Geometrical Interpretation of Product of Vectors

The cross( or vector) product of two vectors results in a vector. Based on this type of product for vectors, we have various applications in geometry, mechanics, and engineering. This application of vectors can be used to find the Area of a Parallelogram and the Area of a Triangle. In real life, we use vector( or cross ) products to find torque on a wrench, magnetic force on a moving electric charge, angular momentum of a rotating object, etc.

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This Story also Contains

1. Geometrical Interpretation of Vector Product
2. 1) Area of Parallelogram
3. 2) Area of Triangle
4. Solved Examples Based on Geometrical Interpretation of Product of VectorsExample 1: Let for a triangle ABC,
5. Summary

In this article, we will cover the concept of Geometrical Interpretation of Vector products. 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 fifteen questions have been asked on this topic in JEE Main from 2013 to 2023 including one in 2020, one in 2021, and three in 2023.

Geometrical Interpretation of Scalar Product
If $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ are two non-zero vectors, then their scalar product (or dot product) is denoted by $\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}$ and is defined as
$\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}=|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}| \cos \theta . \quad(0 \leq \theta \leq \pi) \space{\text {where } \theta}$ is the angle between $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$

Let $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ be two vectors represented by $O A$ and OB, respectively.
Draw $\mathrm{BL} \perp \mathrm{OA}$ and $\mathrm{AM} \perp \mathrm{OB}$.
From triangles $O B L$ and $O A M$ we have $O L=O B \cos \theta$ and $O M=O A \cos \theta$.
Here OL and OM are known as projections of $\overrightarrow{\mathbf{b}}$ on $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{a}}$ on $\overrightarrow{\mathbf{b}}$ respectively.

Now, $\quad \begin{aligned} \vec{a} \cdot \vec{b} & =|\vec{a}||\vec{b}| \cos \theta \\ & =|\vec{a}|(O B \cos \theta) \\ & =|\vec{a}|(O L) \\ & =(\text { magnitude of } \vec{a})(\text { projection of } \vec{b} \text { on } \vec{a}) \\ \text { Again, } \quad \vec{a} \cdot \vec{b} & =|\vec{a}||\vec{b}| \cos \theta \\ & =|\vec{b}|(|\vec{a}| \cos \theta) \\ & =|\vec{b}|(O A \cos \theta) \\ & =|\vec{b}|(O M) \\ & =(\text { magnitude of } \vec{b})(\text { projection of } \vec{a} \text { on } \vec{b})\end{aligned}$

Thus. geometrically interpreted, the scalar product of two vectors is the product of the modulus of either vector and the projection of the other in its direction.

Thus,

Projection of $\overrightarrow{\mathbf{a}}$ on $\overrightarrow{\mathbf{b}}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}=\overrightarrow{\mathbf{a}} \cdot \frac{\vec{b}}{|\vec{b}|}=\overrightarrow{\mathbf{a}} \cdot \hat{\mathbf{b}}$
Projection of $\vec{b}$ on $\overrightarrow{\mathbf{a}}=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}=\vec{b} \cdot \frac{\vec{a}}{|\vec{a}|}=\vec{b} \cdot \hat{\mathbf{a}}$
What is Vector( or cross-product)?
The vector product of two nonzero vectors $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$, is denoted by $\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}$ and defined as,

$
\vec{a} \times \vec{b}=|\vec{a}||\vec{b}| \sin \theta \hat{\mathbf{n}}
$

where $\theta$ is the angle between $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}, 0 \leq \theta \leq \pi$ and $\hat{\mathbf{n}}$ is a unit vector perpendicular to both $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$, such that $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ and $\hat{\mathbf{n}}$ form a right-hand system.

Geometrical Interpretation of Vector Product

The vector product of two nonzero vectors $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$, is denoted by $\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}$ and defined as,

$
\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}=|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}| \sin \theta \hat{\mathbf{n}}
$

where $\theta$ is the angle between $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}, 0 \leq \theta \leq \pi$ and $\hat{\mathrm{n}}$ is a unit vector perpendicular to both $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$, such that $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ and $\hat{\mathrm{n}}$ form a right-hand system.

1) Area of Parallelogram

The area of a parallelogram is the region covered by a parallelogram in the plane. If $\vec{a}$ and $\overrightarrow{\mathbf{b}}$, are two non-zero, non-parallel vectors representing two adjacent sides of the parallelogram then the modulus of cross product of the vector $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$, represents the area of a parallelogram. If $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$, are two non-zero, non-parallel vectors represented by $A D$ and $A B$ respectively and let $\theta$ be the angle between them.

