If the lines \( \vec{r}=\left(\hat{\imath}+\hat{\jmath}-\hat{k}\right)+\lambda\left(3\hat{\imath}-\hat{\jmath}\right) \) and \( \vec{r}=\left(4\hat{\imath}-\hat{k}\right)+\mu\left(2\hat{\imath}+3\hat{k}\right) \) intersect, then their point of intersection is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (4, 0, −1)
A general point of each line is
\( \displaystyle \text{First line: }(1+3\lambda,\ 1-\lambda,\ -1) \)
\( \displaystyle \text{Second line: }(4+2\mu,\ 0,\ -1+3\mu) \)
At the point of intersection the coordinates are equal:
\( \displaystyle y:\ 1-\lambda=0\ \Rightarrow\ \lambda=1 \)
\( \displaystyle z:\ -1=-1+3\mu\ \Rightarrow\ \mu=0 \)
Check the x-coordinates: \( 1+3(1)=4 \) and \( 4+2(0)=4 \). They agree, so the lines do meet, at
\( \displaystyle (4,\ 0,\ -1) \)
Shortcut: Every point of the first line has z = −1, and every point of the second line has y = 0. So the intersection has y = 0 and z = −1, which leaves only Options (1) and (4). With y = 0 on the first line, \( \lambda=1 \) and x = 4, so the point is (4, 0, −1).