Vector Triple Product

The Vector Triple Product means the product of three vectors which result in a Vector. It means taking the cross product of the vectors with the cross product of the other two vectors. In real life, we use Vector Triple Product to solve complex problems related to aircraft design and the volume of parallelepiped.

In this article, we will cover the concept of Vector Triple Product. This topic falls under the broader category of Vector Algebra, which is a crucial chapter in Class 11 Mathematics. This is very important not only for board exams but also for competitive exams, which even include the Joint Entrance Examination Main and other entrance exams: SRM Joint Engineering Entrance, BITSAT, WBJEE, and BCECE. A total of sixteen questions have been asked on this topic in JEE Main from 2013 to 2023 including one in 2019, one in 2020, five in 2021, three in 2022, and four in 2023.

Vector Triple Product: Definition

The Vector Triple Product is defined as the cross product of one of the vectors with the cross product of the other two vectors. It results in the vector. It is expressed as $\overrightarrow{\mathbf{a}} \times(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})$
The resultant of the triple cross vector lies in the plane of the given three vectors.
Formula of Vector Triple Product
For three vectors $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathbf{c}}$ vector triple product is defined as $\overrightarrow{\mathbf{a}} \times(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})$.

$
\vec{a} \times(\vec{b} \times \vec{c})=(\vec{a} \cdot \vec{c}) \cdot \vec{b}-(\vec{a} \cdot \vec{b}) \cdot \vec{c}
$

Derivation of Vector Triple Product

$\vec{p}=\vec{a} \times(\vec{b} \times \vec{c})$ is a vector perpendicular to $\vec{a}$ and $\vec{b} \times \vec{c}$ but $\vec{b} \times \vec{c}$ is a vector perpendicular to the plane of $\vec{b}$ and $\vec{c}$.
Hence, vector $\vec{p}$ must lie in the plane of $\vec{b}$ and $\vec{c}$.
Let $\vec{p}=\vec{a} \times(\vec{b} \times \vec{c})=l \vec{b}+m \vec{c}$
[l,m are scalars]
Taking the dot product of eq (i) with $\vec{a}$, we get

$
\begin{gathered}
\vec{p} \cdot \vec{a}=l(\vec{a} \cdot \vec{b})+m(\vec{a} \cdot \vec{c}) \\
{\left[\begin{array}{l}
\because \vec{a} \times(\vec{b} \times \vec{c}) \text { is } \perp \vec{a} \\
\therefore \vec{a} \times(\vec{b} \times \vec{c}) \cdot \vec{a}=0
\end{array}\right]}
\end{gathered}
$
Therefore,

$
\begin{array}{ll}
\Rightarrow & \vec{p} \cdot \vec{a}=0 \\
\Rightarrow & l(\vec{a} \cdot \vec{b})=-m(\vec{a} \cdot \vec{c}) \\
\Rightarrow & \frac{1}{\vec{a} \cdot \vec{c}}=\frac{-m}{\vec{a} \cdot \vec{b}}=\lambda \\
\Rightarrow & l=\lambda(\vec{a} \cdot \vec{c}) \\
\text { and } & m=-\lambda(\vec{a} \cdot \vec{b})
\end{array}
$

Substituting the value of $l$ and $m$ in Eq . (i), we get

$
\vec{a} \times(\vec{b} \times \vec{c})=\lambda[(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}]
$
Here, the value of $\lambda$ can be determined by taking specific values of $\vec{a}, \vec{b}$ and $\vec{c}$.

The simplest way to determine $\lambda$ is by taking specific vectors $\vec{a}=\hat{i}, \vec{b}=\hat{i}, \vec{c}=\hat{j}$.

We have,
$
\begin{array}{ll}
& \vec{a} \times(\vec{b} \times \vec{c})=\lambda[(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}] \\
& \hat{i} \times(\hat{i} \times \hat{j})=\lambda[(\hat{i} \cdot \hat{j}) \hat{i}-(\hat{i} \cdot \hat{i}) \hat{j}] \\
& \hat{i} \times \hat{k}=\lambda[(0) \hat{i}-(1) \hat{j}] \Rightarrow-\hat{j}=-\lambda \hat{j} \\
\therefore \quad & \lambda=1
\end{array}
$
Hence,

$
\vec{a} \times(\vec{b} \times \vec{c})=(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}
$

1.

