If the angle between the lines,$\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ and $\frac{5 - x}{- 2} = \frac{7y - 14}{p} = \frac{z - 3}{4}$ is $\cos^{-1} \left( \frac{2}{3} \right)$,then $p$ is equal to

  • A
    $-\frac{7}{4}$
  • B
    $\frac{2}{7}$
  • C
    $-\frac{4}{7}$
  • D
    $\frac{7}{2}$

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The line segment joining the points $A(2, 3, 4)$ and $B(-3, 5, -4)$ intersects the $yz$-plane at the point:

Statement-$1$: The distance between the two parallel lines $\frac{x}{2} = \frac{y}{-1} = \frac{z}{2}$ and $\frac{x-1}{4} = \frac{y-1}{-2} = \frac{z-1}{4}$ is $\sqrt{2}$.
Statement-$2$: The distance between two parallel lines is equal to the perpendicular distance from any point on one line to the other line.

The line $L_1$ is parallel to the vector $\vec{a} = -3 \hat{i} + 2 \hat{j} + 4 \hat{k}$ and passes through the point $(7, 6, 2)$,and the line $L_2$ is parallel to the vector $\vec{b} = 2 \hat{i} + \hat{j} + 3 \hat{k}$ and passes through the point $(5, 3, 4)$. The shortest distance between the lines $L_1$ and $L_2$ is:

The point of intersection of the lines $l_1: r(t) = (i - 6j + 2k) + t(i + 2j + k)$ and $l_2: R(u) = (4j + k) + u(2i + j + 2k)$ is

The angle between the lines $\vec{r}=(2 \hat{i}-3 \hat{j}+\hat{k})+\lambda(\hat{i}+4 \hat{j}+3 \hat{k})$ and $\vec{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(\hat{i}+2 \hat{j}-3 \hat{k})$ is

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