The lines \(\vec{r} = (\hat{\imath}+\hat{\jmath}-\hat{k})+\lambda(2\hat{\imath}-\hat{\jmath}+\hat{k})\) and \(\vec{r} = (2\hat{\imath}-2\hat{\jmath}+3\hat{k})+\mu(\hat{\imath}+2\hat{\jmath}-3\hat{k})\) intersect at the point
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (3, 0, 0)
At the point of intersection the two position vectors are equal. Comparing the components:
\(1+2\lambda = 2+\mu\)
\(1-\lambda = -2+2\mu\)
\(-1+\lambda = 3-3\mu\)
From the first equation, \(\mu = 2\lambda-1\). Putting this in the second:
\(1-\lambda = -2+4\lambda-2\)
\(5 = 5\lambda\)
\(\lambda = 1,\quad \mu = 1\)
The third equation also holds: \(-1+1 = 0\) and \(3-3 = 0\).
Point \(= (1+2,\ 1-1,\ -1+1) = (3, 0, 0)\)
Why the other options are incorrect
Option (1) (1, 1, −1) — Incorrect. This is the given point of the first line (\(\lambda = 0\)); on the second line \(2+\mu = 1\) gives \(\mu = -1\), and then \(y = -4 \ne 1\).
Option (2) (2, −2, 3) — Incorrect. This is the given point of the second line (\(\mu = 0\)); on the first line \(1+2\lambda = 2\) gives \(\lambda = \frac{1}{2}\), and then \(y = \frac{1}{2} \ne -2\).
Option (3) (5, −1, 1) — Incorrect. This is the point \(\lambda = 2\) of the first line; on the second line \(2+\mu = 5\) gives \(\mu = 3\), and then \(y = 4 \ne -1\).