Let b = 4i + 3j and c be two vectors perpendi | vedclass

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(D) Let $r = \lambda b + \mu c$. Since $b = 4i + 3j$,$|b| = \sqrt{4^2 + 3^2} = 5$. Since $c$ is perpendicular to $b$ in the $xy$-plane,$c$ must be par

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$2i - j, - \frac{2}{5}i + \frac{11}{5}j$

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(D) Let $r = \lambda b + \mu c$. Since $b = 4i + 3j$,$|b| = \sqrt{4^2 + 3^2} = 5$. Since $c$ is perpendicular to $b$ in the $xy$-plane,$c$ must be parallel to $3i - 4j$. Let $c = k(3i - 4j)$. For projection of $r$ on $c$ to be $2$,we normalize $c$ such that $|c| = 1$,so $c = \pm \frac{1}{5}(3i - 4j)$. Projection of $r$ on $b = \frac{r \cdot b}{|b|} = \lambda |b| = 5\lambda = 1 \Rightarrow \lambda = \frac{1}{5}$. Projection of $r$ on $c = \frac{r \cdot c}{|c|} = \mu |c| = \mu = 2$. Thus,$r = \frac{1}{5}(4i + 3j) \pm 2 \cdot \frac{1}{5}(3i - 4j)$. Case $1$: $r = \frac{1}{5}(4i + 3j + 6i - 8j) = \frac{10i - 5j}{5} = 2i - j$. Case $2$: $r = \frac{1}{5}(4i + 3j - 6i + 8j) = \frac{-2i + 11j}{5} = -\frac{2}{5}i + \frac{11}{5}j$.

Let $b = 4i + 3j$ and $c$ be two vectors perpendicular to each other in the $xy$-plane. All vectors in the same plane having projections $1$ and $2$ along $b$ and $c$ respectively,are given by

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