$
\begin{aligned}
& \vec{a} \times(\vec{b} \times \vec{c})=(\vec{a} \cdot \vec{c}) \cdot \vec{b}-(\vec{a} \cdot \vec{b}) \cdot \vec{c} \\
& (\vec{a} \times \vec{b}) \times \vec{c}=(\vec{c} \cdot \vec{a}) \cdot \vec{b}-(\vec{c} \cdot \vec{b}) \cdot \vec{a}
\end{aligned}
$

2. In general $\vec{a} \times(\vec{b} \times \vec{c}) \neq(\vec{a} \times \vec{b}) \times \vec{c}$
If $\overrightarrow{\mathbf{a}} \times(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})=(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \times \overrightarrow{\mathbf{c}}$ then the vectors $\vec{a}$ and $\vec{c}$

NOTE: are collinear.

Properties of the Vector Triple Product
The vector triple product is a vector quantity.

$
\begin{aligned}
& \vec{a} \times(\vec{b} \times \vec{c}) \neq(\vec{a} \times \vec{b}) \vec{c} \\
& \vec{a} \times(\vec{b} \times \vec{c}) \text { is a vector perpendicular to } \vec{a} \text { and }(\vec{b} \times \vec{c})
\end{aligned}
$

$\vec{a} \times(\vec{b} \times \vec{c})$ is a vector perpendicular to the plane containing $\vec{a}, \vec{b}, \vec{c}$ are three vectors.

Solved Examples Based on Vector Triple Product

Example 1: If $\vec{a}, \vec{b}, \vec{c}$ are three non-zero vectors and $\hat{n}$ is a unit vector perpendicular to $\hat{c}$ such that $\vec{a}=\alpha \vec{b}-\hat{n}(a \neq 0)$ and $\vec{b} \cdot \vec{c}=12$, then $|\vec{c} \times(\vec{a} \times \vec{b})|_{\text {is equal to: }}$
[JEE MAINS 2023]
Solution

$
\begin{aligned}
& \overrightarrow{\mathrm{a}}=\alpha \overrightarrow{\mathrm{b}}-\hat{\mathrm{n}}, \vec{b} \cdot \overrightarrow{\mathrm{c}}=12 \\
& \overrightarrow{\mathrm{c}} \times(\vec{a} \times \vec{b})=(\vec{c} \cdot \vec{b}) \vec{a}-(\vec{c} \cdot \vec{a}) \vec{b} \\
& \overrightarrow{\mathrm{c}} \times(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})=12 \vec{a}-(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{b}}) \\
& \because \overrightarrow{\mathrm{a}}=\alpha \overrightarrow{\mathrm{b}}-\mathrm{n}
\end{aligned}
$
$
\vec{c} \cdot \vec{a}=\alpha \overrightarrow{\mathrm{c}} \cdot \vec{b}-\vec{c} \cdot \mathrm{n}
$
$
\vec{c} \cdot \vec{a}=12 \alpha
$
$
\begin{aligned}
& \vec{c} \times(\vec{a} \times \vec{b})=12 \vec{a}-12 \alpha \vec{b} \\
& |\vec{c} \times(\vec{a} \times \vec{b})|=12|\vec{a}-\alpha \vec{b}| \quad[\because \vec{a}-\alpha \vec{b}=-n \text { then }|\vec{a}-\alpha \vec{b}|=1] \\
& \Rightarrow|\vec{c} \times(\vec{a} \times \vec{b})|=12
\end{aligned}
$
$
|\overrightarrow{\mathrm{c}} \times(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})|=12
$
Hence, the answer is 12

Example 2: Let $\lambda \in \mathbb{R}, \vec{a}=\lambda \hat{\imath}+2 \hat{\jmath}-3 \hat{k}, \vec{b}=\hat{\imath}-\lambda \hat{\jmath}+2 \hat{k}{ }_{\operatorname{If}}((\vec{a}+\vec{b}) \times(\vec{a} \times \vec{b})) \times(\vec{a}-\vec{b})=8 \hat{\imath}-40 \hat{\jmath}-24 \hat{k}$, then $|\lambda(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})|^2$ is equal to
Solution

