Question

The shortest distance between the lines, where the first line passes through \((0,0,0)\) and \((2,0,3)\) and the second line passes through \((2,5,0)\) and \((0,4,0)\) is

• A. \(\frac{24}{7}\) units
• B. \(\frac{9}{7}\) units
• C. \(\frac{1}{7}\) units
• D. \(\frac{36}{7}\) units

Hint:

Find direction vectors of both lines, take their cross product, and project the vector joining points on the lines.

Answer 1

The Correct Option is A

Step 1: Write the lines.
Line 1: $(x, y, z) = s(2, 0, 3)$. Line 2: $(x, y, z) = (2, 5, 0) + t(-2, -1, 0)$.

Step 2: Find the common perpendicular direction.
The normal to both is $\bar{n} = (3, -6, -2)$, found by solving $2n_1 + 3n_3 = 0$ and $-2n_1 - n_2 = 0$ for the components.

Step 3: Plane through line 1.
The plane through the origin with this normal is $3x - 6y - 2z = 0$. Line 2 is parallel to it, since $\bar{d}_2\cdot\bar{n} = -6 + 6 = 0$.

Step 4: Distance of a point of line 2 from the plane.
\[ d = \frac{|3(2) - 6(5) - 2(0)|}{\sqrt{9 + 36 + 4}} = \frac{24}{7} \]

Final Answer:
Option (A). \[ \boxed{\frac{24}{7}} \]