Skip to content
Question of 68

Q.Find the shortest distance between the lines x+17=y+1−6=z+11\dfrac{x+1}{7} = \dfrac{y+1}{-6} = \dfrac{z+1}{1} and x−31=y−5−2=z−71\dfrac{x-3}{1} = \dfrac{y-5}{-2} = \dfrac{z-7}{1}.

Uttar Pradesh UpmspUP Board (UPMSP) Intermediate 2022Subjective· 8mImportance★★★★★
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 →

Applying the skew-lines formula gives shortest distance =116229=229=\dfrac{116}{2\sqrt{29}}=2\sqrt{29}.

Concept. For lines r⃗=a⃗1+λd⃗1\vec r=\vec a_1+\lambda\vec d_1 and r⃗=a⃗2+μd⃗2\vec r=\vec a_2+\mu\vec d_2,

SD=∣(a⃗2−a⃗1)⋅(d⃗1×d⃗2)∣∣d⃗1×d⃗2∣.\text{SD}=\frac{\big|(\vec a_2-\vec a_1)\cdot(\vec d_1\times\vec d_2)\big|}{|\vec d_1\times\vec d_2|}.

Here a⃗1=(−1,−1,−1), d⃗1=(7,−6,1), a⃗2=(3,5,7), d⃗2=(1,−2,1)\vec a_1=(-1,-1,-1),\ \vec d_1=(7,-6,1),\ \vec a_2=(3,5,7),\ \vec d_2=(1,-2,1).

d⃗1×d⃗2=∣i^j^k^7−611−21∣=i^(−6+2)−j^(7−1)+k^(−14+6)=(−4,−6,−8).\vec d_1\times\vec d_2=\begin{vmatrix}\hat i&\hat j&\hat k\\7&-6&1\\1&-2&1\end{vmatrix}=\hat i(-6+2)-\hat j(7-1)+\hat k(-14+6)=(-4,-6,-8). …

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.