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MHTCET 15 Apr 2026 Shift 2 Paper

Question : 143 of 150

Marks: +1, -0

If the lines

\vec{r}=(\hat{i}+m \hat{j}+3 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})

\vec{r}=(4 \hat{i}+\hat{j})+\mu(5 \hat{i}+m \hat{j}+\hat{k})

\mathrm{m}=

Solution:

Concept:

Two lines intersect if they have a common point, so their general point coordinates must be equal.

Explanation:

General point on the first line:

(1+2\lambda,\; m+3\lambda,\; 3+4\lambda)

General point on the second line:

(4+5\mu,\; 1+m\mu,\; \mu)

Equating corresponding coordinates:

1+2\lambda = 4+5\mu

m+3\lambda = 1+m\mu

3+4\lambda = \mu

From the third equation,

\mu = 3+4\lambda

Substitute in the first equation:

1+2\lambda = 4+5(3+4\lambda)

1+2\lambda = 19+20\lambda

-18 = 18\lambda \Rightarrow \lambda = -1

Then,

\mu = 3+4(-1) = -1

Now use

\lambda = -1

and

\mu = -1

in the second equation:

m+3(-1) = 1+m(-1)

m-3 = 1-m

2m = 4 \Rightarrow m = 2

Answer:

m = 2

Correct option: C.

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