Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Let a vector be obtained by rotating the vector <img alt="\sqrt{3} \hat{i}+ \hat{j}" src="/latex-image/?%5Csqrt%7B3%7D%20%5Chat%7Bi

Let a vector \alpha \hat{i}+\beta \hat{j} be obtained by rotating the vector \sqrt{3} \hat{i}+ \hat{j} by an angle 45o about the origin in counterclockwise direction in the first quadrant. Then the area of triangle having vertices \left ( \alpha ,\beta \right ),\left ( 0,\beta \right ) and \left ( 0,0 \right ) is equal to :
Option: 1 2\sqrt{2}
Option: 2 \frac{1}{\sqrt{2}}
Option: 3 \frac{1}{2}
Option: 4 1

Answers (1)

best_answer

 

\begin{aligned} \text { Area of } \Delta\left(\mathrm{OA}^{\prime} \mathrm{B}\right)&=\frac{1}{2} \mathrm{O} \mathrm{A}^{\prime} \cos 15^{\circ} \times \mathrm{OA}^{\prime} \sin 15^{\circ} \\ &=\frac{1}{2}\left(\mathrm{OA}^{\prime}\right)^{2} \frac{\sin 30^{\circ}}{2} \\ &=(3+1) \times \frac{1}{8}=\frac{1}{2} \end{array}

 

Posted by

Suraj Bhandari

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

Amrita BTech Admission 2025: JEE Main-based registration deadline extended to August 30
anam-khan

DASA, CSAB Special round 2 seat allotment result 2025 declared on csab.nic.in

Latest Question