$
\begin{aligned}
& ((\vec{a}+\vec{b}) \times(\vec{a} \times \vec{b}) \times(\vec{a}-\vec{b})-=8 \hat{i}-40 \hat{j}-24 \hat{k} \\
& \Rightarrow(\vec{a} \times(\vec{a} \times \vec{b})+\vec{b} \times(\vec{a} \times \vec{b})) \times(\vec{a}-\vec{b}) \\
& \Rightarrow((\vec{a} \cdot \vec{b}) \vec{a}-(\vec{a} \cdot \vec{a}) \vec{b}+(\vec{b} \cdot \vec{b}) \vec{a}-(\vec{b} \cdot \vec{a}) \vec{b}) \times(\vec{a}-\vec{b}) \\
& \Rightarrow 0-(\vec{a} \cdot \vec{b})(\vec{a} \times \vec{b})-a^2(\vec{b} \times \vec{a})+0-\mathrm{b}^2(\vec{a} \times \vec{b})-(\vec{a} \cdot \vec{b}) \vec{b} \times \vec{a}=8 \hat{\mathrm{i}}-40 \hat{\mathrm{j}}-24 \hat{\mathrm{k}} \\
& \Rightarrow\left(\mathrm{a}^2-\mathrm{b}^2\right)(\vec{a} \times \overrightarrow{\mathrm{b}})=8 \hat{\mathrm{i}}-40 \hat{\mathrm{j}}-24 \hat{\mathrm{k}} \\
& \left(\left(\lambda^2+4+9\right)-\left(1+\lambda^2+4\right)\right)(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \\
& 8(\vec{a} \times \vec{b})=8(\hat{\mathrm{i}}-5 \hat{j}-3 \hat{k}) \\
& \hat{\mathrm{i}}(4-3 \lambda)-\hat{\mathrm{j}}(2 \lambda+3)+\hat{\mathrm{k}}\left(-\lambda^2-2\right)=\hat{\mathrm{i}}-5 \hat{\mathrm{j}}-3 \hat{\mathrm{k}} \\
& \Rightarrow 4-3 \lambda=1 \quad 2 \lambda+3=5 \quad-\lambda^2-2=-3 \\
& 3 \lambda=3 \\
& \lambda=1 \\
& \lambda \\
& \begin{array}{l}
\lambda^2=1
\end{array} \\
& \begin{array}{ll}
\lambda(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})|=|(\vec{a}+\vec{b}) \times\left.(\vec{a}-\vec{b})\right|^2 \\
\Rightarrow|-\vec{a} \times \vec{b}+\vec{b} \times \vec{a}|^2=|2(\vec{a} \times \vec{b})|^2=4(1+25+9)=140
\end{array}
\end{aligned}
$
Hence, the answer is 140

Example 3: Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-zero vectors such that $\vec{b} \cdot \vec{c}=0$ and $\vec{a} \times(\vec{b} \times \vec{c})=\frac{b-\vec{c}}{2}$. If $\vec{d}$ be a vector such that $\vec{b} \cdot \vec{d}=\vec{a} \cdot \vec{b}$, then $(\vec{a} \times \vec{b}) \cdot(\vec{c} \times \vec{d})_{\text {is equal to }}$.
[JEE MAINS 2023]

$
\begin{aligned}
& \text { Solution: }(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{b}}-(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}) \overline{\mathrm{c}}=\frac{\overline{\mathrm{b}}}{2}-\frac{\overline{\mathrm{c}}}{2} \\
& \overline{\mathrm{a}} \cdot \overline{\mathrm{c}}=\frac{1}{2}, \overline{\mathrm{a}} \cdot \overline{\mathrm{b}}=\frac{1}{2} \\
& \overline{\mathrm{b}} \cdot \overline{\mathrm{d}}=\frac{1}{2} \\
& (\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \cdot(\overline{\mathrm{c}} \times \overline{\mathrm{d}})=\overline{\mathrm{a}} \cdot[\overline{\mathrm{b}} \times(\overline{\mathrm{c}} \times \overline{\mathrm{d}})] \\
& =\overline{\mathrm{a}} \cdot[(\overline{\mathrm{b}} \cdot \overline{\mathrm{d}}) \overline{\mathrm{c}}-(\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{d}}] \\
& =\overline{\mathrm{a}} \cdot[\overline{\mathrm{c}} / 2] \\
& =\frac{1}{2}(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \\
& =\frac{1}{4}
\end{aligned}
$
Hence, the answer is $1 / 4$

