Question

ans pls...

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if p+q=r and p=q=r then find the angle between p and q​

Answer 1

Answer:

The Angle Between P & Q is 120°

Explanation:

Answer:

The Angle Between P & Q is 120°

Given:

P + Q = R

|P| = |Q| = |R|

Explanation:

\rule{300}{1.5}

As Given |P| = |Q| = |R|

From vector's Formula,

\large\bigstar\;{\boxed{\tt R = \sqrt{ P^2 + Q^2 + 2\;PQ \cos \theta}}}★

R=

P

2

+Q

2

+2PQcosθ

\begin{gathered}\bold{Here}\begin{cases}\text{R Denotes resultant} \\ \text{P Denotes Vector} \\ \text{Q denotes Vector} \\ \theta \text{ Denotes Angle Between vectors}\end{cases}\end{gathered}

Here

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⎪

⎨

⎪

⎪

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R Denotes resultant

P Denotes Vector

Q denotes Vector

θ Denotes Angle Between vectors

Now,

\large{\boxed{\tt R = \sqrt{ P^2 + Q^2 + 2\;PQ \cos \theta}}}

R=

P

2

+Q

2

+2PQcosθ

Substituting the values,

\longmapsto\large{\tt R = \sqrt{ P^2 + Q^2 + 2\;PQ \cos \theta}}⟼R=

P

2

+Q

2

+2PQcosθ

Let |P| = |Q| = |R| = x [ Only in Magnitude ]

\longmapsto\large{\tt x = \sqrt{ x^2 + x^2 + 2\;x \times x \cos \theta}}⟼x=

x

2

+x

2

+2x×xcosθ

\longmapsto\large{\tt x^2 = x^2 + x^2 + 2\;x \times x \cos \theta}⟼x

2

=x

2

+x

2

+2x×xcosθ

\longmapsto\large{\tt x^2 = x^2 + x^2 + 2\;x^2 \cos \theta}⟼x

2

=x

2

+x

2

+2x

2

cosθ

\longmapsto\large{\tt x^2- x^2 = x^2 + 2\;x^2 \cos \theta}⟼x

2

−x

2

=x

2

+2x

2

cosθ

\longmapsto\large{\tt 0 = x^2 + 2\;x^2 \cos \theta}⟼0=x

2

+2x

2

cosθ

\longmapsto\large{\tt - x^2 = 2\;x^2 \cos \theta}⟼−x

2

=2x

2

cosθ

\longmapsto\large{\tt 2\;x^2 \cos \theta = - x^2}⟼2x

2

cosθ=−x

2

\longmapsto\large{\tt \cos \theta = \dfrac{- x^2}{2x^2}}⟼cosθ=

2x

2

−x

2

\longmapsto\large{\tt \cos \theta = \cancel{\dfrac{- x^2}{2x^2}}}⟼cosθ=

2x

2

−x

2

\longmapsto\large{\tt \cos \theta = \dfrac{- 1}{2}}⟼cosθ=

2

−1

\longmapsto\large{\tt \cos \theta =\cos 120^\circ}⟼cosθ=cos120

∘

\longmapsto\large{\underline{\boxed{\red{\tt \theta = 120^\circ }}}}⟼

θ=120

∘

∴ The Angle Between P & Q is 120°

Answer 2

Answer:

d) 120 degrees

Explanation:

Instead of going for vector algebra, you can solve this question by imagining itself

As magnituude of vectors are equal and by triangle law if they form a triangle then that has to be an equailateral triangle

Angle between sides is 60 degrees so angle between P and Q will be180- 60=120 degrees

OR You can work on the resultant formula

$P$=$\sqrt{P^{2}+P^{2}+2P^{2} cos\alpha }$

Squaring both sides,

$2P^{2}cos\alpha =-P^{2}\\cos \alpha =-1/2\\\alpha = 120$