The vector equation of the line whose Cartesi | vedclass

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Question

(B) Given Cartesian equations are $y=2$ and $4x-3z+5=0$. We can rewrite the second equation as $4x = 3z - 5$,which implies $4x = 3(z - \frac{5}{3})$.

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given Cartesian equations are $y=2$ and $4x-3z+5=0$. We can rewrite the second equation as $4x = 3z - 5$,which implies $4x = 3(z - \frac{5}{3})$. Dividing by $12$,we get $\frac{4x}{12} = \frac{3(z - \frac{5}{3})}{12}$,which simplifies to $\frac{x}{3} = \frac{z - \frac{5}{3}}{4}$. Since $y=2$ is constant,the line can be represented as $\frac{x}{3} = \frac{y-2}{0} = \frac{z - \frac{5}{3}}{4}$. This line passes through the point $(0, 2, \frac{5}{3})$,so its position vector is $\overline{a} = 0\hat{i} + 2\hat{j} + \frac{5}{3}\hat{k}$. The direction ratios of the line are $(3, 0, 4)$,so the direction vector is $\overline{b} = 3\hat{i} + 0\hat{j} + 4\hat{k}$. The vector equation of a line is given by $\overline{r} = \overline{a} + \lambda\overline{b}$. Substituting the values,we get $\overline{r} = (2\hat{j} + \frac{5}{3}\hat{k}) + \lambda(3\hat{i} + 4\hat{k})$.

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

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