Question

The vectors P and Q have equal magnitude and the resultant of these two is also equal to the magnitude of P. What is the angle between P and Q.

Answer 1

Given:

• |P| = |Q|
• |P + Q| = |R| = |P| ( resultant of the vectors is equal to their respective magnitude)

Solution:

$\implies \rm R = \sqrt{ P^2 + Q^2 + 2PQcos\theta}$

$\implies \rm R = \sqrt{ P^2 + P^2 + 2P.Pcos\theta}$

$\implies \rm P = \sqrt{ 2P^2+ 2P^2cos\theta}$

$\implies \rm P = \sqrt{ P^2(2+2cos\theta)}$

$\implies \rm P = P\sqrt{ 2+2cos\theta}$

$\implies \rm 1 = \sqrt{ 2+2cos\theta}$

•• Taking Square on both sides

$\implies \rm 1 = 2+2cos\theta$

$\implies \rm 1 -2=2cos\theta$

$\implies \rm -1=2cos\theta$

$\implies \rm cos\theta = \dfrac{-1}{2}$

$\implies \bf \theta = 120\degree$

★ Angle between vectors P and Q is 120°

Answer 2

Explanation:

Given:

• |P| = |Q|
• |P + Q| = |R| = |P| ( resultant of the vectors is equal to their respective magnitude)

Solution:

$\implies \rm R = \sqrt{ P^2 + Q^2 + 2PQcos\theta}$

$\implies \rm R = \sqrt{ P^2 + P^2 + 2P.Pcos\theta}$

$\implies \rm P = \sqrt{ 2P^2+ 2P^2cos\theta}$

$\implies \rm P = \sqrt{ P^2(2+2cos\theta)}$

$\implies \rm P = P\sqrt{ 2+2cos\theta}$

$\implies \rm 1 = \sqrt{ 2+2cos\theta}$

•• Taking Square on both sides

$\implies \rm 1 = 2+2cos\theta$

$\implies \rm 1 -2=2cos\theta$

$\implies \rm -1=2cos\theta$

$\implies \rm cos\theta = \dfrac{-1}{2}$

$\implies \bf \theta = 120\degree$

★ Angle between vectors P and Q is 120°