Question

find the angle between A=2i+2j-k and B=6i-3j+2k​

Answer 1

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Let,

• θ be the angle between the two given vectors .

☕ A = 2i + 2j - k

☕ B = 6i - 3j + 2k

We know that,

$\bf{\vec{a}\:.\:\vec{b}\:=\:\mid{\vec{a}}\mid\:\mid{\vec{b}}\mid\:\cos{\theta}\:}$

→ $\bf{\vec{a}\:.\:\vec{b}\:}$ = 2 × 6 + 2 × (-3) + (-1) × 2

→ $\bf{\vec{a}\:.\:\vec{b}\:}$ = 4

Now,

• $\bf{\mid{\vec{a}}\mid}$ = 3

• $\bf{\mid{\vec{b}}\mid}$ = 4

☕ $\bf\red{cos{\theta}\:=\:\dfrac{\vec{a}\:.\:\vec{b}}{\mid{\vec{a}}\mid\:\mid{\vec{b}}\mid}\:}$

→ cosθ = 4/(3 × 7) = 4/21

$\bf{\implies\:\theta\:=\:cos^{-1}\:\dfrac{4}{21}\:=\:79°\:}$

Answer 2

$\huge\red{\boxed{\blue{\mathcal{\overbrace{\underbrace{\fcolorbox{blue}{aqua}{\underline{\red{ANSWER}}}}}}}}}$

A - 2i+2j-k

B - 6i-3j+2k

$|a| \\ = 3 \\ \\ \\ \\ |b| = 4$

$\cos = 4 \(3 \times 7) = 4 \21$

$\cos( - 1) \frac{4}{12} = 79 \: hence \: proved$