ExamBro
ExamBro
enEnglishhiहिन्दीguગુજરાતી
JEE Mains · Maths · STD 11 - 9. straight line

દરેક \(p\,>\,0\), સદીશ \(\vec{v}_{2}=2 \hat{i}+(p+1) \hat{j}\) એ સદીશ \(\vec{v}_{1}=\sqrt{3} p \hat{i}+\hat{j}\) ને  \(\theta\) ખૂણે વિષમઘડી દિશામાં ભ્રમણ કરી ને મેળવી શકાય છે. જો  \(\tan \theta=\frac{(\alpha \sqrt{3}-2)}{4 \sqrt{3}+3}\) હોય તો \(\alpha\) ની કિમંત મેળવો.

  1. A \(6\)
  2. B \(5\)
  3. C \(4\)
  4. D \(3\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(6\)

Step-by-step Solution

Detailed explanation

\(\left|\vec{V}_{1}\right|=\left|\overrightarrow{\mathrm{V}}_{2}\right|\) \(3 \mathrm{P}^{2}+1=4+(\mathrm{P}+1)^{2}\) \(2 \mathrm{P}^{2}-2 P-4=0\) \(\Rightarrow \mathrm{P}^{2}-\mathrm{P}-2=0\) \(\mathrm{P}=2,-1\) (rejected)…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app