Find the sin angle between the vectors $A = 3 i - 4 j + 5 k$ and $B = i - j + k$ ?

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Solution

Consider the problem

The resultant of two vectors is the vector sum of two vectors
$\to R = \to a + \to b$
$\begin{matrix} = 3 \to i + 4 \to j + 5 \to k + 5 \to i + 3 \to j + 4 \to k = 8 \to i + 7 \to j + 9 \to k . . . . . . .1 > \end{matrix}$

And magnitude of $R$
$\begin{matrix} ∣ ∣ \to R ∣ ∣ = \sqrt{8^{2} + 7^{2} + 9^{2}} = \sqrt{64 + 49 + 81} = \sqrt{194} = 13.9 \end{matrix}$

Now, using the formula to find the angle between two vectors
$cos θ = \frac{\to a . \to b}{| \to a | ∣ ∣ \to b ∣ ∣}$
$\begin{matrix} cos θ = \frac{(8 \to i + 7 \to j + 9 \to k) . \to i}{∣ ∣ (8 \to i + 7 \to j + 9 \to k) ∣ ∣ ∣ ∣ \to i ∣ ∣} = \frac{8}{13.9 \times 1} = 0.57 \end{matrix}$

$cos θ = 0.57$

$\begin{matrix} \Rightarrow sin θ = \sqrt{1^{2} - {(0.57)}^{2}} = \sqrt{0.6751} = 0.82 \end{matrix}$

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Similar questions

2 \to i - \to j + \to k, \to i - 3 \to j - 5 \to k, - 3 \to i + 4 \to j + 4 \to k

2 \hat{i} - \hat{j} + \hat{k}, \hat{i} - 3 \hat{j} - 5 \hat{k} and 3 \hat{i} - 4 \hat{j} - 4 \hat{k}

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

θ

\to r = 2 i + j - k + (i + j + k) t

\to r \cdot (3 i - 4 j + 5 k) = q

- i + j - k

3 i + 4 j

\vec{a} = \hat{i} + 4 \hat{j} + 2 \hat{k}, \vec{b} = 3 \hat{i} - 2 \hat{j} + 7 \hat{k} and \vec{c} = 2 \hat{i} - \hat{j} + 4 \hat{k} .

\vec{d}

\vec{a} and \vec{b}

and \vec{c} \cdot \vec{d} = 15 .

\vec{a} = 4 \hat{i} + 5 \hat{j} - \hat{k}, \vec{b} = \hat{i} - 4 \hat{j} + 5 \hat{k} and \vec{c} = 3 \hat{i} + \hat{j} - \hat{k}

\vec{d}

\vec{c} and \vec{b} and \vec{d} . \vec{a} = 21

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