Question

Find all real numbers \(x\) such that: a. \(\left[\begin{array}{r}2 \\ -1 \\ 3\end{array}\right]\) and \(\left[\begin{array}{r}x \\ -2 \\ 1\end{array}\right]\) are orthogonal. b. \(\left[\begin{array}{r}2 \\ -1 \\ 1\end{array}\right]\) and \(\left[\begin{array}{l}1 \\ x \\ 2\end{array}\right]\) are at an angle of \(\frac{\pi}{3}\).

Short answer

(a) \(x = -\frac{5}{2}\); (b) \(x = 1\) or \(x = -17\).

Step-by-step solution

Understanding Orthogonality

Two vectors \( \mathbf{a} = \left[ \begin{array}{c} a_1 \ a_2 \ a_3 \end{array} \right] \) and \( \mathbf{b} = \left[ \begin{array}{c} b_1 \ b_2 \ b_3 \end{array} \right] \) are orthogonal if their dot product is zero. This means \( a_1b_1 + a_2b_2 + a_3b_3 = 0 \).

Calculating the Dot Product for Part (a)

For the vectors \( \left[ \begin{array}{c} 2 \ -1 \ 3 \end{array} \right] \) and \( \left[ \begin{array}{c} x \ -2 \ 1 \end{array} \right] \), the dot product is \( 2x + (-1)(-2) + 3 \cdot 1 = 0 \).

Solve the Dot Product Equation for Part (a)

Simplifying the equation: \( 2x + 2 + 3 = 0 \). Solving for \( x \), we get: \( 2x + 5 = 0 \) leading to \( 2x = -5 \), hence \( x = -\frac{5}{2} \).

Understanding Angle Condition for Part (b)

Two vectors \( \mathbf{a} \) and \( \mathbf{b} \) have an angle of \( \theta \) between them if \( \cos(\theta) = \frac{ \mathbf{a} \cdot \mathbf{b} }{ ||\mathbf{a}|| ||\mathbf{b}|| } \). For angle \( \frac{\pi}{3} \), \( \cos(\frac{\pi}{3}) = \frac{1}{2} \).

Calculate Dot Product for Part (b)

The vectors \( \left[ \begin{array}{c} 2 \ -1 \ 1 \end{array} \right] \) and \( \left[ \begin{array}{c} 1 \ x \ 2 \end{array} \right] \) have a dot product \( 2 \cdot 1 + (-1)x + 1 \cdot 2 = 2 - x + 2 = 4 - x \).

Calculate the Norms of the Vectors for Part (b)

The norm of \( \left[ \begin{array}{c} 2 \ -1 \ 1 \end{array} \right] \) is \( \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \). The norm of \( \left[ \begin{array}{c} 1 \ x \ 2 \end{array} \right] \) is \( \sqrt{1^2 + x^2 + 2^2} = \sqrt{1 + x^2 + 4} = \sqrt{x^2 + 5} \).

Set Up the Equation for Cosine

Using the angle condition: \( \frac{4-x}{\sqrt{6} \cdot \sqrt{x^2 + 5}} = \frac{1}{2} \). Clear the fraction by multiplying through, giving: \( 2(4-x) = \sqrt{6} \cdot \sqrt{x^2 + 5} \).

Solve the Equation for \( x \)

Squaring both sides gives \( (8 - 2x)^2 = 6(x^2 + 5) \). Simplify and solve: \( 64 - 32x + 4x^2 = 6x^2 + 30 \). Rearrange to \( 2x^2 + 32x - 34 = 0 \), and solve using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 2 \), \( b = 32 \), \( c = -34 \).

Apply the Quadratic Formula

Calculate the discriminant: \( 32^2 - 4 \cdot 2 \cdot (-34) = 1024 + 272 = 1296 \). \( x = \frac{-32 \pm \sqrt{1296}}{4} = \frac{-32 \pm 36}{4} \). Solve for \( x \): \( \frac{4}{4} = 1 \) and \( \frac{-68}{4} = -17 \).

Key concepts

Orthogonality

Understanding orthogonality in vector mathematics means looking at when two vectors create a right angle, or 90 degrees, with each other. This happens when the dot product of the vectors is equal to zero. For vectors, the dot product is a mathematical operation yielding a single number, calculated as \( a_1b_1 + a_2b_2 + a_3b_3 \).
If the dot product is zero, the vectors are orthogonal. This principle is often applied in physics and engineering, where orthogonal vectors represent mutually perpendicular directions. In the context of matrices or higher-dimensional vectors, this concept ensures directions are independent or can be related to coordinated frames especially important in 3D modeling or computer graphics.

Dot Product

The dot product, also known as the scalar product, is a key operation in vector mathematics.
It combines elements from two vectors to produce a single real number. Given two vectors \( \mathbf{a} = \left[ a_1, a_2, a_3 \right] \) and \( \mathbf{b} = \left[ b_1, b_2, b_3 \right] \), their dot product is calculated as \( a_1b_1 + a_2b_2 + a_3b_3 \).
This product has important geometric implications, as it can tell us about the angle between the vectors. If the dot product is zero, the vectors are orthogonal, as discussed earlier. If not, the magnitude and sign of the dot product can indicate the relative direction and orientation of the vectors, influencing vector projections and transformations used in computer vision and graphics.

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \).
To find the values of \( x \) that satisfy such an equation, the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) can be used.
Inside the formula, \( a \), \( b \), and \( c \) are coefficients from the equation. The term under the square root, known as the discriminant \( b^2 - 4ac \), determines the nature of the solutions. If it's positive, there are two real solutions; if zero, one real solution exists; if negative, no real solutions exist. This formula is foundational in algebra and finds relevance in physics for equations of motion, and in optimizing certain engineering solutions.

Cosine of Angle Between Vectors

The cosine of the angle between two vectors can be found using the dot product and the magnitudes of the vectors themselves. This relationship is expressed with the formula \( \cos(\theta) = \frac{\mathbf{a} \cdot \mathbf{b}}{||\mathbf{a}|| \, ||\mathbf{b}||} \), where \( \theta \) is the angle between the vectors, \( \mathbf{a} \cdot \mathbf{b} \) is the dot product, and \( ||\mathbf{a}|| \) and \( ||\mathbf{b}|| \) are the magnitudes.
The cosine function returns a value between -1 and 1, which corresponds to the angle's relationship; like 1 being 0 degrees (same direction), 0 being 90 degrees (orthogonal), and -1 being 180 degrees (opposite direction).
This method of finding angles is crucial in fields like computer graphics, where determining the orientation of objects can significantly affect rendering techniques and animations.