The vector equation of the line whose Cartesi | vedclass

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(D) Given Cartesian equations are $y=2$ and $4x-3z+5=0$. From $4x-3z+5=0$,we have $4x = 3z-5$,which implies $4x = 3(z - \frac{5}{3})$. This can be wri

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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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(D) Given Cartesian equations are $y=2$ and $4x-3z+5=0$. From $4x-3z+5=0$,we have $4x = 3z-5$,which implies $4x = 3(z - \frac{5}{3})$. This can be written as $\frac{x}{3} = \frac{z - 5/3}{4}$. Since $y=2$,we can write the equation as $\frac{x-0}{3} = \frac{y-2}{0} = \frac{z-5/3}{4}$. This line passes through the point $(0, 2, 5/3)$ and has direction ratios $(3, 0, 4)$. The vector equation of a line passing through point $\vec{a}$ and parallel to vector $\vec{b}$ is $\vec{r} = \vec{a} + \lambda \vec{b}$. Here,$\vec{a} = 0\hat{i} + 2\hat{j} + \frac{5}{3}\hat{k}$ and $\vec{b} = 3\hat{i} + 0\hat{j} + 4\hat{k}$. Thus,the vector equation is $\vec{r} = (2\hat{j} + \frac{5}{3}\hat{k}) + \lambda(3\hat{i} + 4\hat{k})$.

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

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