$ and (ii) by $3$: $16x + 12y = 20$ and $9x - 12y = 30$.Adding these,$25x = 50 \implies x = 2$.Substituting $x = 2$ into $(i)$: $4(2) + 3y = 5 \implies 8 + 3y = 5 \implies 3y = -3 \implies y = -1$.Thus,$c = 2 \hat{i} - \hat{j}$.

$a = 4 \hat{i} + 3 \hat{j}$ and $b$ are two vectors in the $XOY$ plane,and $a$ is perpendicular to $b$. $A$ vector $c$ lying in the same plane and having projections $1$ and $2$ on $a$ and $b$ respectively is:

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