UPES B.Tech Admissions 2025
Ranked #42 among Engineering colleges in India by NIRF | Highest Package 1.3 CR , 100% Placements | Last Date to Apply: 28th April
Scalar Triple Product is considered one of the most asked concept.
58 Questions around this concept.
If then
is equal to:
$\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]$ is not equal to
$\begin{bmatrix} \hat{i}-\hat{j} & 2\hat{i}+3\hat{j} & -3\hat{i}-\hat{j} \end{bmatrix}=$
JEE Main 2025: Session 2 Result Out; Direct Link
JEE Main 2025: College Predictor | Marks vs Percentile vs Rank
New: JEE Seat Matrix- IITs, NITs, IIITs and GFTI | NITs Cutoff
Latest: Meet B.Tech expert in your city to shortlist colleges
$\begin{bmatrix} \vec{a} & \vec{b} & \vec{c} \end{bmatrix}=\: \: ?\: \: \: \: ;\: \: if \: \: \vec{a}=\hat{i}-\hat{j},\: \vec{b}=2\hat{i}-2\hat{j}+\hat{k},\: \: \vec{c}=2\hat{i}+\hat{j}-3\hat{k}$
$\begin{bmatrix} \vec{a}-\vec{b} & \vec{b}-\vec{c} & \vec{c}-\vec{a} \end{bmatrix}=$
$\begin{bmatrix} \vec{a}+\vec{b} & \hat{j} & \hat{i}-\hat{k} \end{bmatrix}=$
$\begin{bmatrix} \vec{a}+\vec{b} &\vec{b}+\vec{c} &\vec{c}+\vec{a} \end{bmatrix}=$
Ranked #42 among Engineering colleges in India by NIRF | Highest Package 1.3 CR , 100% Placements | Last Date to Apply: 28th April
Merit Scholarships | NAAC A+ Accredited | Top Recruiters : E&Y, CYENT, Nvidia, CISCO, Genpact, Amazon & many more
$\begin{bmatrix} 2\hat{i} & 3\hat{j} & -5\hat{k} \end{bmatrix}=$
The value of $\mathrm{i} \cdot(\mathrm{j} \times \mathrm{k})+\mathrm{j} \cdot(\mathrm{i} \times \mathrm{k})+\mathrm{k} \cdot(\mathrm{i} \times \mathrm{j})$ is
Find volume of tetrahedron formed by vectors $\hat{i}-\hat{j}+2 \hat{k}, 2 \hat{i}+\hat{j}$ and $\hat{j}-2 \hat{k}$ is :
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 $=\frac{1}{6}\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]$
"Stay in the loop. Receive exam news, study resources, and expert advice!"