JEE Mathematics 2024 Question with Solution

Given: The shortest distances $d_1$ and $d_2$ are to be found for two pairs of skew lines.

Find: The value of $\frac{32\sqrt{3}d_1}{d_2}$.

Use the formula for shortest distance between two skew lines:

$$d = \frac{\left| (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) \right|}{\left| \vec{b}_1 \times \vec{b}_2 \right|}$$

For the first pair,

$$L_1 : \frac{x + 1}{2} = \frac{y - 1}{-12} = \frac{z}{1}, \qquad L_2 : \frac{x - 1}{-7} = \frac{y + 8}{2} = \frac{z - 4}{5}$$

From the extracted working, we get:

$$d_1 = 2$$

For the second pair,

$$L_3 : \frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 6}{-3}, \qquad L_4 : \frac{x + 2}{1} = \frac{y + 2}{1} = \frac{z - 1}{6}$$

From the extracted working, we get:

$$d_2 = \frac{12}{\sqrt{3}} = 4\sqrt{3}$$

Now substitute in the required expression:

$$\frac{32\sqrt{3}d_1}{d_2} = \frac{32\sqrt{3} \times 2}{12/\sqrt{3}} = 16$$

Therefore, the value of the expression is $16$.

Given: Two shortest distances $d_1$ and $d_2$ between pairs of skew lines.

Find: $\frac{32\sqrt{3}d_1}{d_2}$.

For the first pair of lines, the solution identifies point and direction vectors and evaluates the scalar triple product.

For $d_1$:

$$\vec{a}_2 - \vec{a}_1 = \hat{i} - 2\hat{j} + \hat{k}$$ $$\vec{b}_1 \times \vec{b}_2 = 12\hat{i} - 18\hat{j} + 36\hat{k}$$

Then,

$$\left| (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) \right| = |12 + 36 + 36| = 84$$ $$\left| \vec{b}_1 \times \vec{b}_2 \right| = \sqrt{12^2 + (-18)^2 + 36^2} = 42$$

Hence,

$$d_1 = \frac{84}{42} = 2$$

For $d_2$, the extracted solution gives:

$$\vec{a}_4 - \vec{a}_3 = 10\hat{j} + 2\hat{k}$$ $$\vec{b}_3 \times \vec{b}_4 = 16\hat{i} + 16\hat{j} + 16\hat{k}$$

So,

$$\left| (\vec{a}_4 - \vec{a}_3) \cdot (\vec{b}_3 \times \vec{b}_4) \right| = 192$$ $$\left| \vec{b}_3 \times \vec{b}_4 \right| = 16\sqrt{3}$$

Therefore,

$$d_2 = \frac{192}{16\sqrt{3}} = \frac{12}{\sqrt{3}} = 4\sqrt{3}$$

Now,

$$\frac{32\sqrt{3}d_1}{d_2} = \frac{32\sqrt{3} \times 2}{4\sqrt{3}} = \frac{64\sqrt{3}}{4\sqrt{3}} = 16$$

Therefore, the required numerical value is $16$.