If $A(3,4,5)$,$B(4,6,3)$,$C(-1,2,4)$,and $D(1,0,5)$ are points such that the angle between the lines $DC$ and $AB$ is $\theta$,then $\cos \theta$ is equal to

  • A
    $\frac{7}{9}$
  • B
    $\frac{2}{9}$
  • C
    $\frac{4}{9}$
  • D
    $\frac{5}{9}$

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Similar Questions

Assertion $(A)$: For the lines $\overline{r}=\overline{a}+t \overline{b}$ and $\overline{r}=\overline{p}+s \overline{q}$,if $(\bar{a}-\bar{p}) \cdot(\bar{b} \times \bar{q}) \neq 0$,then the two lines are coplanar.
Reason $(R)$: $|(\bar{a}-\bar{p}) \cdot(\bar{b} \times \bar{q})|$ is $|\bar{b} \times \bar{q}|$ times the shortest distance between the lines $\overline{r}=\overline{a}+t\bar{b}$ and $\overline{r}=\overline{p}+s \overline{q}$.

$ABC$ is a triangle with vertices $A(2, 3, 5)$,$B(-1, 3, 2)$,and $C(\lambda, 5, \mu)$. If the median through $A$ is equally inclined to the coordinate axes,then the value of $(\lambda^3 + \mu^3 + 5)$ is equal to:

If the shortest distance between the line joining the points $(1, 2, 3)$ and $(2, 3, 4)$,and the line $\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-2}{0}$ is $\alpha$,then $28 \alpha^2$ is equal to $........$.

The point of intersection of the lines represented by $\overline{r}=(\overline{i}-6 \overline{j}+2 \overline{k})+t(\overline{i}+2 \overline{j}+\overline{k})$ and $\overline{r}=(4 \overline{j}+\overline{k})+s(2 \overline{i}+\overline{j}+2 \overline{k})$ is

$\triangle ABC$ is formed by $A(1, 8, 4)$,$B(0, -11, 4)$,and $C(2, -3, 1)$. If $D$ is the foot of the perpendicular drawn from $A$ to $BC$,then the coordinates of $D$ are

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