Find the volume of the parallelepiped with adjacent edges PQ,

Anish Buchanan 2021-08-07 Answered
Find the volume of the parallelepiped with adjacent edges PQ, PR, and PS.
P(-2, 1, 0), Q(2, 3, 2), R(1, 4, -1), S(3, 6, 1)

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Expert Answer

Maciej Morrow
Answered 2021-08-08 Author has 4398 answers

Solution:
Principles:
Knowing, any three vectors \(\displaystyle{a}={<}{a}_{{1}},{a}_{{2}},{a}_{{3}}{>},{b}={<}{b}_{{1}},{b}_{{2}},{b}_{{3}}{>}\) and \(\displaystyle{c}={<}{c}_{{1}},{c}_{{2}},{c}_{{3}}{>},\) we can find the volume of the parallelogram that the three vectors are shaping, using the triple product of these three vectors \(\displaystyle{a}\cdot{\left({b}\times{c}\right)}\) where the volume is given by the following formula
\(\displaystyle\text{Volume}={\left[{a}\cdot{\left({b}\times{c}\right)}\right]}\) (1)
Thus, knowing any three vectors side of a parallelogram we can use equation (1)
The triple product can be expressed in determiants as follows
\([a\cdot(b\times c)]=\begin{vmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{vmatrix}\)
And the value \(\displaystyle\triangle\) of the determined can be computed, using Leibniz as follows
\(\begin{vmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{vmatrix}=a_1\begin{vmatrix}b_2&b_3\\c_2&c_3\end{vmatrix}-a_2\begin{vmatrix}b_1&b_3\\c_1&c_3\end{vmatrix}+a_3\begin{vmatrix}b_1&b_2\\c_1&c_2\end{vmatrix}\)
\(\displaystyle\triangle={a}_{{1}}{\left({\left({b}_{{2}}{c}_{{3}}\right)}-{\left({b}_{{3}}{c}_{{2}}\right)}\right)}-{a}_{{2}}{\left({\left({b}_{{1}}{c}_{{3}}\right)}-{\left({b}_{{3}}{c}_{{1}}\right)}\right)}+{a}_{{3}}{\left({\left({b}_{{1}}{c}_{{2}}\right)}-{\left({b}_{{2}}{c}_{{1}}\right)}\right)}\)
Or the value of the determinant can be found using Sarrus's rule \(\displaystyle{3}\times{3}\) determinant "This work for \(\displaystyle{3}\times{3}\) determinant Only", where by duplicating the first two columuns of the matrix after the third columns, a \(\displaystyle{3}\times{5}\) determinant is formed as follows
\(\begin{vmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{vmatrix}\begin{vmatrix}a_1&a_2\\b_1&b_2\\c_1&c_2\end{vmatrix}\)
Then according to Sarrus's rule, the value of the determinant is the sum of the product all the South-East diagonals, minus the sum of the product of all the South-West diagonals, such that
\(\displaystyle\triangle={\left({\left({a}_{{1}}{b}_{{2}}{c}_{{3}}\right)}+{\left({a}_{{2}}{b}_{{3}}{c}_{{1}}\right)}+{\left({a}_{{3}}{b}_{{1}}{c}_{{2}}\right)}\right)}-{\left({\left({a}_{{2}}{b}_{{1}}{c}_{{3}}\right)}+{\left({a}_{{1}}{b}_{{3}}{c}_{{2}}\right)}+{\left({a}_{{3}}{b}_{{2}}{c}_{{1}}\right)}\right)}\)
One by finding the value of the triple product "the value of the determinant", we take the absolute value in order to find the magnitude for the triple product which would represent the volume of the parallelogram.
Since, we are given points only we can calculate the vector between any two points \(\displaystyle{P}={<}{x}_{{i}},{y}_{{i}},{z}_{{i}}{>}\) and \(\displaystyle{Q}={<}{x}_{{f}},{y}_{{f}},{z}_{{f}}{>}\), as follows
\(\displaystyle{P}{Q}={<}{x}_{{f}}-{x}_{{i}},{y}_{{f}}-{y}_{{i}}.{z}_{{f}}-{z}_{{i}}{>}\)
Calculations:
We find three vectors PQ, PR and PS, as follows
PQ=Q-P
=<2+2,3-1,2-0>
=<4,2,2>
PS=S-P
=<3+2,6-1,1-0>
=<5,5,1>
PR=R-P
=<1+2,4-1,-1-0>
=<3,3,-1>
\(PS\cdot(PQ\times PR)=\begin{vmatrix}5&5&1\\4&2&2\\3&3&-1\end{vmatrix}\)
\(\triangle=\begin{vmatrix}5&5&1\\4&2&2\\3&3&-1\end{vmatrix}\begin{vmatrix}5&5\\4&2\\3&3\end{vmatrix}\)
\(\displaystyle={\left({\left({\left({5}\times{2}\times-{1}\right)}+{\left({5}\times{2}\times{3}\right)}+{\left({1}\times{4}\times{3}\right)}\right)}-{\left({\left({5}\times{4}\times-{1}\right)}+{\left({5}\times{2}\times{3}\right)}+{\left({1}\times{2}\times{3}\right)}\right)}\right)}\)
\(\displaystyle={\left({\left(-{10}+{30}+{12}\right)}-{\left(-{20}+{30}+{6}\right)}\right)}\)
\(\displaystyle={\left({\left({32}\right)}-{\left({16}\right)}\right)}\)
\(\displaystyle={16}\)

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