ExamBro
ExamBro
enEnglishhiहिन्दीguગુજરાતી
JEE Mains · Maths · STD 12 - 11. three dimension geometry

રેખાઓ \(\vec{r}_{1}=\alpha \hat{i}+2 \hat{j}+2 \hat{k}+\lambda(\hat{i}-2 \hat{j}+2 \hat{k}), \lambda \in R, \alpha>0\) અને \(\vec{r}_{2}=-4 \hat{i}-\hat{k}+\mu(3 \hat{i}-2 \hat{j}-2 \hat{k}), \mu \in R\) વચ્ચે  નું ન્યૂનતમ અંતર \(9,\) હોય તો \(\alpha\) ની કિમંત મેળવો.

  1. A \(21\)
  2. B \(4\)
  3. C \(66\)
  4. D \(6\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(6\)

Step-by-step Solution

Detailed explanation

If \(\vec{r}=\bar{a}+\lambda \vec{b}\) and \(\vec{r}=\vec{c}+\lambda \vec{d}\) then shortest distance between two lines is \(L=\frac{(\bar{a}-\vec{c}) \cdot(\bar{b} \times \bar{d})}{|b \times d|}\) \(\therefore \vec{a}-\vec{c}=((\alpha+4) \hat{i}+2 \hat{j}+3 k)\)…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app