The Cartesian equation of a line is $2x - 2 = 3y + 1 = 6z - 2$. The vector equation of the line is:

  • A
    $\bar{r} = \left(\hat{i} - \frac{1}{3}\hat{j} + \frac{1}{3}\hat{k}\right) + \lambda(3\hat{i} + 2\hat{j} + \hat{k})$
  • B
    $\bar{r} = \left(-\hat{i} + \frac{1}{3}\hat{j} - \frac{1}{3}\hat{k}\right) + \lambda\left(\frac{1}{2}\hat{i} + \frac{1}{3}\hat{j} + \frac{1}{6}\hat{k}\right)$
  • C
    $\bar{r} = (3\hat{i} - \hat{j} - \hat{k}) + \lambda(3\hat{i} + 2\hat{j} + \hat{k})$
  • D
    $\bar{r} = (\hat{i} - \hat{j} + \hat{k}) + \lambda\left(\frac{1}{2}\hat{i} + \frac{1}{3}\hat{j} + \frac{1}{6}\hat{k}\right)$

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Find the equations of the two lines passing through the origin which intersect the line $\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}$ at angles of $\frac{\pi}{3}$ each.

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The length of the perpendicular from the point $(0, 2, 3)$ to the line $\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3}$ is:

The shortest distance between lines $\overline{r}=(2 \hat{i}-\hat{j})+\lambda(2 \hat{i}+\hat{j}-3 \hat{k})$ and $\overline{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(2 \hat{i}+\hat{j}-5 \hat{k})$ is

The distance of the point $(2, 4, 0)$ from the point of intersection of the lines $\frac{x+6}{3} = \frac{y}{2} = \frac{z+1}{1}$ and $\frac{x-7}{4} = \frac{y-9}{3} = \frac{z-4}{2}$ is

The $xy$-plane divides the line segment joining the points $(2, 4, 5)$ and $(-4, 3, -2)$ in the ratio:

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