$\begin{aligned} & \text { In } \triangle \mathrm{ADE}, \sin \theta=\frac{D E}{A D} \\ & \Rightarrow \quad D E=A D \sin \theta=|\overrightarrow{\mathbf{a}}| \sin \theta \\ & \text { Area of parallelogram } \mathrm{ABCD}=\mathrm{AB} \cdot \mathrm{DE} \\ & \text { Thus, } \\ & \text { Area of parallelogram } \mathrm{ABCD}=|\vec{b}||\vec{a}| \sin \theta=\mid \vec{a} \times \vec{b}\end{aligned}$

2) Area of Triangle

The region covered by a triangle in a plane is called the area of a Triangle. If $\vec{a}$ and $\overrightarrow{\mathbf{b}}$, are two non-zero, non-parallel vectors represented as the adjacent sides of a triangle then the area of a triangle is half of the modulus of the vector product of the vector $\vec{a}$ and $\vec{b}$. If $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$, are two non-zero, non-parallel vectors represented as the adjacent sides of a triangle then its area is given as $\frac{1}{2}|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|$

The area of a triangle is $1 / 2$ (Base) $\times$ (Height)
From the figure,
Area of triangle $\mathrm{ABC}=\frac{1}{2} \mathrm{AB} \cdot \mathrm{CD}$
But $\mathrm{AB}=|\overrightarrow{\mathbf{b}}|$ (as given), and $\mathrm{CD}=|\overrightarrow{\mathbf{a}}| \sin \theta$
Thus, Area of triangle $\mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathbf{b}}||\overrightarrow{\mathbf{a}}| \sin \theta=\frac{1}{2}|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|$
NOTE:
1. The area of a parallelogram with diagonals $\overrightarrow{\mathbf{d}}_1$ and $\overrightarrow{\mathbf{d}}_2$ is $\frac{1}{2}\left|\overrightarrow{\mathbf{d}}_1 \times \overrightarrow{\mathbf{d}}_2\right|$.
2. The area of a plane quadrilateral $A B C D$ with AC and BD as diagonal is $\frac{1}{2}|\overrightarrow{\mathbf{A C}} \times \overrightarrow{\mathrm{BD}}|$.
3. If $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ are position vectors of a $\triangle A B C$, then its area is $\frac{1}{2}|(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}})+(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})+(\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}})|$

Recommended Video Based on Geometrical Interpretation of Product of Vectors

Solved Examples Based on Geometrical Interpretation of Product of Vectors
Example 1: Let for a triangle ABC,

$\begin{aligned} & \overrightarrow{\mathrm{AB}}=-2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{CB}}=\alpha \hat{\mathrm{i}}+\beta \hat{\mathrm{j}}+\gamma \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{CA}}=4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\delta \hat{\mathrm{k}}\end{aligned}$

If $\delta>0$ and the area of the triangle ABC is $5 \sqrt{6}$, then $\overrightarrow{\mathrm{CB}} \cdot \overrightarrow{\mathrm{CA}}$ is equal to [JEE MAINS 2023]

Solution

$\begin{aligned} & \overrightarrow{\mathrm{CA}}+\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{CB}} \\ & \langle 4,3, \delta\rangle \cdot+\langle-2,1,3\rangle=\overrightarrow{\mathrm{CB}} \\ & \Rightarrow \overrightarrow{\mathrm{CB}}=\langle 2,4,3+\delta\rangle\end{aligned}$

$\begin{aligned} & \Rightarrow|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|^2=600 \\ & \Rightarrow 5 \delta^2+30 \delta-275=0 \\ & \Rightarrow \mathrm{S}^2+6 \delta-55=0 \\ & \Rightarrow(\delta+11)(\delta-5)=0 \\ & \delta=5 \\ & \overrightarrow{\mathrm{CB}}=<2,3,8> \\ & \overrightarrow{\mathrm{CB}} \cdot \overrightarrow{\mathrm{CA}} \cdot=<2,4,8>\cdot<4,3,5> \\ & =8+12+40=60\end{aligned}$

Hence, the answer is 60

Example 2: Let $\vec{a}=4 \hat{i}+3 \hat{j}+5 \hat{k}$ and $\vec{\beta}=\hat{i}+2 \hat{j}-4 \hat{k}$, Let $\vec{\beta}_1$ be parallel to $\vec{a}$ and $\vec{\beta}_2$ be perpendicular to $\vec{a}$.If $\vec{\beta}=\vec{\beta}_1+\vec{\beta}_2$, then the value of $5 \vec{\beta}_2 \cdot(\hat{i}+\hat{j}+\hat{k})$ is [JEE MAINS 2023]

