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}.

Haryana BsehBSEH Intermediate Board 2023Subjective· 6mImportance★★★★★
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 the skew-line shortest-distance formula d=∣(b⃗−a⃗)⋅(d⃗1×d⃗2)∣∣d⃗1×d⃗2∣d=\dfrac{|(\vec b-\vec a)\cdot(\vec d_1\times\vec d_2)|}{|\vec d_1\times\vec d_2|}.

Line 1: passes through (−1,−1,−1)(-1,-1,-1), direction d⃗1=(7,−6,1)\vec d_1=(7,-6,1).

Line 2: passes through (3,5,7)(3,5,7), direction d⃗2=(1,−2,1)\vec d_2=(1,-2,1).

Vector connecting the two points: b⃗−a⃗=(3−(−1), 5−(−1), 7−(−1))=(4,6,8)\vec b-\vec a=(3-(-1),\,5-(-1),\,7-(-1))=(4,6,8)

d⃗1×d⃗2=∣i^j^k^7−611−21∣=i^[(−6)(1)−(1)(−2)]−j^[(7)(1)−(1)(1)]+k^[(7)(−2)−(−6)(1)]\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)(1)-(1)(-2)]-\hat j[(7)(1)-(1)(1)]+\hat k[(7)(-2)-(-6)(1)]

=i^(−6+2)−j^(7−1)+k^(−14+6)=(−4,−6,−8)=\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.