Question

Find the volume of the parallelepiped with adjacent edges t = 3j – 4j – 6k, u = 3i – 7j – 7k and v = 5i + j + 3k.

Answer 1

Answer:

94 cubic units

Step-by-step explanation:

The volume of this parallelepiped (the product of area of its base and height) is equal to the absolute value of the scalar triple product $|\vec{t}\cdot (\vec{u}\times \vec{v})|$

Since

$\vec{t}=3\vec{i}-4\vec{j}-6\vec{k}=(3,-4,-6)\\ \\\vec{u}=3\vec{i}-7\vec{j}-7\vec{k}=(3,-7,-7)\\ \\\vec{v}=5\vec{i}+\vec{j}+3\vec{k}=(5,1,3)$

we have

$|\vec{t}\cdot (\vec{u}\times \vec{v})|=\left|\left|\begin{array}{ccc}3&-4&-6\\3&-7&-7\\5&1&3\end{array}\right|\right|=|3\cdot (-7)\cdot 3+(-4)\cdot (-7)\cdot 5+3\cdot 1\cdot (-6)-5\cdot (-7)\cdot (-6)-3\cdot (-7)\cdot 1-3\cdot (-4)\cdot 3|=|-63+140-18-210+21+36|=|-94|=94\ un^3.$