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Mathematics Question on Equation of a Line in Space

The length of the shortest distance between the lines r⃗=3i^+5j^+7k^+λ(i^−2j^+k^)\vec{r}=3\hat{i}+5\hat{j}+7\hat{k}+\lambda(\hat{i}-2\hat{j}+\hat{k}) and r⃗=−i^−j^−k^+μ(7i^−6j^+k^)\vec{r} = -\hat{i}-\hat{j}-\hat{k}+\mu (7\hat{i}-6\hat{j}+\hat{k}) is

A

83 units

B

6\sqrt{6} units

C

3\sqrt{3} units

D

2292\sqrt{29} units

Answer

2292\sqrt{29} units

Explanation

Solution

Shortest distance PQ = ∣(b1⃗×b2⃗) . (a2⃗−a1⃗)∣(b1⃗×b2⃗)∣∣\bigg| \frac{(\vec{b_1} \times \vec{b_2} ) \ . \ (\vec{a_2} - \vec{a_1})}{|(\vec{b_1} \times \vec{b_2})|} \bigg|
Now, a2⃗ − a1⃗ = −i^−j^−k^−3i^−5j^−7k^\vec{a_2} \ - \ \vec{a_1} \ = \ -\hat{i}-\hat{j}-\hat{k}-3\hat{i}-5\hat{j}-7\hat{k}
⇒   a2⃗−a1⃗ = −4i^−6j^−8k^\Rightarrow \ \ \ \vec{a_2} - \vec{a_1} \ = \ -4\hat{i}-6\hat{j}-8\hat{k}

And b1⃗ × b2⃗=∣i^j^k^ 1−21 7−61∣\vec{b_1} \ \times \ \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\\ 1 & -2 & 1 \\\ 7 & -6 & 1 \end{vmatrix}
⇒  b1⃗×b2⃗=i^(−2+6)−j^(1−7)+k^(−6+14)\Rightarrow \ \ \vec{b_1} \times \vec{b_2} = \hat{i} (-2+6)-\hat{j}(1-7)+\hat{k}(-6+14)
⇒  b1⃗×b2⃗ = 4i^+6j^+8k^\Rightarrow \ \ \vec{b_1} \times \vec{b_2} \ = \ 4\hat{i}+6\hat{j}+8\hat{k}
∴\therefore Shortest distance
PQ=∣(4i^+6j^+8k^) . (−4i^−6j^−8k^)16+36+64∣PQ= \bigg| \frac{(4\hat{i} + 6\hat{j} +8 \hat{k}) \ . \ (-4\hat{i}-6\hat{j}-8\hat{k})}{\sqrt{16+36+64}}\bigg|
⇒  PQ = ∣−16−36−64116∣\Rightarrow \ \ PQ \ = \ \bigg| \frac{-16-36-64}{\sqrt{116}} \bigg|
=∣−116116∣ =116 =229=|\frac{-116}{\sqrt{116}}| \ =\sqrt{116} \ = 2\sqrt{29}
∴  PQ=229 units\therefore \ \ PQ=2\sqrt{29} \ units