The length of the altitude through the point | vedclass

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(D) The volume of a tetrahedron with vertices $A, B, C, D$ is given by $V = \frac{1}{6} |(\vec{a}-\vec{d}) \cdot ((\vec{b}-\vec{d}) 	imes (\vec{c}-\v

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$11$

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(D) The volume of a tetrahedron with vertices $A, B, C, D$ is given by $V = \frac{1}{6} |(\vec{a}-\vec{d}) \cdot ((\vec{b}-\vec{d}) 	imes (\vec{c}-\vec{d})))|$.Alternatively,$V = \frac{1}{3} 	imes 	ext{Area}(	riangle ABC) 	imes h$,where $h$ is the altitude from $D$.First,find vectors $\vec{AB} = (4-2, 1-3, -2-1) = (2, -2, -3)$ and $\vec{AC} = (6-2, 3-3, 7-1) = (4, 0, 6)$.The cross product $\vec{n} = \vec{AB} 	imes \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -2 & -3 \ 4 & 0 & 6 \end{vmatrix} = \hat{i}(-12) - \hat{j}(12+12) + \hat{k}(8) = (-12, -24, 8)$.The area of $	riangle ABC = \frac{1}{2} |\vec{n}| = \frac{1}{2} \sqrt{(-12)^2 + (-24)^2 + 8^2} = \frac{1}{2} \sqrt{144 + 576 + 64} = \frac{1}{2} \sqrt{784} = \frac{28}{2} = 14$ sq units.The volume $V = \frac{1}{6} |(\vec{AD}) \cdot (\vec{AB} 	imes \vec{AC})|$. $\vec{AD} = (-5-2, -4-3, 8-1) = (-7, -7, 7)$.$V = \frac{1}{6} |(-7, -7, 7) \cdot (-12, -24, 8)| = \frac{1}{6} |84 + 168 + 56| = \frac{308}{6} = \frac{154}{3}$.Using $V = \frac{1}{3} 	imes 14 	imes h = \frac{154}{3}$,we get $14h = 154$,so $h = 11$ units.

The length of the altitude through the point $D$ of a tetrahedron with vertices $A(2,3,1)$,$B(4,1,-2)$,$C(6,3,7)$,and $D(-5,-4,8)$ is: (in $units$)

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