Question

two vectors of equal magnitude are inclined at an angle theta. when one of them is halved the angle with the resultent makes with other angle is also halved then find theta

Answer 1

Given:

Two vectors are of equal magnitude and are inclined at an angle θ

When one of the vectors is halved, the angle which the resultant makes with the other is also halved

To find:

The value of theta "θ"

Solution:

From the figure, we can see

• A and B are the two vectors and both are inclined at an angle θ

• R is the resultant vector

• $\vec{R}\:$ makes an angle $\phi$ with $\vec{A}$

We know the formula for the angle subtended by the resultant vector the vector A is given by,

$\boxed{tan\:\phi\:=\:\frac{B\sin\theta}{A\:+\:Bcos\theta} }$

Case 1:- Finding the value of θ when the magnitude of the vectors are equal i.e., A = B:

We will substitute A = B in the formula

$tan\:\phi\:=\:\frac{A\sin\theta}{A\:+\:Acos\theta} = \:\frac{A\sin\theta}{A(1\:+\:cos\theta)} \:=\:\frac{sin\theta}{1\:+\:cos\theta}$  

∵ according to the half angle formula ⇒ $tan(\frac{\theta}{2} ) = \frac{sin\theta}{(1\:+\:cos\theta)}$

∴ $tan\:\phi\:=\frac{sin\theta}{1\:+\:cos\theta} = tan (\frac{\theta}{2} )$

⇒ $tan\:\phi\:= tan (\frac{\theta}{2} )$

cancelling tan from both sides

⇒ $\phi=\frac{\theta}{2}$

⇒ $\bold{\theta = 2\phi}$ .......... (i)

Case 2:- Finding the value of θ, when one of the two equal magnitude vectors is halved and the angle between the resultant and that vector is halved

Let us assume B = $\frac{A}{2}$  so the angle $\phi$ will also be halved

Now, we will substitute B = $\frac{A}{2}$ and the angle as $\frac{\phi}{2}$ in the formula

$tan(\frac{\phi}{2})\:=\:\frac{\frac{A}{2} \sin\theta}{A\:+\:\frac{A}{2} cos\theta} = \:\frac{\frac{1}{2}A \sin\theta}{\frac{2A + Acos\theta}{2} } = \frac{A\:sin\:\theta}{2A\: +\: Acos\theta} = \frac{sin\theta}{2\:+\:cos\theta}$

∵ according to the half angle formula  we have ⇒ $tan(\frac{\phi}{2} ) = \frac{(1\:-\:cos\phi)}{sin\phi}$

∴ $\frac{(1\:-\:cos\phi)}{sin\phi} = tan(\frac{\phi}{2} ) = \frac{sin\theta}{2\:+\:cos\theta}$

⇒ $\frac{(1\:-\:cos\phi)}{sin\phi} = \frac{sin\theta}{2\:+\:cos\theta}$

substituting $\bold{\theta = 2\phi}$ from (i), we get

⇒ $\frac{(1\:-\:cos\phi)}{sin\phi} = \frac{sin\:2\phi}{2\:+\:cos\:2\phi}$

we have the formula:⇒ **sin 2θ = (2 sinθ cosθ)** & **cos 2θ = (2 cos²θ - 1)**

⇒ $\frac{(1\:-\:cos\phi)}{sin\phi}= \frac{2 \:sin\phi\: cos\phi}{2\:+\:2 cos^2\phi - 1}$

⇒ $\frac{(1\:-\:cos\phi)}{sin\phi} = \frac{2 \:sin\phi\: cos\phi}{1\:+\:2 cos^2\phi }$

now let's substitute sin$\phi$ = x and cos $\phi$ = y and

⇒ $\frac{(1\:-\:y)}{x} = \frac{2 \:x\: y}{1\:+\:2 y^2 }$

⇒ (1 - y)(1 + 2y²) = 2x²y

⇒ 1 + 2y² - y - 2y³ = 2x²y

we can say ⇒ x² + y² = 1 ...(∵ sin²θ + cos²θ = 1)

⇒ 1 + 2y² - y - 2y³ = 2(1 - y²)y

⇒ 1 + 2y² - y - 2y³ = 2y - 2y³

⇒ 1 + 2y² - y - 2y³ - 2y + 2y³ = 0

⇒ 1 + 2y² - y - 2y = 0

⇒ 2y² - 3y + 1 = 0

⇒ 2y² - 2y - y + 1 = 0

⇒ 2y(y - 1) - 1(y - 1) = 0

⇒ (2y - 1)(y - 1) = 0

⇒ y = ½ or 1

Therefore,

(1). cos $\phi$ = 1

⇒ $\phi$ = cos⁻¹ (1)

⇒ $\phi$ = 0 ← this value is not possible

(2). cos $\phi$ = ½

⇒ $\phi$ = cos⁻¹ (½)

⇒ $\phi$ = 60°

Now, substituting $\phi$ = 60° in eq. (i), we get

θ = 2 $\phi$ = 2 × 60° = 120°

Thus, the value of theta is 120°.

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Answer 2

Answer:

60 degree

Explanation:

Case 1:- Finding the value of θ when the magnitude of the vectors are equal i.e., A = B:

We will substitute A = B in the formula

 

∵ according to the half angle formula ⇒

∴

⇒

cancelling tan from both sides

⇒

⇒  .......... (i)

Case 2:- Finding the value of θ, when one of the two equal magnitude vectors is halved and the angle between the resultant and that vector is halved

Let us assume B =   so the angle  will also be halved

Now, we will substitute B =  and the angle as  in the formula

∵ according to the half angle formula  we have ⇒

∴

⇒

substituting  from (i), we get

⇒

we have the formula:⇒ **sin 2θ = (2 sinθ cosθ)** & **cos 2θ = (2 cos²θ - 1)**

⇒

⇒

now let's substitute sin = x and cos  = y and

⇒

⇒ (1 - y)(1 + 2y²) = 2x²y

⇒ 1 + 2y² - y - 2y³ = 2x²y

we can say ⇒ x² + y² = 1 ...(∵ sin²θ + cos²θ = 1)

⇒ 1 + 2y² - y - 2y³ = 2(1 - y²)y

⇒ 1 + 2y² - y - 2y³ = 2y - 2y³

⇒ 1 + 2y² - y - 2y³ - 2y + 2y³ = 0

⇒ 1 + 2y² - y - 2y = 0

⇒ 2y² - 3y + 1 = 0

⇒ 2y² - 2y - y + 1 = 0

⇒ 2y(y - 1) - 1(y - 1) = 0

⇒ (2y - 1)(y - 1) = 0

⇒ y = ½ or 1

Therefore,

(1). cos  = 1

⇒  = cos⁻¹ (1)

⇒  = 0 ← this value is not possible

(2). cos  = ½

⇒  = cos⁻¹ (½)

⇒  = 60°