- Identify the condition for orthogonality: the dot product of \(\vec{u}\) and \(\vec{v}\) must be zero.
- Calculate the dot product: \((-5)(x) + (5)(3) = -5x + 15\).
- Set the expression to zero: \(-5x + 15 = 0\).
- Solve for \(x\) to get \(x = 3\).
- The final answer is \(\boxed{3}\).
Explanation
- Understanding Orthogonality and the Dot Product
We are given two vectors in a 2D plane: \(\vec{u} = \langle -5, 5 \rangle\) and \(\vec{v} = \langle x, 3 \rangle\). Two vectors are said to be orthogonal if their dot product equals zero. The dot product of two vectors \(\vec{a} = \langle a_1, a_2 \rangle\) and \(\vec{b} = \langle b_1, b_2 \rangle\) is defined as \(\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2\). We need to determine the value of \(x\) that satisfies the orthogonality condition \(\vec{u} \cdot \vec{v} = 0\).
- Setting Up and Simplifying the Orthogonality Equation
Applying the definition of the dot product to our specific vectors \(\vec{u}\) and \(\vec{v}\), we have:
\[\vec{u} \cdot \vec{v} = (-5)(x) + (5)(3)\]
Simplifying this expression:
\[\vec{u} \cdot \vec{v} = -5x + 15\]
Since the vectors are orthogonal, we set the dot product to zero:
\[-5x + 15 = 0\]
- Solving for x and Verifying the Result
To isolate \(x\), we subtract 15 from both sides of the equation:
\[-5x = -15\]
Next, divide both sides by \(-5\):
\[x = \frac{-15}{-5} = 3\]
Comparing this result to the given options (15, -3, 3, -15, 3/2, None of the above), we find that \(x = 3\) is the correct value.
Examples
Orthogonality is a foundational concept in physics and engineering. For example, if you are designing a structural beam or an object like a shelf, the structural integrity often depends on whether components are perfectly perpendicular to each other. By calculating the dot product of the direction vectors of these components, engineers can verify if they meet at a precise \(90^\circ\) angle, ensuring the construction is stable and balanced. This principle is also used in computer graphics to determine how light reflects off a surface, where the "normal vector" (perpendicular to the surface) is used to calculate lighting intensities.