Skip to content
Question of 68

Q.Find the shortest distance between the lines r⃗=(i^+2j^+k^)+λ(i^−j^+k^)\vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k}) and r⃗=(2i^−j^−k^)+μ(2i^+j^+2k^)\vec{r} = (2\hat{i} - \hat{j} - \hat{k}) + \mu(2\hat{i} + \hat{j} + 2\hat{k}).

Assam AhsecAHSEC Higher Secondary (HS) Final Examination 2018Subjective· 4mImportance★★★★★
0% · 0/68 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Use d=∣(a⃗2−a⃗1)⋅(b⃗1×b⃗2)∣∣b⃗1×b⃗2∣d = \dfrac{|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)|}{|\vec b_1\times\vec b_2|}; here it equals 32\dfrac{3}{\sqrt2}.

The lines are r⃗=a⃗1+λb⃗1\vec r = \vec a_1 + \lambda\vec b_1 and r⃗=a⃗2+μb⃗2\vec r = \vec a_2 + \mu\vec b_2 with a⃗1=(1,2,1)\vec a_1 = (1,2,1), b⃗1=(1,−1,1)\vec b_1 = (1,-1,1), a⃗2=(2,−1,−1)\vec a_2 = (2,-1,-1), b⃗2=(2,1,2)\vec b_2 = (2,1,2).

a⃗2−a⃗1=(1,−3,−2)\vec a_2 - \vec a_1 = (1, -3, -2).

b⃗1×b⃗2=∣i^j^k^1−11212∣=i^(−1⋅2−1⋅1)−j^(1⋅2−1⋅2)+k^(1⋅1−(−1)⋅2)=−3i^+0j^+3k^\vec b_1\times\vec b_2 = \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{vmatrix} = \hat i(-1\cdot 2 - 1\cdot 1) - \hat j(1\cdot 2 - 1\cdot 2) + \hat k(1\cdot 1 - (-1)\cdot 2) = -3\hat i + 0\hat j + 3\hat k.

∣b⃗1×b⃗2∣=(−3)2+02+32=18=32|\vec b_1\times\vec b_2| = \sqrt{(-3)^2 + 0^2 + 3^2} = \sqrt{18} = 3\sqrt{2}.

(a⃗2−a⃗1)⋅(b⃗1×b⃗2)=(1)(−3)+(−3)(0)+(−2)(3)=−3−6=−9(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2) = (1)(-3) + (-3)(0) + (-2)(3) = -3 - 6 = -9.

Shortest distance =∣−9∣32=932=32=322= \dfrac{|-9|}{3\sqrt2} = \dfrac{9}{3\sqrt2} = \dfrac{3}{\sqrt2} = \dfrac{3\sqrt2}{2}.

…

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.