Question

1. find the volume of the parallelepiped with edges $\vec{u}$, $\vec{v}$, and $\vec{w}$:

(a) $\vec{u} = \langle 0, 1, -2 \
angle$, $\vec{v} = \langle -3, -4, 5 \
angle$, $\vec{w} = \langle -6, 7, 8 \
angle$.

(b) $\vec{u} = \langle 0, -3, -6 \
angle$, $\vec{v} = \langle 1, -4, 7 \
angle$, $\vec{w} = \langle -2, 5, 8 \
angle$.

(c) $\vec{u} = \langle 2, -1, -3 \
angle$, $\vec{v} = \langle 4, -2, 1 \
angle$, $\vec{w} = \langle 3, -4, 5 \
angle$.

(d) $\vec{u} = \langle 3, 0, 0 \
angle$, $\vec{v} = \langle 0, 4, 0 \
angle$, $\vec{w} = \langle 0, 0, 5 \
angle$.

(e) $\vec{u} = \langle 3, 135, 246 \
angle$, $\vec{v} = \langle 0, 4, 159 \
angle$, $\vec{w} = \langle 0, 0, 5 \
angle$.

Explanation:

Part (a)

Step1: Recall the formula for the volume of a parallelepiped formed by vectors \(\vec{u}\), \(\vec{v}\), \(\vec{w}\) is the absolute value of the scalar triple product \(|\vec{u} \cdot (\vec{v} \times \vec{w})|\). First, find \(\vec{v} \times \vec{w}\).

Given \(\vec{v}=\langle - 3,-4,5
angle\) and \(\vec{w}=\langle - 6,7,8
angle\), the cross product \(\vec{v}\times\vec{w}=

$$\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\-3&-4&5\\-6&7&8\end{vmatrix}$$

\)
\[

$$\begin{align*} \vec{v}\times\vec{w}&=\vec{i}((-4)\times8 - 5\times7)-\vec{j}((-3)\times8-5\times(-6))+\vec{k}((-3)\times7-(-4)\times(-6))\\ &=\vec{i}(-32 - 35)-\vec{j}(-24 + 30)+\vec{k}(-21-24)\\ &=\langle - 67,-6,-45 angle \end{align*}$$

\]

Step2: Now find the dot product of \(\vec{u}=\langle0,1, - 2

angle\) and \(\vec{v}\times\vec{w}=\langle - 67,-6,-45
angle\)
\(\vec{u}\cdot(\vec{v}\times\vec{w})=0\times(-67)+1\times(-6)+(-2)\times(-45)=- 6 + 90 = 84\)

Step3: Take the absolute value to get the volume.

\(|\vec{u}\cdot(\vec{v}\times\vec{w})| = |84|=84\)

Step1: Find \(\vec{v}\times\vec{w}\) where \(\vec{v}=\langle1,-4,7

angle\) and \(\vec{w}=\langle - 2,5,8
angle\)
\(\vec{v}\times\vec{w}=

$$\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\1&-4&7\\-2&5&8\end{vmatrix}$$

\)
\[

$$\begin{align*} \vec{v}\times\vec{w}&=\vec{i}((-4)\times8-7\times5)-\vec{j}(1\times8 - 7\times(-2))+\vec{k}(1\times5-(-4)\times(-2))\\ &=\vec{i}(-32-35)-\vec{j}(8 + 14)+\vec{k}(5 - 8)\\ &=\langle - 67,-22,-3 angle \end{align*}$$

\]

Step2: Find the dot product of \(\vec{u}=\langle0,-3,-6

angle\) and \(\vec{v}\times\vec{w}=\langle - 67,-22,-3
angle\)
\(\vec{u}\cdot(\vec{v}\times\vec{w})=0\times(-67)+(-3)\times(-22)+(-6)\times(-3)=66 + 18=84\)

Step3: Take the absolute value.

\(|\vec{u}\cdot(\vec{v}\times\vec{w})|=|84| = 84\)

Step1: Find \(\vec{v}\times\vec{w}\) where \(\vec{v}=\langle4,-2,1

angle\) and \(\vec{w}=\langle3,-4,5
angle\)
\(\vec{v}\times\vec{w}=

$$\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\4&-2&1\\3&-4&5\end{vmatrix}$$

\)
\[

$$\begin{align*} \vec{v}\times\vec{w}&=\vec{i}((-2)\times5-1\times(-4))-\vec{j}(4\times5 - 1\times3)+\vec{k}(4\times(-4)-(-2)\times3)\\ &=\vec{i}(-10 + 4)-\vec{j}(20 - 3)+\vec{k}(-16 + 6)\\ &=\langle - 6,-17,-10 angle \end{align*}$$

\]

Step2: Find the dot product of \(\vec{u}=\langle2,-1,-3

angle\) and \(\vec{v}\times\vec{w}=\langle - 6,-17,-10
angle\)
\(\vec{u}\cdot(\vec{v}\times\vec{w})=2\times(-6)+(-1)\times(-17)+(-3)\times(-10)=-12 + 17+30 = 35\)

Step3: Take the absolute value.

\(|\vec{u}\cdot(\vec{v}\times\vec{w})|=|35| = 35\)

Answer:

84

Part (b)