Question

please with explaination.​

Answer 1

Let ∠BOA=ϕ

ln ΔOBE

BE=OB

∠BEO=∠BOE

=(θ−ϕ)

lnΔOBE

∠BEO+∠BOE+∠EBO=180

∘

(θ−ϕ)+2ϕ+(θ−ϕ)=180

∘

2θ+2ϕ−2ϕ=180

∘

2θ=180

∘

θ=90

∘

So, vector (2

P

+

Q

) will make 90

∘

with

Q

.

Ans =90

please mark as the brainliest

please

Answer 2

Question :-

The Vector sum of two vectors $\sf \vec{P}$ and $\sf \vec{Q}$ is $\sf \vec{R}$ .If vector$\sf \vec{Q}$ is reversed, the resultant becomes $\sf \vec{S}$ then prove that R² + S² = 2(P² + Q²)

AnSwer :-

According to the question...

Let θ be the angle between vectors $\sf \vec{P} </p><p>$ and$\sf \vec{Q} is$

Then,

R² = P² + Q² + 2PQcosθ ⠀⠀⠀⠀⠀⠀.......(1)

When Vector $\sf \vec{Q}$ is reversed, angle between the vector $\sf \vec{P}$ and $\sf -\vec{Q}$ will become 180° - θ

Thus,

S² =P² + Q² + 2PQ (180°−θ)

S² =P² + Q² + 2PQ × (−cosθ)

S² =P² + Q² − 2PQcosθ ⠀⠀⠀⠀.......(2)

Adding the equation (1) and (2)

$\sf {R}^{2} + {S}^{2} = {P}^{2} + {Q}^{2} \cancel{+ 2PQcos \theta} + {P}^{2} + {Q}^{2} \cancel{- 2PQcos \theta}$

R² + S² =2( P² + Q² )

hence proved....