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Let the equation of lines be as L1:r1⃗=a1⃗+λb1⃗L_1: \vec{r_1} = \vec{a_1} + \lambda\vec{b_1} and L2:r2⃗=a2⃗+λb2⃗L_2: \vec{r_2} = \vec{a_2} + \lambda\vec{b_2} such that a1⃗−a2⃗=2i^+4j^+4k^\vec{a_1} - \vec{a_2} = 2\hat{i} + 4\hat{j} + 4\hat{k} and b1⃗×b2⃗=8i^−4k^\vec{b_1} \times \vec{b_2} = 8\hat{i} - 4\hat{k}. Then the shortest distance between L1L_1 and L2L_2 is

Solution

✅ Correct Option: 3

To find the shortest distance (dd) between two skew lines, we use the standard formula:

d=∣(a2⃗−a1⃗)⋅(b1⃗×b2⃗)∣b1⃗×b2⃗∣∣d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right|

Identify the given values

From the problem, we have:

  • a1⃗−a2⃗=2i^+4j^+4k^\vec{a_1} - \vec{a_2} = 2\hat{i} + 4\hat{j} + 4\hat{k}
  • Note: For the formula, a2⃗−a1⃗=−(a1⃗−a2⃗)=−2i^−4j^−4k^\vec{a_2} - \vec{a_1} = -(\vec{a_1} - \vec{a_2}) = -2\hat{i} - 4\hat{j} - 4\hat{k}
  • b1⃗×b2⃗=8i^−4k^\vec{b_1} \times \vec{b_2} = 8\hat{i} - 4\hat{k} (or 8i^+0j^−4k^8\hat{i} + 0\hat{j} - 4\hat{k})

Calculate the Dot Product

(a2⃗−a1⃗)⋅(b1⃗×b2⃗)=(−2)(8)+(−4)(0)+(−4)(−4)(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (-2)(8) + (-4)(0) + (-4)(-4)

=−16+0+16=0= -16 + 0 + 16 = 0

Calculate the Magnitude of the Cross Product

∣b1⃗×b2⃗∣=82+02+(−4)2=64+16=80=45|\vec{b_1} \times \vec{b_2}| = \sqrt{8^2 + 0^2 + (-4)^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5}

Calculate Shortest Distance

d=∣045∣=0d = \left| \frac{0}{4\sqrt{5}} \right| = 0

The shortest distance between the lines L1L_1 and L2L_2 is 0. This implies that the two lines are intersecting.

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