Example 4: Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors mutually perpendicular to each other and have the same magnitude. If a vector $\vec{r}$ satisfies $\vec{a} \times\{(\vec{r}-b) \times \vec{a}\}+b \times\{(\vec{r}-\vec{c}) \times b\}+\vec{c} \times\{(\vec{r}-\vec{a}) \times \vec{c}\}=0$,then $\vec{r}$ is equal to:
Solution: $|\vec{a}|=|\vec{b}|=|\vec{c}|$ and $\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{a}=0$
Let $\vec{r}=x \vec{a}+y \vec{b}+z \vec{c}$
where $\vec{r} \cdot \vec{a}=x|\vec{a}|^2, \vec{r} \cdot \vec{b}=y|\vec{b}|^2, \vec{r} \cdot \vec{c}=z|\vec{c}|^2$
Give expression is

$
\begin{aligned}
& (\vec{a} \times(\vec{r} \times \vec{a}))-(\vec{a} \times(\vec{b} \times \vec{a}))+\vec{b} \times(\vec{r} \times \vec{b})-\vec{b} \times(\vec{c} \times \vec{b})+ \\
& \vec{c} \times(\vec{r} \times \vec{c})-(\vec{c} \times(\vec{a} \times c))=0 \\
& \Rightarrow(\vec{a} \cdot \vec{r}) \vec{a}-|\vec{a}|^2 \vec{r}-(\vec{a} \cdot \vec{b}) \vec{a}+|\vec{a}|^2 \vec{b}+(\vec{b} \cdot \vec{r}) \vec{b}-|\vec{b}|^2 \vec{r}- \\
& (\vec{b} \cdot \vec{c}) \vec{b}+|\vec{b}|^2 \vec{c}+(\vec{c} \cdot \vec{r}) \vec{c}-|\vec{c}|^2 \vec{r}-(\vec{c} \cdot \vec{a}) \vec{a}+|\vec{c}|^2 \vec{a}=0 \\
& \Rightarrow x|\vec{a}|^2 \vec{a}+y|\vec{b}|^2 \vec{b}+z|\vec{c}|^2 \vec{c}-\vec{r}\left(|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2\right)+ \\
& |\vec{a}|^2 \vec{b}+|\vec{b}|^2 \vec{c}+|\vec{c}|^2 \vec{a}=0 \\
& \Rightarrow|\vec{a}|^2(x \vec{a}+y \vec{b}+z \vec{c})-3|\vec{a}|^2 \vec{r}+|\vec{a}|^2(\vec{a}+\vec{b}+\vec{c})=0 \\
& \Rightarrow 3 \vec{r}-\vec{r}=\vec{a}+\vec{b}+\vec{c} \\
& \Rightarrow \vec{r}=\frac{1}{2}(\vec{a}+\vec{b}+\vec{c})
\end{aligned}
$
Hence, the answer is $\frac{1}{2}(\vec{a}+\vec{b}+\vec{c})$

Example 5: Let three vector $\vec{a}, \vec{b}$ and $\vec{c}$ be such that $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}, \vec{a} \cdot \vec{b}=7$ and $\vec{b}$ is perpendicular to $\vec{c}$, where $\vec{a}=-\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=2 \hat{i}+\hat{k}$. Then the value of $2|\vec{a}+\vec{b}+\vec{c}|_{\text {is }}^2$ $\qquad$
[JEE MAINS 2021]
Solution

$
\begin{aligned}
\vec{c} & =\lambda(\vec{b} \times(\vec{a} \times \vec{b})) \\
& =\lambda((\vec{b} \cdot \vec{b}) \vec{b}-(\vec{b} \cdot \vec{a}) \vec{b}) \\
& =\lambda(5(-\hat{i}+\hat{j}+\hat{k})+2 \hat{i}+\hat{k}) \\
& =\lambda(-3 \hat{i}+5 \hat{j}+6 \hat{k}) \\
\vec{c} & \cdot \vec{a}=7 \Rightarrow 3 \lambda+5 \lambda+6 \lambda=7 \\
\Rightarrow & \lambda=\frac{1}{2} \\
\therefore & 2\left|\left(\frac{-3}{2}-1+2\right) \hat{i}+\left(\frac{5}{2}+1\right) \hat{j}+(3+1+1) \hat{k}\right|^2 \\
& =2\left(\frac{1}{4}+\frac{49}{4}+25\right)=25+50=75
\end{aligned}
$
Hence, the answer is 75

Summary

The vector triple product helps us understand how three vectors interact in three-dimensional space. It's used in various fields like mechanics, electromagnetism, and geometry to calculate important quantities like torque, magnetic fields, and volumes. Understanding vector triple product helps us to analyze and solve the problems related to real life.