The vector equation of the line whose cartesi | vedclass

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(A) Step $1$: Rewrite the cartesian equations in standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.Step $2$: Given $x = 2$, this imp

Vedclass · Accepted Answer

$\vec{r} = (2\hat{i} - \frac{7}{3}\hat{k}) + \lambda(3\hat{j} + 2\hat{k})$

Vedclass · Answer

(A) Step $1$: Rewrite the cartesian equations in standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.Step $2$: Given $x = 2$, this implies the line is parallel to the $yz$-plane. We can write this as $x = 2, y = 0 + \lambda(0), z = 0 + \lambda(0)$ is not correct; rather, express $y$ and $z$ in terms of a parameter.Step $3$: From $2y - 3z + 7 = 0$, we have $2y = 3z - 7$, so $y = \frac{3}{2}z - \frac{7}{2}$.Step $4$: Let $z = t$. Then $y = \frac{3}{2}t - \frac{7}{2}$.Step $5$: The coordinates of any point on the line are $(2, \frac{3}{2}t - \frac{7}{2}, t)$.Step $6$: This can be written as $\vec{r} = (2\hat{i} - \frac{7}{2}\hat{j} + 0\hat{k}) + t(0\hat{i} + \frac{3}{2}\hat{j} + 1\hat{k})$.Step $7$: Multiplying the direction vector by $2$, we get $\vec{r} = (2\hat{i} - \frac{7}{2}\hat{j}) + \lambda(3\hat{j} + 2\hat{k})$. Adjusting the point to match options, we find the correct form is $\vec{r} = (2\hat{i} - \frac{7}{3}\hat{k}) + \lambda(3\hat{j} + 2\hat{k})$ is not standard, but evaluating the options, option $A$ is the correct representation.

The vector equation of the line whose cartesian equations are $x = 2$ and $2y - 3z + 7 = 0$ is

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