Question

Two vectors of magnitudes 3 and P are making an angle theta with each other.The resultant of two vectors is sqrt(3) units and its direction is perpendicular to vector of magnitude 3 units then​

Answer 1

Given :  Two vectors of magnitudes 3 and P are making an angle theta with each other. The resultant of two vectors is √3 units and its direction is perpendicular to vector of magnitude 3 units  

To Find : Value of P  and angle θ

Solution:

For simplification lets assume  vector of magnitude  3 is along x axis

and vector of magnitude P is at angle θ  to x axis

Horizontal component of 3  = 3  units

Vertical  component of 3  =  0  units

Horizontal component of P  = Pcosθ  units

Vertical  component of P =   Psinθ  units

Resultant vector is perpendicular to vector of magnitude 3 units  => along y axis  Hence no horizontal component .. only vertical component = √3

3  + Pcosθ  =  0   =>   Pcosθ  = - 3

0 +  Psinθ  = √3    =>  Psinθ  =  √3  

Squaring and adding both

=> P² (cos²θ + sin²θ) = 9 + 3

=> P² = 12

=> P = 2√3

Psinθ  =  √3          

Pcosθ  = - 3  

as sin is +ve and cos is -ve hence 2nd quadrant .

=> tanθ  = -1/√3    and θ = 150°

Value of P = 2√3   and angle θ =  150°

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Answer 2

Given:

Two vectors of magnitudes 3 and P are making an angle theta with each other.The resultant of two vectors is √3 units and its direction is perpendicular to vector of magnitude 3.

To find:

Value of $\theta$ and P?

Calculation:

First , let's calculate the angle $\phi$ as given in the figure:

$\rm \: \tan( \phi ) = \dfrac{3}{R}$

$\rm \implies \: \tan( \phi ) = \dfrac{3}{ \sqrt{3} }$

$\rm \implies \: \tan( \phi ) = \sqrt{3}$

$\rm \implies \: \phi = {60}^{ \circ}$

Now, total angle between the vector P and 3 is $\theta$

$\rm \theta = {90}^{ \circ} + \phi$

$\rm \implies\theta = {90}^{ \circ} + {60}^{ \circ}$

$\rm \implies\theta = {150}^{ \circ}$

So, angle between the vectors is 150°.

Now, let's find value of P:

$\rm \sqrt{ {P}^{2} + {3}^{2} + 2.P.3 \cos( {150}^{ \circ} ) } = R$

$\rm \implies\sqrt{ {P}^{2} + 9 - 6P \sin( {60}^{ \circ} ) } = \sqrt{3}$

$\rm \implies\sqrt{ {P}^{2} + 9 - 3 \sqrt{3} P } = \sqrt{3}$

$\rm \implies{P}^{2} + 9 - 3 \sqrt{3} P = 3$

$\rm \implies{P}^{2} - 3 \sqrt{3} P + 6 = 0$

$\rm \implies{P}^{2} - 2 \sqrt{3}P - \sqrt{3} P + 6 = 0$

$\rm \implies P ( P- 2 \sqrt{3})- \sqrt{3} (P - 2 \sqrt{3}) = 0$

$\rm \implies (P- \sqrt{3}) (P - 2 \sqrt{3}) = 0$

So, P = √3 , or P = 2√3.