Understanding the Vector Dot Product in a Right Angled Triangle

This question asks us to evaluate a vector expression involving dot products of vectors formed by the sides of a right-angled triangle ABC. We are given that AC is the hypotenuse and its length is p. In a right-angled triangle ABC where AC is the hypotenuse, the right angle must be at vertex B. This means the sides AB and BC are perpendicular to each other.

The expression we need to evaluate is: \[\overrightarrow {{\rm{AB}}} \cdot \overrightarrow {{\rm{AC}}} + \overrightarrow {{\rm{BC}}} \cdot \overrightarrow {{\rm{BA}}} + \overrightarrow {{\rm{CA}}} \cdot \overrightarrow {{\rm{CB}}}\]

Let's analyze each term using vector properties.

Key Vector Properties for Dot Products

• The dot product of two vectors \(\vec{u}\) and \(\vec{v}\) is given by \(\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos \theta\), where \(\theta\) is the angle between the vectors.
• The dot product of a vector with itself is the square of its magnitude: \(\vec{u} \cdot \vec{u} = |\vec{u}|^2\).
• The dot product is commutative: \(\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}\).
• If two vectors are perpendicular, their dot product is zero. Since the right angle is at B, \(\overrightarrow {{\rm{AB}}}\) is perpendicular to \(\overrightarrow {{\rm{BC}}}\). Thus, \(\overrightarrow {{\rm{AB}}} \cdot \overrightarrow {{\rm{BC}}} = 0\).
• For any two points X and Y, \(\overrightarrow {{\rm{YX}}} = -\overrightarrow {{\rm{XY}}}\).
• Vector addition in a triangle: \(\overrightarrow {{\rm{AB}}} + \overrightarrow {{\rm{BC}}} = \overrightarrow {{\rm{AC}}}\).

Evaluating Each Term of the Expression

Let's evaluate each term separately:

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

Using the vector addition \(\overrightarrow {{\rm{AC}}} = \overrightarrow {{\rm{AB}}} + \overrightarrow {{\rm{BC}}}\):

Since \(\overrightarrow {{\rm{AB}}}\) is perpendicular to \(\overrightarrow {{\rm{BC}}}\) (right angle at B), \(\overrightarrow {{\rm{AB}}} \cdot \overrightarrow {{\rm{BC}}} = 0\).

So, \(\overrightarrow {{\rm{AB}}} \cdot \overrightarrow {{\rm{AC}}} = {\rm{AB}}^2\).

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

We know that \(\overrightarrow {{\rm{BA}}} = -\overrightarrow {{\rm{AB}}}\). So:

Since the dot product is commutative, \(\overrightarrow {{\rm{BC}}} \cdot \overrightarrow {{\rm{AB}}} = \overrightarrow {{\rm{AB}}} \cdot \overrightarrow {{\rm{BC}}}\). As established, \(\overrightarrow {{\rm{AB}}} \cdot \overrightarrow {{\rm{BC}}} = 0\).

So, \(\overrightarrow {{\rm{BC}}} \cdot \overrightarrow {{\rm{BA}}} = 0\). This makes sense geometrically because the angle between \(\overrightarrow {{\rm{BC}}}\) and \(\overrightarrow {{\rm{BA}}}\) is the angle at B, which is 90 degrees.

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

We know that \(\overrightarrow {{\rm{CA}}} = -\overrightarrow {{\rm{AC}}}\) and \(\overrightarrow {{\rm{CB}}} = -\overrightarrow {{\rm{BC}}}\). So:

Using the vector addition \(\overrightarrow {{\rm{AC}}} = \overrightarrow {{\rm{AB}}} + \overrightarrow {{\rm{BC}}}\) again:

Since \(\overrightarrow {{\rm{AB}}} \cdot \overrightarrow {{\rm{BC}}} = 0\):

So, \(\overrightarrow {{\rm{CA}}} \cdot \overrightarrow {{\rm{CB}}} = {\rm{BC}}^2\).

Summing the Terms

Now, let's sum the results of the three terms:

Applying the Pythagorean Theorem

In a right-angled triangle ABC with the right angle at B, the Pythagorean theorem states that the square of the hypotenuse is equal to the sum of the squares of the other two sides.

We are given that the hypotenuse AC = p. Therefore, \({\rm{AC}}^2 = p^2\).

Substituting this into our expression:

Thus, the value of the expression is \(p^2\).

Conclusion

The value of \(\overrightarrow {{\rm{AB}}} \cdot \overrightarrow {{\rm{AC}}} + \overrightarrow {{\rm{BC}}} \cdot \overrightarrow {{\rm{BA}}} + \overrightarrow {{\rm{CA}}} \cdot \overrightarrow {{\rm{CB}}}\) in the given right-angled triangle is \(p^2\).

Sum = \({\rm{AB}}^2 + 0 + {\rm{BC}}^2 = {\rm{AB}}^2 + {\rm{BC}}^2\). By Pythagorean theorem, \({\rm{AB}}^2 + {\rm{BC}}^2 = {\rm{AC}}^2 = p^2\).

Revision Table: Vector Dot Products & Right Triangles

Additional Information: Geometric Interpretation of Dot Product

The dot product \(\overrightarrow{a} \cdot \overrightarrow{b}\) can be interpreted as the magnitude of \(\overrightarrow{a}\) multiplied by the projection of \(\overrightarrow{b}\) onto \(\overrightarrow{a}\) (or vice versa).

• \(\overrightarrow {{\rm{AB}}} \cdot \overrightarrow {{\rm{AC}}}\): This is \(|\overrightarrow {{\rm{AB}}}| \times (\text{projection of } \overrightarrow {{\rm{AC}}} \text{ onto } \overrightarrow {{\rm{AB}}})\). In the right triangle ABC, the projection of \(\overrightarrow {{\rm{AC}}}\) onto \(\overrightarrow {{\rm{AB}}}\) is simply \(\overrightarrow {{\rm{AB}}}\) itself (since B is the foot of the perpendicular from C to the line containing AB). So the magnitude of the projection is AB. Thus, \(\overrightarrow {{\rm{AB}}} \cdot \overrightarrow {{\rm{AC}}} = {\rm{AB}} \times {\rm{AB}} = {\rm{AB}}^2\).
• \(\overrightarrow {{\rm{BC}}} \cdot \overrightarrow {{\rm{BA}}}\): These vectors are perpendicular (\(\angle {\rm{CBA}} = 90^\circ\)), so their dot product is 0.
• \(\overrightarrow {{\rm{CA}}} \cdot \overrightarrow {{\rm{CB}}}\): This is \(|\overrightarrow {{\rm{CB}}}| \times (\text{projection of } \overrightarrow {{\rm{CA}}} \text{ onto } \overrightarrow {{\rm{CB}}})\). The projection of \(\overrightarrow {{\rm{CA}}}\) onto \(\overrightarrow {{\rm{CB}}}\) is \(\overrightarrow {{\rm{CB}}}\) itself (since A is the foot of the perpendicular from A to the line containing BC). So the magnitude of the projection is CB = BC. Thus, \(\overrightarrow {{\rm{CA}}} \cdot \overrightarrow {{\rm{CB}}} = {\rm{CB}} \times {\rm{CB}} = {\rm{CB}}^2 = {\rm{BC}}^2\).

Adding these geometric interpretations gives \({\rm{AB}}^2 + 0 + {\rm{BC}}^2\), which, by the Pythagorean theorem (\({\rm{AB}}^2 + {\rm{BC}}^2 = {\rm{AC}}^2 = p^2\)), equals \(p^2\).