Question

If A ((2, 1, 3), B(5, 3, 9), C(1, -1, 3) and D(2, 3, 11), find the angle between the lines AB and CD.

Answer 1

$A(2,1,3);B(5,3,9);C(1,-1,3);D(2,3,11)$

The equation of the line given by points $A$ and $B$ is $\overrightarrow{r_1}=\overrightarrow{a_1}+\lambda (\overrightarrow{b_1}-\overrightarrow{b_1})$

where $\overrightarrow{a}$ and $\overrightarrow{b}$ are the position vectors, $\lambda$ being $a$ constant.

$\overrightarrow{a_1}=2\hat{i}+\hat{j}+3\hat{k}$ [ formed from coordinates of $A$]

$\overrightarrow{b_1}=5\hat{i}+3\hat{j}+9\hat{k}$ [ formed from coordinates of $B$]

Now, $\overrightarrow{r_1}=2\overrightarrow{i}+\overrightarrow{j}+3\hat{k}+\lambda (5\hat{i}+3\hat{j}+9\hat{k}-2\hat{i}-\hat{j}-3\hat{k})$

$\overrightarrow{r_1}=2\hat{i}+\hat{j}+3\hat{k}+\lambda (3\hat{i}+2\hat{j}+6\hat{k})$........(1)

The equation of line formed from points $c(1,-1,3)$ & $D(2,3,11)$

$\overrightarrow{r_2}=\overrightarrow{a_2}+\lambda (\overrightarrow{b_2}-\overrightarrow{b_2})$ $[\overrightarrow{a_2}$ and $\overrightarrow{b_2}$ are position vectors]

$\overrightarrow{r_2}=\overrightarrow{i}-\overrightarrow{j}+3\hat{k}+\lambda (2\hat{i}+3\hat{j}+11\hat{k}-\hat{i}+\hat{j}-3\hat{k})$

$\overrightarrow{r_2}=\hat{i}-\hat{j}+3\hat{k}+\lambda (\hat{i}+4\hat{j}+8\hat{k})$........(2)

The angle between two lines are given by

$\cos \theta =\frac{\overrightarrow{b_1}.\overrightarrow{b_2}}{|\overrightarrow{b_1}||\overrightarrow{b_2}|}$.........(3)

where $\overrightarrow{b_1}$ and $\overrightarrow{b_2}$ is the position vector of the line parallel to the given line

from equation (1) $b_1=3\hat{i}+2\hat{j}+6\hat{k}$

from equation (2) $b_2=\hat{i}+4\hat{j}+8\hat{k}$

Now using formula (3)

$\cos \theta =\frac{(3\hat{i}+2\hat{j}+6\hat{k}).(\hat{i}+4\hat{j}+8\hat{k})}{\left(\sqrt{(3{)}^2+(2{)}^2+(6{)}^2}\right)\sqrt{(1{)}^2+(4{)}^2+(3{)}^2}}$

$\cos \theta =\frac{3+8+48}{7+9}=\frac{59}{16},\theta ={\cos }^{-1}\left(\frac{59}{16}\right)$