Question

If two vetors and there resultant have same value than find out the agnle between two vector.

Answer 1

Solution:

Given:

=> It two vectors and there resultant have same value.

To Find:

=> Angle between two vectors.

So,

If two vectors are $\sf{\overrightarrow{P}}$ and $\sf{\overrightarrow{Q}}$ and resultant is $\sf{\overrightarrow{R},}$ then

$\implies \sf{\overrightarrow{P} = \overrightarrow{Q}=\overrightarrow{R}=A\;\;\;\;..........(1)}$

Now,

$\sf{\implies \overrightarrow{R} = \sqrt{\overrightarrow{P}^{2} +\overrightarrow{Q}^{2} + 2\overrightarrow{P}\overrightarrow{Q}\cos \theta}}$

From Equation (1),

$\sf{\implies A = \sqrt{A^{2}+A^{2}+2A^{2}\cos \theta}}$

$\sf{\implies A = \sqrt{2A^{2}+2A^{2}\cos\theta}}$

$\sf{\implies A=\sqrt{2A^{2}(1+\cos\theta)}}$

Square both sides, we get

$\sf{\implies A^{2}=2A^{2}(1+\cos\theta)}$

$\sf{\implies \dfrac{1}{2}=1+\cos\theta}$

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

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

$\large{\boxed{\boxed{\red{\sf{\implies \theta = 120^{\circ}}}}}}$

Answer 2

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore Angle=120\degree}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• In the given question information given about two vetors and there resultant have same value.

• We have to find out the agnle between two vector.

$\underline \bold{Given : } \\ \implies \vec{P} = \vec{Q} = \vec{R} = N \\ \\ \underline \bold{To \: Find: }\\ \implies Angle \: between \: \vec{P} \: and \: \vec{Q} = ?$

• According to given question :

$\bold{According \: to \: Vector \: formula : } \\ \implies \vec{R} = \sqrt{ (\vec{P})^{2} + ({ \vec{Q}})^{2} + 2 \: \vec{P} \: \vec{Q} \: cos \theta } \\ \\ \implies \vec{R} = \sqrt{ {N}^{2} + {N}^{2} + 2 \times N \times N \times cos \theta } \\ \\ \implies N= \sqrt{ {2N}^{2} + 2 {N}^{2} \times cos \theta} \\ \\ \bold{Squaring \: both \: side : }\\ \implies {N}^{2} = ({ \sqrt{2{N}^{2} (1 + cos \theta} })^{2} \\ \\ \implies {N}^{2} = 2 {N}^{2} (1 + cos \theta) \\ \\ \implies \frac{ \cancel{ {N}^{2} }}{2 \cancel{{N}^{2} }} = 1 + cos \theta \\ \\ \implies cos \theta = \frac{1}{2} - 1 \\ \\ \implies cos \theta = \frac{1 - 2}{2} \\ \\ \implies {cos} \theta = - \frac{ 1}{2} \\ \\ \implies\cancel{cos} \theta =\cancel{cos} 120 \degree \: \: \: \: \: \bold{(cos 120 \degree = - \frac{1}{2} )} \\ \\ \bold{\implies \theta = 120 \degree}$