QJEEM24Ma0163: JEE Main 2024 Mathematics - Class 12 JEEM24Ma0163 Solution: The shortest distance between the lines x − 3 2 = y + 15 − 7 =…

Given two lines are represented as:

$\frac{x-3}{2} = \frac{y+15}{-7} = \frac{z-9}{5}$

and

$\frac{x+1}{2} = \frac{y-1}{1} = \frac{z-9}{-3}$

The formula for the shortest distance between two lines is:

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

From the given lines:

$\vec{a}_1 = (3, -15, 9)$

$\vec{a}_2 = (-1, 1, 9)$

$\vec{b}_1 = (2, -7, 5)$

$\vec{b}_2 = (2, 1, -3)$

Calculate the difference:

$\vec{a}_2 - \vec{a}_1 = (-1-3, 1+15, 9-9) = (-4, 16, 0)$

Next, compute the cross product:

$\vec{b}_1 \times \vec{b}_2 = <mjx-container class="MathJax CtxtMenu_Attached_0" jax="CHTML" display="true" tabindex="0" ctxtmenu_counter="6" style="font-size: 130.9$

The magnitude of the cross product is:

$|\vec{b}_1 \times \vec{b}_2| = |(16\hat{i} + 16\hat{j} + 16\hat{k})| = 16\sqrt{3}$

Calculate the dot product:

$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (-4\hat{i} + 16\hat{j} + 0\hat{k}) \cdot (16\hat{i} + 16\hat{j} + 16\hat{k}) = -64 + 256 + 0 = 192$

Finally, find the shortest distance:

$d = \frac{|192|}{16\sqrt{3}} = \frac{192}{16\sqrt{3}} = 4\sqrt{3}$

Thus, the shortest distance is:

$\boxed{4 \sqrt{3}}$