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Question Number 158081 by Eric002 last updated on 30/Oct/21

$$determinetheanglebetweentwovectors$$$$A=4ax+ay-3azandB=2ax+4ay-3az$$

Answered by ajfour last updated on 30/Oct/21

$$\cos \theta =\frac{4\left(2\right)+1\left(4\right)-3(-3)}{\sqrt{4^2+1^2+{(-3)}^2}\sqrt{2^2+4^2+{(-3)}^2}}$$$$=\frac{21}{\sqrt{26}\sqrt{29}}$$$$\theta ={\cos }^{-1}\left(\frac{21}{\sqrt{26}\sqrt{29}}\right).$$

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