The sine of the angle between the two vectors | vedclass

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(A) Let $\vec{a} = 3i + 2j - k$ and $\vec{b} = 12i + 5j - 5k$.The cross product $\vec{a} 	imes \vec{b}$ is given by:$\vec{a} 	imes \vec{b} = \begin{

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$\frac{\sqrt{115}}{\sqrt{14}\sqrt{194}}$

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(A) Let $\vec{a} = 3i + 2j - k$ and $\vec{b} = 12i + 5j - 5k$.The cross product $\vec{a} 	imes \vec{b}$ is given by:$\vec{a} 	imes \vec{b} = \begin{vmatrix} i & j & k \ 3 & 2 & -1 \ 12 & 5 & -5 \end{vmatrix} = i(-10 - (-5)) - j(-15 - (-12)) + k(15 - 24) = -5i + 3j - 9k$.The magnitude of the cross product is $|\vec{a} 	imes \vec{b}| = \sqrt{(-5)^2 + 3^2 + (-9)^2} = \sqrt{25 + 9 + 81} = \sqrt{115}$.The magnitudes of the vectors are $|\vec{a}| = \sqrt{3^2 + 2^2 + (-1)^2} = \sqrt{9 + 4 + 1} = \sqrt{14}$ and $|\vec{b}| = \sqrt{12^2 + 5^2 + (-5)^2} = \sqrt{144 + 25 + 25} = \sqrt{194}$.Since $|\vec{a} 	imes \vec{b}| = |\vec{a}| |\vec{b}| \sin 	heta$,we have $\sin 	heta = \frac{|\vec{a} 	imes \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{\sqrt{115}}{\sqrt{14}\sqrt{194}}$.

The sine of the angle between the two vectors $3i + 2j - k$ and $12i + 5j - 5k$ will be

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