Question

Determine the angle between the directions of vector a = 3.00i 1.00j and vector b = 1.00i 3.00j .

Answer 1

The angle between the two vectors is 126° 52' 11".

The given parameters;

vector A = 3.00i + 1.00j

vector B = 1.00i + 3.00j

The angle between the two vectors is calculated as follows;

$cos \theta = \frac{A.B}{|A|.|B|}$

The dot product of vectors A and B is calculated as;

A.B = ( 3i + 1j ) . ( 1i +3j )

      = ( 3 × 1 ) + ( 1  × 3 )

      = 3 + 3

      = 6

The magnitude of vectors A and B is calculated as;

|A|  = $\sqrt[]{3^2 + 1^2} = \sqrt[]{10}$

|B| = $\sqrt[]{1^2 + 3^2} = \sqrt[]{10}$

The angle between in two vectors is calculated as;

$Cos \theta = \frac{6}{\sqrt[]{10}\sqrt[]{10} } \\\\Cos \theta = \frac{6}{\ 10 }\\\\Cos \theta = 0.6\\\\\theta = cos^-^1 (0.6)\\\\\theta = 126$

Therefore, the angle between the two vectors is 126° 52' 11".

Learn more about vectors here:

brainly.com/question/25705666

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