The acute angle theta between the vector vec{ | vedclass

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(A) Step $1$: Find the normal vector $\vec{n}$ to the plane containing $\vec{b}$ and $\vec{c}$ by calculating the cross product $\vec{n} = \vec{b} 	i

Vedclass · Accepted Answer

$\sin^{-1}(4/\sqrt{42})$

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(A) Step $1$: Find the normal vector $\vec{n}$ to the plane containing $\vec{b}$ and $\vec{c}$ by calculating the cross product $\vec{n} = \vec{b} 	imes \vec{c}$.$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & -1 \ 1 & -1 & 2 \end{vmatrix} = \hat{i}(6 - 1) - \hat{j}(4 + 1) + \hat{k}(-2 - 3) = 5\hat{i} - 5\hat{j} - 5\hat{k}$.Step $2$: The angle $\alpha$ between the vector $\vec{a}$ and the normal $\vec{n}$ is given by $\cos \alpha = \frac{|\vec{a} \cdot \vec{n}|}{|\vec{a}| |\vec{n}|}$.$\vec{a} \cdot \vec{n} = (2)(5) + (1)(-5) + (-3)(-5) = 10 - 5 + 15 = 20$.$|\vec{a}| = \sqrt{2^2 + 1^2 + (-3)^2} = \sqrt{4 + 1 + 9} = \sqrt{14}$.$|\vec{n}| = \sqrt{5^2 + (-5)^2 + (-5)^2} = \sqrt{25 + 25 + 25} = \sqrt{75} = 5\sqrt{3}$.Step $3$: $\cos \alpha = \frac{20}{\sqrt{14} \cdot 5\sqrt{3}} = \frac{4}{\sqrt{42}}$.Step $4$: The angle $	heta$ between the vector and the plane is $	heta = 90^\circ - \alpha$. Thus, $\sin 	heta = \cos \alpha = \frac{4}{\sqrt{42}}$, which implies $	heta = \sin^{-1}(4/\sqrt{42})$.

The acute angle $\theta$ between the vector $\vec{a} = 2\hat{i} + \hat{j} - 3\hat{k}$ and the plane containing the vectors $\vec{b} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{c} = \hat{i} - \hat{j} + 2\hat{k}$ is:

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