Question

By evaluating their dot product, find the values of the scalar \(s\) for which the two vectors \(\mathbf{b}=\hat{\mathbf{x}}+s \hat{\mathbf{y}}\) and \(\mathbf{c}=\hat{\mathbf{x}}-s \hat{\mathbf{y}}\) are orthogonal. (Remember that two vectors are orthogonal if and only if their dot product is zero.) Explain your answers with a sketch.

Short answer

The values of \(s\) for which the vectors are orthogonal are \(s = 1\) and \(s = -1\).

Step-by-step solution

Define Dot Product Formula

The dot product of two vectors \(\mathbf{b} = b_1 \hat{\mathbf{x}} + b_2 \hat{\mathbf{y}}\) and \(\mathbf{c} = c_1 \hat{\mathbf{x}} + c_2 \hat{\mathbf{y}}\) is calculated as \( b_1c_1 + b_2c_2 \).

Substitute the Components of Vectors

For \(\mathbf{b} = \hat{\mathbf{x}} + s \hat{\mathbf{y}}\), we have \(b_1 = 1\) and \(b_2 = s\). For \(\mathbf{c} = \hat{\mathbf{x}} - s \hat{\mathbf{y}}\), we have \(c_1 = 1\) and \(c_2 = -s\).

Write the Dot Product Expression

Substitute the components into the dot product formula: \(1 \cdot 1 + s \cdot (-s) = 1 - s^2\). The dot product becomes \(1 - s^2\).

Set the Dot Product to Zero

To find when the vectors are orthogonal, set the dot product equation \(1 - s^2 = 0\).

Solve the Equation for s

Rearranging the equation \(1 - s^2 = 0\), we have \(s^2 = 1\). Solving this gives \(s = \pm 1\).

Key concepts

Orthogonal Vectors

In the world of vector mathematics, orthogonal vectors represent a pair of vectors that are at a right angle to each other. This right angle corresponds to a 90-degree angle. Vectors are orthogonal when their dot product equals zero. This occurs because when two vectors form a right angle, their projection onto each other becomes zero.

Think of orthogonal vectors like the x and y axes on a graph. They meet at a right angle, just as orthogonal vectors do. In our exercise, we're dealing with vectors \( \mathbf{b} \) and \( \mathbf{c} \). Our goal is to determine the scalar \( s \) such that these vectors are orthogonal, which in turn means their dot product, to be discussed next, is zero.

Scalar Calculation

Scalar calculations are essential when finding specific conditions such as orthogonality. In the given exercise, we were required to evaluate a particular scalar \( s \).

The steps to achieve this involve computing the dot product and setting it to zero.

• Start by writing down the vectors: \( \mathbf{b} = \hat{\mathbf{x}} + s \hat{\mathbf{y}} \) and \( \mathbf{c} = \hat{\mathbf{x}} - s \hat{\mathbf{y}} \).
• Using the dot product formula: \( b_1c_1 + b_2c_2 \), substitute the vector components.
• For \( \mathbf{b} \), this gives \( b_1 = 1 \) and \( b_2 = s \). For \( \mathbf{c} \), \( c_1 = 1 \) and \( c_2 = -s \).
• Inserting these into the dot product formula, you'll get: \( 1 \cdot 1 + s \cdot (-s) = 1 - s^2 \).

The task then is to find the value of \( s \) that makes this expression zero, indicating orthogonality: solving \( 1 - s^2 = 0 \), gives \( s = \pm 1 \). Hence, for \( s = 1 \) or \( s = -1 \), the vectors are orthogonal.

Vector Mathematics

Vector mathematics involves various operations including vector addition, subtraction, and the dot product. In this exercise, the focus is on the dot product, a fundamental operation that helps in understanding the angle between two vectors.

The dot product, also called the scalar product, is calculated using the formula \( b_1c_1 + b_2c_2 \). This operation is crucial for identifying whether two vectors are orthogonal, as we have seen.

• Firstly, recognize that vectors are directed quantities with both magnitude and direction. They're typically represented as arrows in a coordinate space.
• The dot product is a measure that outputs a scalar (a single number) rather than another vector.
• In mathematical and physical applications, orthogonality implies independence. Thus, knowing when vectors are orthogonal is useful.

Understanding vector mathematics paves the way to solving geometric and physical problems, such as finding angles, projections, or even computing work done by a force. It’s a versatile tool that opens many doors in various scientific and engineering fields.