Question

what is the angle between 2 vectors if the ratio between their dot product the magnitude of their cross product is square root of 3.

Answer 1

Answer:

The angle between the vectors is 30°.

Dot product of two vectors A and B: A.B = |A|.|B|cos(Θ).

Cross product of two vectors A and B: AxB = |A|.|B|sin(Θ)

Explanation:

Given that the ratio of the dot product to cross product is $\sqrt{3}$.

this implies that cos(Θ)/sin(Θ) = $\sqrt{3}$ .

We will get the angle between the vectors as 30°. Since cot(30°) = $\sqrt{3}$.

Answer 2

Answer:

The correct answer is $30^{\circ}$.

Explanation:

To find the angle between 2 vectors if the ratio between their dot product the magnitude of their cross product is the square root of 3.

Dot product

The dot product exists as one method of multiplying two or more given vectors. The last result of the dot product of vectors exists as a scalar quantity. Therefore, the dot product exists also recognized as a scalar product.

Cross product

A vector perpendicular to two given vectors, u and v, and containing a magnitude similar to the product of the magnitudes of the two provided vectors multiplied by the sine of the angle between the two provided vectors, usually defined by u × v.

Step 1

Let $A$ and $B$ are two vectors, dot product of $A$ and $B$

$A.B=|A||B| cosQ$

the cross product of $A$ and $B$

$A*B=|A||B|sinQ$

Step 2

now the ratio of these

$\frac{A.B}{A*B} =\frac{|A||B|cosQ}{|A||B|sinQ}$

$\frac{|A||B|cosQ}{|A||B|sinQ} = \sqrt{3}$

$\frac{cosQ}{sinQ} =cotQ$

Step 3

now according to your question ratio is equal to $\sqrt{3}$.

$\Rightarrow \cot Q=\sqrt{3}$

$&\Rightarrow \tan Q=\frac{1}{\sqrt{3}}$

$\ Q =\tan ^{-1}\left(\frac{1}{3}\right)=30^{\circ}$

Therefore, the correct answer is $30^{\circ}$.

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