Question

The point of intersection of the lines
$\overrightarrow{r}=(1\overrightarrow{i}+2\overrightarrow{j}+3\overrightarrow{k})+t(-2\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k})$ and $\overrightarrow{r}=(2\overrightarrow{i}+3\overrightarrow{j}+5\overrightarrow{k})+s(\overrightarrow{i}+2\overrightarrow{j}+3\overrightarrow{k})$ is

• A. $(2,1,1)$
• B. $(1,2,1)$
• C. $(1,1,2)$
• D. $(1,1,1)$

Answer 1

Answer: (C)

$(1,1,2)$

A point on the first line can be written as $(1-2t,2+t,3+t)$ while a point on the second line can be written as $(2+s,3+2s,5+3s)$

Since we need the point of intersection of two lines, we equate the two coordinates.

$\therefore 1-2t=2+s...\left(1\right)$

$2+t=3+2s...\left(2\right)$

$3+t=5+3s...\left(3\right)$

Subtracting equation (2) from (3), we get

$s+2=1$ i.e. $s=-1$

The point thus becomes $(2-1,3-2,5-3)=(1,1,2)$