Solution

$
\begin{aligned}
& \vec{\beta}_1=\frac{(\vec{\alpha} \cdot \vec{\beta})}{|\vec{\alpha}|} \hat{\alpha} \\
& =\left(\frac{4+6-20}{\sqrt{16+9+25}}\right) \frac{(4,3,5)}{\sqrt{50}} \\
& =\frac{-10}{50}(4,3,5) \\
& \vec{\beta}_1=\frac{(-4,-3,-5)}{5} \\
& \vec{\beta}_1+\vec{\beta}_2-=(1,2,-4) \\
& \beta_2=\left(1+\frac{4}{5}, 2+\frac{3}{5},-4+1\right) \\
& \beta_2=\left(\frac{9}{5}, \frac{13}{5},-3\right) \\
& \therefore 5 \beta_2=(9,13,-15) \\
& \therefore 5 \beta_2 \cdot(1,1,1)=9+13-15 \\
& =7
\end{aligned}
$
Hence, the answer is 7

Example 3: If $\mathrm{a}^{\prime}=\hat{\imath}+2 k, b=\hat{\imath}+\hat{\jmath}+k, \vec{c}=7 \hat{\imath}-3 \hat{\jmath}+4 k, \mathrm{r}^{\prime} \times \mathrm{b}+\mathrm{b} \times \mathrm{c}^{\prime}=0$ and $\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{a}}=0$. Then $\vec{r} \cdot \vec{c}$ is equal to
Solution

$
\begin{aligned}
& \vec{r} \times \vec{b}+\vec{b} \times \vec{c}=0 \\
& \Rightarrow \vec{r} \times \vec{b}-\vec{c} \times \vec{b}=0 \\
& \Rightarrow(\vec{r}-\vec{c}) \times \vec{b}=0 \\
& \vec{r}-\vec{c} \| \vec{b} \\
& \vec{r}-\vec{c}=\lambda \vec{b} \\
& \vec{r}=\lambda \vec{b}+\vec{c} \\
& =\lambda(i+j+k)+(7 i-3 j+4 k) \\
& =\mathrm{i}(\lambda+7)+j(\lambda-3)+\mathrm{k}(\lambda+4) \\
& \vec{r} \cdot \vec{a}=0 \\
& \Rightarrow(7+\lambda)+2(\lambda+4)=0 \\
& \Rightarrow 3 \lambda=-15 \Rightarrow \lambda=-5 \\
& \therefore \vec{r}=2 \mathrm{i}-8 \mathrm{j}-\mathrm{k} \\
& \overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{c}}=(2 \mathrm{i}-8 \mathrm{j}-\mathrm{k}) \cdot(7 \mathrm{i}-3 \mathrm{j}+4 \mathrm{k}) \\
& =14+24-4=34
\end{aligned}
$
Hence, the answer is 34

Example 4: Let $\vec{a}=\hat{i}+2 \hat{j}-\hat{k}, \vec{b}=\hat{i}-2 \hat{j}$ and $\vec{c}=\hat{i}-\hat{j}-\hat{k}$ be three given vectors. If $\vec{r}$ is a vector such that $\vec{r}$ is a vector such that $\vec{r} \times \vec{a}=\vec{c} \times \vec{a}$ and $\vec{r} \cdot \vec{b}=0$, then $\vec{r} \cdot \vec{a}$ is equal to
Solution

$
\begin{aligned}
& (\vec{r}-\vec{c}) \times \vec{a}=0 \\
& \Rightarrow \vec{r}=\vec{c}+\lambda \vec{a}
\end{aligned}
$
Now, $0=\vec{b} \cdot \vec{c}+\lambda \vec{a} \cdot \vec{b}$

$
\Rightarrow \lambda=\frac{-\vec{b} \cdot \vec{c}}{\vec{a} \cdot \vec{b}}=-\frac{2}{-1}=2
$
So, $\vec{r} \cdot \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}+2 \mathrm{a}^2=12$
Hence, the answer is 12

Example 5: The area (in sq. units) of the parallelogram whose diagonals are along the vectors $8 \hat{i}-6 \hat{j}$ and $3 \hat{i}+4 \hat{j}-12 \hat{k}$,
[JEE MAINS 2017]
Solution: Area of parallelogram $=\frac{1}{2}|\vec{a} \times \vec{b}|$
where $\vec{a}$ and $\vec{b}$ are diagonals

$
\begin{aligned}
& =\frac{1}{2}\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
8 & -6 & 0 \\
3 & 4 & -12
\end{array}\right| \\
& =\left|\frac{1}{2}(72 \hat{i}+96 \hat{j}+50 \hat{k})\right| \\
& =|36 \hat{i}+48 \hat{j}+25 \hat{k}| \\
& \text { magnitude }=\sqrt{36^2+48^2+25^2} \\
& =65
\end{aligned}
$
Hence, the answer is 65

Summary

Vector (or cross product) of vector algebra is an important concept with a variety of applications. Its ability to describe rotational dynamics in physics, and analyze forces in engineering tells about its versatility and importance. Its properties and application help us to analyze and solve various real-life problems.