Question

if a and b have same magnitude angle between them is 60°and their scalar product is 1/2 magnitude of a is​

Answer 1

Step-by-step explanation:

the value of a is 1 .

hope this will help you....

Answer 2

Given:

a and b have same magnitude angle between them is 60°and their scalar product is 1/2.

To find:

Magnitude of a.

Diagram:

$\boxed{\setlength{\unitlength}{1cm}\begin{picture}(6,6)\put(2,2){\vector(1,0){2}}\put(2,2){\vector(1,1){2}}\put(4.25,2){ \vec{a} }\put(4.25,4){ \vec{b} }\put(3.25,2.5){ {60}^{\circ} }\end{picture}}$

Calculation:

Scalar product or dot product between the supplied vectors can be represented as follows:

$\therefore \: \vec{a} \: . \: \vec{b} = \dfrac{1}{2}$

$= > | \vec{a}| \times | \vec{b}| \times \cos( \theta) = \dfrac{1}{2}$

$= > | \vec{a}| \times | \vec{b}| \times \cos( {60}^{ \circ} ) = \dfrac{1}{2}$

$= > | \vec{a}| \times | \vec{a}| \times \cos( {60}^{ \circ} ) = \dfrac{1}{2}$

$= > | \vec{a}| \times | \vec{a}| \times \dfrac{1}{2} = \dfrac{1}{2}$

$= > | \vec{a}| \times | \vec{a}| = 1$

$= > \: { | \vec{a}| }^{2} = 1$

$= > \: | \vec{a}| = 1$

So, final answer is:

$\boxed{ \bf{ \: | \vec{a}| = 1}}$