Question

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

• A. $(2,1,1)$
• B. $(1,2,1)$
• C. $(-3,1,2)$
• D. $(1,1,1)$

Answer 1

Answer: (C)

$(-3,1,2)$

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

Equating the two coordinates, we have

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

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

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

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

$1=2+t$ or $t=-1$

The point of intersection is therefore $(-2-1,3-2,5-3)=(-3,1,2)$