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Q7.4:
Show that the area of the triangle contained between the vectors $a$ and $b$ is one half of the magnitude of $a \times b$.

Solution :

Step 1: Geometric Representation of Vector Cross Product

Consider two vectors $\vec{a}$ and $\vec{b}$ originating from a common point $O$. Let $\theta$ be the angle between these vectors. By the definition of the cross product of two vectors:

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

where $\hat{n}$ is a unit vector perpendicular to the plane containing $\vec{a}$ and $\vec{b}$. [Per the Geometric Definition of Cross Product].

O $\vec{a}$ $\vec{b}$ Height $h = |\vec{b}| \sin \theta$

Step 2: Defining the Area of a Triangle

The area ($A$) of a triangle is defined as:

$A = \frac{1}{2} \times \text{base} \times \text{height}$

In the triangle formed by vectors $\vec{a}$ and $\vec{b}$, we set the base equal to the magnitude of vector $\vec{a}$:

$\text{base} = |\vec{a}|$

The height ($h$) is the perpendicular projection of $\vec{b}$ onto the direction perpendicular to $\vec{a}$. From trigonometry in the right-angled triangle formed by the height:

$h = |\vec{b}| \sin \theta$

Step 3: Derivation and Substitution

Substitute the expressions for base and height into the area formula:

$A = \frac{1}{2} |\vec{a}| (|\vec{b}| \sin \theta)$

By the algebraic property of the cross product magnitude:

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

Substituting this identity into the area equation:

$A = \frac{1}{2} |\vec{a} \times \vec{b}|$

Step 4: Analytical Verification

Component Magnitude Relation
Base $|\vec{a}|$
Height $|\vec{b}| \sin \theta$
Area $\frac{1}{2} |\vec{a}| |\vec{b}| \sin \theta$

Since the magnitude of the cross product represents the area of a parallelogram defined by $\vec{a}$ and $\vec{b}$, and a triangle formed by the same vectors constitutes exactly half of that parallelogram, the relation holds true [By Parallelogram Law of Vector Geometry].

Final Solution: The area of the triangle formed by vectors $\vec{a}$ and $\vec{b}$ is $A = \frac{1}{2} |\vec{a} \times \vec{b}|$.


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