JEE Main 2025 Apr 4 Shift 2, Mathematics Q6: Three Dimensional Geometry

JEE Main2025Apr 4, Shift 2Mathematics

Let the values of $p$ , for which the shortest distance between the lines $\frac{x + 1}{3} = \frac{y}{4} = \frac{z}{5}$ and $r = (p \hat{i} + 2 \hat{j} + \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 4 \hat{k})$ is $\frac{1}{6}$ , be $a, b$ , ( $a < b$ ). Then the length of the latus rectum of the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ is :-

A
9
B
$\frac{3}{2}$
C
$\frac{2}{3}$
D
18

Solution

Take a point on each line and the direction vectors:

$L_{1}$ : point $A = (- 1, 0, 0)$ , direction $p = (3, 4, 5)$ .

$L_{2}$ : point $B = (p, 2, 1)$ , direction $q = (2, 3, 4)$ .

$p \times q = (4 \cdot 4 - 5 \cdot 3, 5 \cdot 2 - 3 \cdot 4, 3 \cdot 3 - 4 \cdot 2) = (1, - 2, 1)$ , so $∣ p \times q ∣ = 6$ .

$B - A = (p + 1, 2, 1)$ . The shortest distance formula gives

$d = \frac{∣ ( B - A ) \cdot ( p \times q ) ∣}{∣ p \times q ∣} = \frac{∣ ( p + 1 ) - 4 + 1∣}{6} = \frac{∣ p - 2∣}{6}$ .

Setting $d = \frac{1}{6}$ gives $∣ p - 2∣ = 1$ , so $p = 1$ or $p = 3$ . Thus $a = 1$ and $b = 3$ .

For $\frac{x ^{2}}{1} + \frac{y ^{2}}{9} = 1$ the major axis is along $y$ with semi-axes $b^{'} = 3$ (major) and $a^{'} = 1$ (minor). Length of latus rectum $= \frac{2 \cdot ( minor ) ^{2}}{major} = \frac{2 \cdot 1}{3} = \frac{2}{3}$ .

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