Geometrical Interpretation of Product of Vectors

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