Let \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) be three vectors. Then the scalar \((\vec{a}\times \vec{b}).\vec{c}\) is called the scalar triple product of \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) and is denoted by [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)].
Thus, we have
[\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)] = \((\vec{a}\times \vec{b}).\vec{c}\)
For three vectors \(\vec{a}\), \(\vec{b}\) & \(\vec{c}\), it is also defined as : (\(\vec{a}\times\vec{b}\)).\(\vec{c}\) = \(|\vec{a}||\vec{b}||\vec{c}|sin\theta cos\phi\) where \(\theta\) is the angle between \(\vec{a}\) & \(\vec{b}\) and \(\phi\) is the angle between \(\vec{a}\) \(\times\) \(\vec{b}\) & \(\vec{c}\).
Note – It geometrically represents the volume of the parallelopiped whose three coterminous edges are represented by \(\vec{a}\), \(\vec{b}\) & \(\vec{c}\)
V = [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)]
Properties of Scalar Triple Product :
1). If \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are cyclically permuted the value of scalar triple product remains same.
i.e. \((\vec{a}\times \vec{b}).\vec{c}\) = \((\vec{b}\times \vec{c}).\vec{a}\) =\((\vec{c}\times \vec{a}).\vec{b}\)
or [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)] = [\(\vec{b}\) \(\vec{c}\) \(\vec{a}\)] = [\(\vec{c}\) \(\vec{a}\) \(\vec{b}\)]
2). The change of cyclic order of vectors in scalar triple product changes the sign of scalar triple product but not the magnitude.
i.e. [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)] = – [\(\vec{b}\) \(\vec{a}\) \(\vec{c}\)] = – [\(\vec{c}\) \(\vec{b}\) \(\vec{a}\)] = – [\(\vec{a}\) \(\vec{c}\) \(\vec{b}\)]
3). In scalar triple product the position of dot and cross can be interchanged provided that the cyclic order of the vectors remain same.
i.e. \((\vec{a}\times \vec{b}).\vec{c}\) = \(\vec{a}.(\vec{b}\times \vec{c}\)
4). The scalar triple product of three vectors is zero if any two of them are equal or if any two of them are parallel or collinear.
5). For any three vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) and scalar \(\lambda\), we have
[\(\lambda\) \(\vec{a}\) \(\vec{b}\) \(\vec{c}\)] = \(\lambda\) [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)]
6). For any three vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) and three scalars l, m, n
[\(l\vec{a}\) \(m\vec{b}\) \(n\vec{c}\)] = lmn [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)]
7). The necessary and sufficient condition for three non-zero, non-collinear vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) to be coplanar is that [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)] = 0
i.e. \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are coplanar \(\iff\) [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)] = 0
8). If \(\vec{a}\), \(\vec{b}\), \(\vec{c}\), \(\vec{d}\) are four vectors, then
[\(\vec{a}\) + \(\vec{b}\) \(\vec{c}\) \(\vec{d}\)] = [\(\vec{a}\) \(\vec{c}\) \(\vec{d}\)] + [\(\vec{b}\) \(\vec{c}\) \(\vec{d}\)]
9). Let \(\vec{a}\) = \(a_1\hat{i} + a_2\hat{j} + a_3\hat{k}\), \(\vec{b}\) = \(b_1\hat{i} + b_2\hat{j} + b_3\hat{k}\) and \(\vec{c}\) = \(c_1\hat{i} + c_2\hat{j} + c_3\hat{k}\) be three vectors. Then
[\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)] = \(\begin{vmatrix}
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3 \\
\end{vmatrix}\)
10). (Distributivity of vector product over vector addition) For any three vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\),
w have \(\vec{a}\times (\vec{b} + \vec{c})\) = \(\vec{a}\times \vec{b}\) + \(\vec{a}\times \vec{c}\)
11). If \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are three non-coplanar vectors and \(\vec{u}\), \(\vec{v}\), \(\vec{w}\) are three vectors such that
\(\vec{u}\) = \(x_1\hat{i} + y_1\hat{j} + z_1\hat{k}\)
\(\vec{v}\) = \(x_2\hat{i} + y_2\hat{j} + z_2\hat{k}\)
\(\vec{w}\) = \(x_3\hat{i} + y_3\hat{j} + z_3\hat{k}\)
Then, [\(\vec{u}\) \(\vec{v}\) \(\vec{w}\)] = \(\begin{vmatrix}
x_1 & y_1 & z_1 \\
x_2 & y_2 & z_2 \\
x_3 & y_3 & z_3 \\
\end{vmatrix}\) [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)]
Example : Show that \(\vec{a}\) = \(-2\hat{i} – 2\hat{j} + 4\hat{k}\), \(\vec{b}\) = \(-2\hat{i} + 4\hat{j} – 2\hat{k}\) and \(\vec{c}\) = \(4\hat{i} – 2\hat{j} -2 \hat{k}\) are coplanar.
Solution : We know that three vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are coplanar if [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)] = 0.
Here, [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)] = \(\begin{vmatrix}
-2 & -2 & 4 \\
-2 & 4 & -2 \\
4 & -2 & -2 \\
\end{vmatrix}\)
-2(-8 – 4) + 2(4 + 8) + 4(4 – 16) = 24 + 24 – 48 = 0
Hence, the given vectors are coplanar.
Hope you learnt what is scalar triple product and its properties, learn more concepts of vectors and practice more questions to get ahead in the competition. Good luck!