Evaluating Vector Dot Product in a Right Triangle

The problem asks for the value of a specific vector expression involving the sides of a right-angled triangle ABC, where AC is the hypotenuse with length p.

The expression is: \(\overrightarrow {AB} \cdot \overrightarrow {AC} + \overrightarrow {BC} \; \cdot \overrightarrow {BA} + \overrightarrow {CA} \cdot \overrightarrow {CB} \)

Let's consider a right-angled triangle ABC with the right angle at vertex B. AC is the hypotenuse, and its length is given as p, so \(|\overrightarrow{AC}| = p\).

Let's denote the lengths of the sides as follows:

AB = c, so \(|\overrightarrow{AB}| = c\)

BC = a, so \(|\overrightarrow{BC}| = a\)

AC = p, so \(|\overrightarrow{AC}| = p\)

By the Pythagorean theorem in the right triangle ABC (with angle B = 90 degrees), we have \(c^2 + a^2 = p^2\).

Now, let's evaluate each term in the given vector expression using the definition of the dot product: \(\overrightarrow{u} \cdot \overrightarrow{v} = |\overrightarrow{u}| |\overrightarrow{v}| \cos \theta\), where \(\theta\) is the angle between the vectors.

Term 1: \(\overrightarrow {AB} \cdot \overrightarrow {AC}\)

This is the dot product of vector \(\overrightarrow{AB}\) and vector \(\overrightarrow{AC}\). The angle between these two vectors is \(\angle CAB\). Let's call this angle \(\alpha\).

\(\overrightarrow {AB} \cdot \overrightarrow {AC} = |\overrightarrow {AB}| |\overrightarrow {AC}| \cos (\angle CAB)\)

In the right triangle ABC, \(\cos (\angle CAB) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{c}{p}\).

So, \(\overrightarrow {AB} \cdot \overrightarrow {AC} = c \cdot p \cdot \frac{c}{p} = c^2\).

Term 2: \(\overrightarrow {BC} \; \cdot \overrightarrow {BA}\)

This is the dot product of vector \(\overrightarrow{BC}\) and vector \(\overrightarrow{BA}\). These vectors are directed along the sides BC and BA. Since \(\angle ABC = 90^\circ\), the angle between the vectors \(\overrightarrow{BC}\) and \(\overrightarrow{BA}\) is \(90^\circ\).

\(\overrightarrow {BC} \; \cdot \overrightarrow {BA} = |\overrightarrow {BC}| |\overrightarrow {BA}| \cos (\angle CBA)\)

The angle \(\angle CBA = 90^\circ\). We know that \(\cos (90^\circ) = 0\).

So, \(\overrightarrow {BC} \; \cdot \overrightarrow {BA} = a \cdot c \cdot 0 = 0\).

Term 3: \(\overrightarrow {CA} \cdot \overrightarrow {CB}\)

This is the dot product of vector \(\overrightarrow{CA}\) and vector \(\overrightarrow{CB}\). The angle between these two vectors is \(\angle ACB\). Let's call this angle \(\gamma\).

\(\overrightarrow {CA} \cdot \overrightarrow {CB} = |\overrightarrow {CA}| |\overrightarrow {CB}| \cos (\angle ACB)\)

Note that \(|\overrightarrow{CA}| = |\overrightarrow{AC}| = p\) and \(|\overrightarrow{CB}| = |\overrightarrow{BC}| = a\).

In the right triangle ABC, \(\cos (\angle ACB) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{a}{p}\).

So, \(\overrightarrow {CA} \cdot \overrightarrow {CB} = p \cdot a \cdot \frac{a}{p} = a^2\).

Summing the terms:

Now, we add the results of the three dot products:

Expression = \(\overrightarrow {AB} \cdot \overrightarrow {AC} + \overrightarrow {BC} \; \cdot \overrightarrow {BA} + \overrightarrow {CA} \cdot \overrightarrow {CB} = c^2 + 0 + a^2\)

Expression = \(c^2 + a^2\)

Using the Pythagorean theorem (\(c^2 + a^2 = p^2\)), we substitute the value:

Expression = \(p^2\).

Thus, the value of the given vector expression is equal to \(p^2\).