Q82 / 150 · MHT CET 2026 (15 April Shift 2) · Mathematics
If the lines $\vec{r} = (\hat{i} + m\hat{j} + 3\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 4\hat{k})$ and $\vec{r} = (4\hat{i} + \hat{j}) + \mu(5\hat{i} + m\hat{j} + \hat{k})$ intersect each other, then m =
- A$1$
- B$-1$
- Correct answerC$2$
- D$-2$
Solution
Equating the position vectors of the two lines for intersection:
$1 + 2\lambda = 4 + 5\mu$
$m + 3\lambda = 1 + m\mu$
$3 + 4\lambda = \mu$
From the third equation, $\mu = 4\lambda + 3$.
Substituting this into the first equation:
$2\lambda - 5(4\lambda + 3) = 3$
$-18\lambda - 15 = 3$
$-18\lambda = 18 \Rightarrow \lambda = -1$
Substituting $\lambda = -1$ into $\mu = 4\lambda + 3$ gives $\mu = -1$.
Now, substituting $\lambda = -1$ and $\mu = -1$ into the second equation:
$m + 3(-1) = 1 + m(-1)$
$m - 3 = 1 - m$
$2m = 4 \Rightarrow m = 2$
Answer: $2$
$1 + 2\lambda = 4 + 5\mu$
$m + 3\lambda = 1 + m\mu$
$3 + 4\lambda = \mu$
From the third equation, $\mu = 4\lambda + 3$.
Substituting this into the first equation:
$2\lambda - 5(4\lambda + 3) = 3$
$-18\lambda - 15 = 3$
$-18\lambda = 18 \Rightarrow \lambda = -1$
Substituting $\lambda = -1$ into $\mu = 4\lambda + 3$ gives $\mu = -1$.
Now, substituting $\lambda = -1$ and $\mu = -1$ into the second equation:
$m + 3(-1) = 1 + m(-1)$
$m - 3 = 1 - m$
$2m = 4 \Rightarrow m = 2$
Answer: $2$
Hence, the correct answer is option: 'C'.