Question

The vector equations of two lines $L_1$ and $L_2$ are respectivly
$\overrightarrow{r}=17\hat{i}-9\hat{j}+9\hat{k}+\lambda (3\hat{i}+\hat{j}+5\hat{k})$ and $\overrightarrow{r}=15\hat{i}-8\hat{j}-\hat{k}+\mu (4\hat{i}+3\hat{j})$
I $L_1$ and $L_2$ are skew lines
II $(11,-11,-1)$ is the point of intersection of $L_1$ and $L_2$
III $(-11,-11,1)$ is the point of intersection of $L_1$ and $L_2$
IV $co{s}^{-1}\left(3/\sqrt{35}\right)$ is the acute angle between $L_1$ and $L_2$
then, which of the following is true ?

• A. II and IV
• B. I and IV
• C. IV only
• D. III and IV

Answer 1

The correct option is A II and IV

$L_1=2\hat{i}+9\hat{j}+13\hat{k}+\gamma (\hat{i}+2\hat{j}+3\hat{k})$ and

$L_2=-3\hat{i}+7\hat{j}+P\hat{k}+H(\widehat{-i}+2\hat{j}+3\hat{k})$

direction ratio for first line $(1,2,3)$ and direction ratio for second line $(-1,2,3)$

Cortesianform of line $L_1\to \frac{x-2}{1}=\frac{y-9}{2}=\frac{z-13}{3}=K$

$\Rightarrow x=k+2$......(i)

$y=2k+9$.....(ii)

$z=3k+12$.....(iii)

$L_2\to \frac{x+3}{-1}=\frac{y-T}{2}=\frac{z-P}{-3}$

if they are interesting

from $L_2\to \frac{K+2+3}{-1}=\frac{2k+9-7}{2}$

$\frac{k+5}{-1}=\frac{2k+2}{2}\Rightarrow 2K+10=-2k$

$4K=12$

$K=-3$

and from $\left(i\right),\left(ii\right)$ and $\left(iii\right)$

$x=-1$

$y=3$

$z=4$

Putting in $L_2$

$\to \frac{3-7}{2}=\frac{4-P}{-3}$

$\to -2=\frac{4-P}{-3}$

$\to +6=4-P$

$\to P=-2$

here for $P=-2$ line intersecting