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Applications of Dot Product

The angle bet...

Question

The angle between the vectors $3 i + j + 2 k$ and $2 i - 2 j + 4 k$ is

{cos}^{- 1} \sqrt{\frac{3}{28}}

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{sin}^{- 1} \frac{2}{\sqrt{7}}

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{cos}^{- 1} \frac{2}{\sqrt{5}}

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{sin}^{- 1} \frac{2}{\sqrt{5}}

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Solution

The correct option is B ${cos}^{- 1} \sqrt{\frac{3}{28}}$
Let $\to a =$ $3 i + j + 2 k$ and $\to b =$ $2 i - 2 j + 4 k$ .

Now angle between $\to a$ and $\to b$ be

${cos}^{- 1} (\frac{\to a . \to b}{| \to a | | \to b |}) =$ ${cos}^{- 1} (\frac{6 - 2 + 8}{\sqrt{14} . \sqrt{24}}) =$ ${cos}^{- 1} (\sqrt{\frac{3}{28}})$ .

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Similar questions

\to A = 3 i + j + 2 k

\to B = 2 i - 2 j + 4 k

θ

s i n θ

2 ¯ i - 3 ¯ j + 4 ¯ ¯ ¯ k, ¯ i - 2 ¯ j + 3 ¯ ¯ ¯ k

3 ¯ i + ¯ j - 2 ¯ ¯ ¯ k

2 \hat{i} - 2 \hat{j} + 4 \hat{k} and 3 \hat{i} + \hat{j} + 2 \hat{k},

\frac{2}{3}

\frac{2}{\sqrt{7}}

\frac{\sqrt{2}}{7}

\sqrt{\frac{2}{7}}

\vec{r} \cdot (2 \hat{i} - 3 \hat{j} + 4 \hat{k}) = 1 and \vec{r} \cdot (- \hat{i} + \hat{j}) = 4

\vec{r} \cdot (2 \hat{i} - \hat{j} + 2 \hat{k}) = 6 and \vec{r} \cdot (3 \hat{i} + 6 \hat{j} - 2 \hat{k}) = 9

\vec{r} \cdot (2 \hat{i} + 3 \hat{j} - 6 \hat{k}) = 5 and \vec{r} \cdot (\hat{i} - 2 \hat{j} + 2 \hat{k}) = 9

2 \sin \frac{5 π}{12} \sin \frac{π}{12} = \frac{1}{2}

2 \cos \frac{5 π}{12} \cos \frac{π}{12} = \frac{1}{2}

2 \sin \frac{5 π}{12} \cos \frac{π}{12} = \frac{\sqrt{3} + 2}{2}

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