Question

The resultant of two vectors at an angle 150 degree is 10 units and is perpendicular to one vector. The magnitude of the smaller vector is

Answer 1

Ans. is 10✓3 units .... solution is in the pic :)

Answer 2

Answer:

The magnitude of the smaller vector is ${10}{\sqrt{3}}$ units.

Explanation:

Given the resultant of two vectors, R = 10 units

Angle between the two vectors, $\theta = 150^o$

And the resultant is perpendicular to one vector.    

Let us draw the vectors as shown in the figure.

If we join the vectors, $\vec A, \vec B ~\&~ \vec R$, that makes a parallelogram.

And the smaller vector is vector A.

Extending the vector in opposite direction towards E, then the angle made between vector B and vector E is $180-150=30^{o}$

Therefore, $\angle ABC =30^{o}$

And triangle ABC is a right angled triangle, with vector B along BC.

Therefore, using the definition of tan θ, we have

$tan\theta = \dfrac{opp}{adj}$

Substituting the given values,

$tan\ 30^{o} = \dfrac{10}{|\vec A|}\implies \dfrac{1}{\sqrt{3} } = \dfrac{10}{|\vec A|}$

                       $\implies |\vec A| = {10}{\sqrt{3}} units$

Therefore, the magnitude of the smaller vector is ${10}{\sqrt{3}}$ units.

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