Scalar Triple Product of Vectors - Practice Questions & MCQ

The scalar triple product (also called the mixed or box product) is defined as the dot product of one of the vectors with the cross product of the other two.
If $\vec{a}, \vec{b}$ and $\vec{c}$ are any three vectors, then their scalar product is defined as
$
\overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}) \text { and it is denoted as }\left[\begin{array}{lll}
\overrightarrow{\mathbf{a}} & \overrightarrow{\mathbf{b}} & \overrightarrow{\mathbf{c}}
\end{array}\right]
$

The scalar triple product can be evaluated numerically using any one of the following
$
\begin{array}{ll} 
& (\vec{a} \times \vec{b}) \cdot \vec{c}=\vec{a} \cdot(\vec{b} \times \vec{c})=\vec{b} \cdot(\vec{c} \times \overrightarrow{\mathbf{a}})=\overrightarrow{\mathbf{c}} \cdot(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \\
\text { i.e. } & {\left[\begin{array}{lll}
\overrightarrow{\mathbf{a}} & \vec{b} & \overrightarrow{\mathbf{c}}
\end{array}\right]=\left[\begin{array}{lll}
\vec{b} & \overrightarrow{\mathbf{c}} & \overrightarrow{\mathbf{a}}
\end{array}\right]=\left[\begin{array}{lll}
\overrightarrow{\mathbf{c}} & \overrightarrow{\mathbf{a}} & \vec{b}
\end{array}\right]=-\left[\begin{array}{lll}
\vec{b} & \overrightarrow{\mathbf{a}} & \overrightarrow{\mathbf{c}}
\end{array}\right]=-\left[\begin{array}{lll}
\overrightarrow{\mathbf{c}} & \vec{b} & \overrightarrow{\mathbf{a}}
\end{array}\right]}
\end{array}
$

If $\overrightarrow{\mathbf{a}}=a_1 \hat{\mathbf{i}}+a_2 \hat{\mathbf{j}}+a_3 \hat{\mathbf{k}}, \quad \overrightarrow{\mathbf{b}}=b_1 \hat{\mathbf{i}}+b_2 \hat{\mathbf{j}}+b_3 \hat{\mathbf{k}} \quad$ and $\quad \overrightarrow{\mathbf{c}}=c_1 \hat{\mathbf{i}}+c_2 \hat{\mathbf{j}}+c_3 \hat{\mathbf{k}}$ then
$
\begin{aligned}
{\left[\begin{array}{lll}
\overrightarrow{\mathbf{a}} & \overrightarrow{\mathbf{b}} & \overrightarrow{\mathbf{c}}
\end{array}\right] } & =(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \cdot \overrightarrow{\mathbf{c}}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
a_1 & a_2 & a_2 \\
b_1 & b_2 & b_3
\end{array}\right| \cdot\left(c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}\right) \\
& =\left\lvert\, \begin{array}{ccc}
\hat{i} \cdot\left(c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}\right) & \hat{j} \cdot\left(\begin{array}{c}
\left.c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}\right) \\
a_1 \\
b_1
\end{array}\right. & \hat{k} \cdot\left(c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}\right) \\
a_2 & a_2 \\
b_2
\end{array}\right.
\end{aligned}
$

$
=\left|\begin{array}{lll}
c_1 & c_2 & c_3 \\
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3
\end{array}\right|=\left|\begin{array}{lll}
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3
\end{array}\right|
$

Also,
$
\begin{aligned}
{[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}] } & =\overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})=\left(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\right) \cdot\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
b_1 & b_2 & b_2 \\
c_1 & c_2 & c_3
\end{array}\right| \\
& =\left|\begin{array}{lll}
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3
\end{array}\right|
\end{aligned}
$

NOTE :
1. $\left[\begin{array}{lll}m \vec{a} & \vec{b} & \vec{c}\end{array}\right]=m\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]$, where $m$ is a scalar..
2. $\left[m_1 \vec{a} \quad m_2 \vec{b} \quad m_3 \vec{c}\right]=m_1 m_2 m_3\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]$, where $m_1, m_2, m_3$ are scalares.
3. $\left[\begin{array}{llll}\vec{a}+\vec{b} & \vec{c} & \vec{d}\end{array}\right]=\left[\begin{array}{lll}\vec{a} & \vec{c} & \vec{d}\end{array}\right]+\left[\begin{array}{lll}\vec{b} & \vec{c} & \vec{d}\end{array}\right]$

The necessary and sufficient condition for three non-zero, non-collinear vectors $\vec{a}, \vec{b}$ and $\vec{c}$ is coplanar is that $\left[\begin{array}{ll}\vec{a} & \vec{b} \\ \vec{c}\end{array}\right]=0$.

Let vectors $\overrightarrow{\mathbf{a}}, \quad \overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ represent the sides of a parallelepiped OA, OB and OC respectively. Then, $\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}$ is a vector perpendicular to the plane of $\overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$. Let $\theta$ be the angle between vectors $\overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ and $\boldsymbol{\alpha}$ be the angle between $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}$.

If $\hat{\mathbf{n}}$ is a unit vector along $\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}$, then $\alpha$ is the angle between $\hat{\mathbf{n}}$ and $\overrightarrow{\mathbf{a}}$.

$\begin{aligned} {\left[\begin{array}{ll}\overrightarrow{\mathbf{a}} \mathbf{b} & \overrightarrow{\mathbf{c}}]\end{array}\right.} & =\overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}) \\ & =\overrightarrow{\mathbf{a}} \cdot(\mathbf{b} \mathbf{c} \sin \theta \hat{\mathbf{n}}) \\ & =(\mathbf{b} \mathbf{c} \sin \theta)(\overrightarrow{\mathbf{a}} \cdot \hat{\mathbf{n}}) \\ & =(\mathbf{b c} \sin \theta)(\mathbf{a} \cdot \mathbf{1} \cdot \cos \alpha) \\ & =(\mathbf{a} \cdot \cos \alpha)(\mathbf{b} \mathbf{c} \sin \theta) \\ & =(\text { Height }) \cdot(\text { Area of Base }) \\ & =\text { Volume of parallelepiped }\end{aligned}$

Volume of Tetrahedron

Tetrahedron is a pyramid having a triangular base. Therefore

$\therefore \quad$ Volume $=\frac{1}{6}\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]$