Given the vectors vec x = 3i - 6j - k,vec y = | vedclass

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(B) The projection of vector $\vec a$ onto vector $\vec b$ is given by $\frac{\vec a \cdot \vec b}{|\vec b|}$.Here,$\vec a = \vec x 	imes \vec y$ and

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$-14$

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(B) The projection of vector $\vec a$ onto vector $\vec b$ is given by $\frac{\vec a \cdot \vec b}{|\vec b|}$.Here,$\vec a = \vec x 	imes \vec y$ and $\vec b = \vec z$.So,the projection is $\frac{(\vec x 	imes \vec y) \cdot \vec z}{|\vec z|} = \frac{[\vec x \vec y \vec z]}{|\vec z|}$.First,calculate the scalar triple product $[\vec x \vec y \vec z]$:$[\vec x \vec y \vec z] = \begin{vmatrix} 3 & -6 & -1 \ 1 & 4 & -3 \ 3 & -4 & -12 \end{vmatrix} = 3(-48 - 12) + 6(-12 + 9) - 1(-4 - 12) = 3(-60) + 6(-3) - 1(-16) = -180 - 18 + 16 = -182$.Next,calculate the magnitude of $\vec z$:$|\vec z| = \sqrt{3^2 + (-4)^2 + (-12)^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$.Finally,the projection is $\frac{-182}{13} = -14$.

Given the vectors $\vec x = 3i - 6j - k$,$\vec y = i + 4j - 3k$,and $\vec z = 3i - 4j - 12k$,find the projection of the vector $\vec x \times \vec y$ onto the vector $\vec z$.

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