0

Guest

Courses New

Good afternoon. Please provide the solution for the attachment

Good afternoon. Please provide the solution for the attachment

Question Image
Grade:11

2 Answers

Sanket Kulkarni
11 Points
6 years ago
Let P and Q be the magnitude of vectors.P:Q=1:2 => 2P=QNow let a be the angle made by resultant vector R with vector P. And b be angle between P and Qa= tan–¹(√3/2)tan a = √3/2By vector formula tan a= Qsinb / (P + Qcosb)So √3/2= Qsinb / (P + Qcosb)Put Q = 2P√3/2= 2Psinb/ (P+2Pcosb)= 2sinb/ 1+2cosbSquaring both sides3/4 = 4 sin²b /( 1 + 4 cos²b + 4cosb)Cross multiply & Substitute sin²b=1- cos²b3+12cos²b+12cosb= 16(1-cos²b)Simplify into28 cos²b+ 12cosb -13=0cos b= {-12 ± √[144– 4×28×(-13)] } / 2×28 cos b= (-12 ± 40 )/ 2×28 cos b = -52/56 or 28/56cos b = -13/14 or 1/2b = cos ^-1( -13/14 or 1/2)The negative can be neglectedTherefore b = cos ^-1(1/2)b = 60° or [π/3]
Gaurav Gupta
askIITians Faculty 686 Points
6 years ago
564-433_60.jpg

Think You Can Provide A Better Answer ?

Other Related Questions on General Physics

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More