Skip to content
Question of 68

Q.Find the shortest distance between the lines :
→r = î + ĵ + λ (2î − ĵ + k̂)
→r = 2î + ĵ − k̂ + μ (3î − 5ĵ + 2k̂)

Kerala DhseKerala DHSE Plus Two Board 2025Subjective· 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 →

For two skew lines 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, the shortest distance is ∣(a⃗2−a⃗1)⋅(b⃗1×b⃗2)∣b⃗1×b⃗2∣∣\left|\dfrac{(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)}{|\vec b_1\times\vec b_2|}\right|.

Line 1: a⃗1=i^+j^\vec a_1=\hat i+\hat j, b⃗1=2i^−j^+k^\vec b_1=2\hat i-\hat j+\hat k.

Line 2: a⃗2=2i^+j^−k^\vec a_2=2\hat i+\hat j-\hat k, b⃗2=3i^−5j^+2k^\vec b_2=3\hat i-5\hat j+2\hat k.

Step 1: a⃗2−a⃗1=(2−1)i^+(1−1)j^+(−1−0)k^=i^−k^=(1,0,−1)\vec a_2-\vec a_1 = (2-1)\hat i+(1-1)\hat j+(-1-0)\hat k = \hat i-\hat k = (1,0,-1).

Step 2: b⃗1×b⃗2\vec b_1\times\vec b_2:

b⃗1×b⃗2=∣i^j^k^2−113−52∣=i^[(−1)(2)−(1)(−5)]−j^[(2)(2)−(1)(3)]+k^[(2)(−5)−(−1)(3)]\vec b_1\times\vec b_2 = \begin{vmatrix}\hat i & \hat j & \hat k\\ 2 & -1 & 1\\ 3 & -5 & 2\end{vmatrix} = \hat i[(-1)(2)-(1)(-5)] - \hat j[(2)(2)-(1)(3)] + \hat k[(2)(-5)-(-1)(3)]

=i^(−2+5)−j^(4−3)+k^(−10+3)=3i^−j^−7k^=(3,−1,−7)= \hat i(-2+5) - \hat j(4-3) + \hat k(-10+3) = 3\hat i - \hat j - 7\hat k = (3,-1,-7)

…

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.