91Ó°ÊÓ

A unit vector is a vector with a magnitude of 1. It is often used to indicate direction while ignoring other aspects, like the strength or length of the vector. Unit vectors are typically represented with a hat symbol over a letter, such as \( \mathbf{\hat{a}} \). To find the unit vector in the direction of a given vector \( \mathbf{v} \), you simply divide the vector by its magnitude:

• If \( \mathbf{v} = (x, y, z) \), then the unit vector \( \mathbf{\hat{v}} = \left( \frac{x}{\|\mathbf{v}\|}, \frac{y}{\|\mathbf{v}\|}, \frac{z}{\|\mathbf{v}\|} \right) \).

A unit vector maintains the same direction as the original vector.
But it has no physical dimension making it perfect for scenarios where only direction matters.
In our exercise, \( \mathbf{a} \) is a unit vector, ensuring that any manipulations involving it focus purely on directional relationships.

The cross product is a special vector multiplication operation used in vector calculus. It takes two vectors in three-dimensional space and returns a vector that is perpendicular to both. Given two vectors \( \mathbf{a} \) and \( \mathbf{b} \), the cross product is denoted as \( \mathbf{a} \times \mathbf{b} \).

• Mathematically, the cross product is calculated as:\[ \mathbf{a} \times \mathbf{b} = (a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1) \]
• The result is a vector orthogonal to both \( \mathbf{a} \) and \( \mathbf{b} \).

An interesting property of the cross product is its connection to area.
The magnitude of the resulting vector equals the area of the parallelogram spanned by the two vectors.
In the problem, \( \sqrt{3}(\mathbf{a} \times \mathbf{b}) \) creates a side of the triangle illustrating how this perpendicular vector contributes to the triangle's structure.

The dot product is another form of vector multiplication resulting in a scalar rather than a vector.
Given two vectors \( \mathbf{a} \) and \( \mathbf{b} \), the dot product is expressed as \( \mathbf{a} \cdot \mathbf{b} \).

• The mathematical expression for the dot product is: \[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \]
• It provides a measure of the extent to which the two vectors point in the same direction.

This scalar product depends on both the magnitudes of the vectors and the cosine of the angle between them.
For example, if the two vectors are perpendicular, their dot product is zero because \( \cos(90^{\circ}) = 0 \).
In the given exercise, \( \mathbf{v}_2 = \mathbf{b} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{a} \) uses the dot product to remove the component of \( \mathbf{b} \) in the direction of \( \mathbf{a} \), simplifying the relationships within the triangle.

A right triangle is a triangle with one angle measuring exactly \( 90^{\circ} \).
In geometry, the side opposite the right angle is known as the hypotenuse, and it is the longest side of the triangle.

• Such triangles follow Pythagoras' theorem, \( a^2 + b^2 = c^2 \), where \( c \) is the hypotenuse.
• Properties of right triangles mean the sum of the other two angles totals \( 90^{\circ} \).

Understanding right triangles is crucial in this exercise since the calculated angles were \( 45^{\circ}, 45^{\circ}, 90^{\circ} \), suggesting that the triangle is an isosceles right triangle.
This property reveals the geometry formed by the vectors \( \sqrt{3}(\mathbf{a} \times \mathbf{b}) \) and \( \mathbf{b} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{a} \), where these vectors act as the perpendicular legs forming the right angle.