Question

Find the angle between the vectors i + 2j + 3 k and 3i - j + 2 k.

Answer 1

Answer:

Angle between vectors 60° or π/3

Step-by-step explanation:

let

$\vec a=\hat i+2\hat j+3\hat k\\\\\\\vec b=3\hat i-\hat j+2\hat k$

Angle between to vectors

$cos\:\theta=\frac{\vec a.\vec b}{|\vec a||\vec b|} \\\\cos\:\theta=\frac{(\hat i+2\hat j+3\hat k).(3\hat i-\hat j+2\hat k)}{|\hat i+2\hat j+3\hat k|\:|3\hat i-\hat j+2\hat k|}$

$cos\:\theta=\frac{3-2+6}{\sqrt{1^{2}+2^{2} +3^{2}}\sqrt{3^{2}+1^{2} +2^{2}  } } \\\\\\cos\:\theta=\frac{7}{\sqrt{14}\sqrt{14} }\\ \\\\cos\:\theta=\frac{1}{2} \\\\\\\theta=cos^{-1} \frac{1}{2} \\\\\\\theta=cos^{-1} cos(\frac{\pi }{3}) \\\\\\\theta =\frac{\pi }{3}$