The vector equation of a line whose Cartesian | vedclass

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(C) The given Cartesian equations of the line are $y=2$ and $4x-3z+5=0$. We can rewrite these as $y=2$ and $4x=3z-5$. Dividing by $12$ to get the stan

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$\overline{r}=(2 \hat{j}+\frac{5}{3} \hat{k})+\lambda(3 \hat{i}+4 \hat{k})$

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(C) The given Cartesian equations of the line are $y=2$ and $4x-3z+5=0$. We can rewrite these as $y=2$ and $4x=3z-5$. Dividing by $12$ to get the standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$: $4x = 3(z - \frac{5}{3}) \implies \frac{x}{3} = \frac{z - 5/3}{4}$. Since $y=2$ is a constant,the direction ratio for $y$ is $0$. Thus,the line can be written as $\frac{x-0}{3} = \frac{y-2}{0} = \frac{z-5/3}{4}$. The line passes through the point $(0, 2, 5/3)$ and has direction ratios $(3, 0, 4)$. The vector equation is $\overline{r} = \vec{a} + \lambda \vec{b}$,where $\vec{a} = 0\hat{i} + 2\hat{j} + \frac{5}{3}\hat{k}$ and $\vec{b} = 3\hat{i} + 0\hat{j} + 4\hat{k}$. Therefore,$\overline{r} = (2\hat{j} + \frac{5}{3}\hat{k}) + \lambda(3\hat{i} + 4\hat{k})$.

The vector equation of a line whose Cartesian equations are $y=2$ and $4x-3z+5=0$ is

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