Question

If A (1,2,-1), B (3,5,0), C (5,11,22) and D (5,8,6) are the vertices. Find the volume of the parallelepiped with segment AB, AC and AD as concurrent edges.

Answer 1

To solve this question, we will use the concept of position vectors and volume of parallelepiped with concurrent edges. The volume of parallelepiped with concurrent edges $\vec a,\vec b$ and $\vec c$ is given as scalar triple product of $\vec a,\vec b$ and $\vec c$ and is denoted by \left[ {\vec a{\text{ }}\vec b{\text{ }}\vec c} \right] = \left( {\vec a \times \vec b} \right).\vec c = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}&{{a_3}} \\\ {{b_1}}&{{b_2}}&{{b_3}} \\\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right| .

Complete step by step answer:
Given that,
A (1,2,-1), B (3,5,0), C (5,11,22) and D (5,8,6) are the vertices.
Now, we have to find the segments AB, AC and AD.
We know that,
$\overrightarrow {AB}$= Position vector of B – Position vector of A.
So,
Position vector of A = $\hat i + 2\hat j - \hat k$
Position vector of B = $3\hat i + 5\hat j + 0\hat k$
Position vector of C = $5\hat i + 11\hat j + 22\hat k$
Position vector of D = $5\hat i + 8\hat j + 6\hat k$
Therefore,
$\overrightarrow {AB} = \left( {3\hat i + 5\hat j + 0\hat k} \right) - \left( {\hat i + 2\hat j - \hat k} \right) \\\ \overrightarrow {AB} = 2\hat i + 3\hat j + \hat k \\\$
Similarly,
$\overrightarrow {AC} = \left( {5\hat i + 11\hat j + 22\hat k} \right) - \left( {\hat i + 2\hat j - \hat k} \right) \\\ \overrightarrow {AC} = 4\hat i + 9\hat j + 23\hat k \\\$
And,
$\overrightarrow {AD} = \left( {5\hat i + 8\hat j + 6\hat k} \right) - \left( {\hat i + 2\hat j - \hat k} \right) \\\ \overrightarrow {AD} = 4\hat i + 6\hat j + 7\hat k \\\$
Now we have got all the segments AB, AC and AD.
We know that,
The volume of parallelepiped with concurrent edges = scalar triple product of those edges.
We also know that, the scalar triple product in terms of components is,
If $\overrightarrow a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k,$ $\overrightarrow b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k$ and $\overrightarrow c = {c_1}\hat i + {c_2}\hat j + {c_3}\hat k$ be three vectors. Then,
\left[ {\overrightarrow a \overrightarrow {{\text{ }}b} {\text{ }}\overrightarrow c } \right] = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}&{{a_3}} \\\ {{b_1}}&{{b_2}}&{{b_3}} \\\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right|.
Therefore,

\left[ {\overrightarrow {AB} \overrightarrow {{\text{ AC}}} {\text{ }}\overrightarrow {AD} } \right] = \left| {\begin{array}{*{20}{c}} 2&3&1 \\\ 4&9&{23} \\\ 4&6&7 \end{array}} \right| \\\ \left[ {\overrightarrow {AB} \overrightarrow {{\text{ AC}}} {\text{ }}\overrightarrow {AD} } \right] = 2\left( {63 - 138} \right) - 3\left( {28 - 92} \right) + 1\left( {24 - 36} \right) \\\ \left[ {\overrightarrow {AB} \overrightarrow {{\text{ AC}}} {\text{ }}\overrightarrow {AD} } \right] = 2\left( { - 75} \right) - 3\left( { - 64} \right) + 1\left( { - 12} \right) \\\ \left[ {\overrightarrow {AB} \overrightarrow {{\text{ AC}}} {\text{ }}\overrightarrow {AD} } \right] = - 150 + 192 - 12 \\\ \left[ {\overrightarrow {AB} \overrightarrow {{\text{ AC}}} {\text{ }}\overrightarrow {AD} } \right] = 30 \\\

Hence, Volume of parallelepiped = $\left[ {\overrightarrow {AB} \overrightarrow {{\text{ AC}}} {\text{ }}\overrightarrow {AD} } \right]$ = 30.

Note: Whenever we ask such types of questions, we should remember some basic points of vectors like positioning of vectors, scalar triple product etc. First, we have to find out the position vectors of the segments by using the given points. Then we will use those vectors in the formula of volume of parallelepiped, i.e. scalar triple product and after that by finding the scalar triple product of those segments, we will get the required answer.