Question

Find the angle between the vectors : vector a =3i - 4j and vector b = -2i+3k

Answer 1

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$\underline{\large\mathcal\red{solution}}$

a =3i - 4j and vector b = -2i+3k

now ...|a|=√[(3)²+(-4)²]=5

and.....|b|=√[(-2)²+(3)²]=√13

now the dot product of a and b is

a.b=-6-12=-18

the angle is

$\cos {}^{ - 1} ( \frac{ - 18 }{ 5 \sqrt{13} } )$

=3.09°

$\large\mathcal\red{hope\: this \: helps \:you......}$

Answer 2

Explanation:

$Given \: it \: \\ a = 3i - 4j \: \: \:, \: \: b = - 2i + 3k \\ let \: \alpha \: be \: an \: angle \: between \: a \: and \: b \\ cos \alpha = \frac{(3i - 4j).( - 2i + 3k)}{ |3i - 4j| | - 2i + 3k| } \\ cos \alpha = \frac{ - 6}{ 5 \sqrt{13} } \\ \alpha = {cos}^{ - 1} \frac{ - 6}{5 \sqrt{13} } \\ \\ \\ \\ \\ hope \: it \: will \: help \: you...$