Three Dimensional Geometry – Intersection of Lines and Distance Formula – JEE Main 24 January 2025 Shift 1

Solution:

Let $L_1$ be the line passing through $(-1, 2, 1)$ and parallel to the vector $\vec{d} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.
The equation of $L_1$ is:
$$ \frac{x+1}{2} = \frac{y-2}{3} = \frac{z-1}{4} = \lambda $$
General point on $L_1$: $P(2\lambda-1, 3\lambda+2, 4\lambda+1)$.

Let $L_2$ be the second line:
$$ \frac{x+2}{3} = \frac{y-3}{2} = \frac{z-4}{1} = \mu $$
General point on $L_2$: $(3\mu-2, 2\mu+3, \mu+4)$.

For intersection, the coordinates must be equal:
1) $2\lambda – 1 = 3\mu – 2 \Rightarrow 2\lambda – 3\mu = -1$
2) $3\lambda + 2 = 2\mu + 3 \Rightarrow 3\lambda – 2\mu = 1$
3) $4\lambda + 1 = \mu + 4 \Rightarrow 4\lambda – \mu = 3 \Rightarrow \mu = 4\lambda – 3$

Substitute $\mu$ from (3) into (2):
$$ 3\lambda – 2(4\lambda – 3) = 1 $$
$$ 3\lambda – 8\lambda + 6 = 1 $$
$$ -5\lambda = -5 \Rightarrow \lambda = 1 $$

Check with equation (1):
$$ 2(1) – 3(4(1)-3) = 2 – 3(1) = -1 $$
(Satisfied).

The coordinates of point P are found by putting $\lambda = 1$ in $L_1$:
$$ P = (2(1)-1, 3(1)+2, 4(1)+1) = (1, 5, 5) $$

Now, find the distance between $P(1, 5, 5)$ and $Q(4, -5, 1)$:
$$ PQ = \sqrt{(4-1)^2 + (-5-5)^2 + (1-5)^2} $$
$$ PQ = \sqrt{3^2 + (-10)^2 + (-4)^2} $$
$$ PQ = \sqrt{9 + 100 + 16} = \sqrt{125} $$
$$ PQ = \sqrt{25 \times 5} = 5\sqrt{5} $$

Ans. (4)