Chapter 14: Problem 29
Let \(u=i+m j+k\) and \(v=2 i-j+n k\) Compute all values of \(m\) and \(n\) for which \(\mathbf{u} \perp \mathbf{v}\) and \(|\mathbf{u}|=|\mathbf{v}|\)
Short Answer
Expert verified
m=7/4, n=-1/4
Step by step solution
01
Analyze the Perpendicular Condition
Two vectors \( \mathbf{u} \) and \( \mathbf{v} \) are perpendicular if their dot product is zero. Given \( \mathbf{u} = i + m j + k \) and \( \mathbf{v} = 2i - j + nk \), the dot product is \( \mathbf{u} \cdot \mathbf{v} = 1 \cdot 2 + m \cdot (-1) + 1 \cdot n = 2 - m + n = 0 \). Therefore, the equation \( 2 - m + n = 0 \) must be satisfied.
02
Analyze the Equal Magnitude Condition
Vectors \( \mathbf{u} \) and \( \mathbf{v} \) have equal magnitudes when \( |\mathbf{u}| = |\mathbf{v}| \). This means \( \sqrt{1^2 + m^2 + 1^2} = \sqrt{2^2 + (-1)^2 + n^2} \). Simplifying both sides gives \( \sqrt{2 + m^2} = \sqrt{5 + n^2} \). Squaring both sides gives the equation: \( 2 + m^2 = 5 + n^2 \).
03
Solve the System of Equations
We have two equations: 1. \( 2 - m + n = 0 \) 2. \( 2 + m^2 = 5 + n^2 \)From equation 1, solve for \( n \): \( n = m - 2 \).Substitute \( n = m - 2 \) into equation 2:\( 2 + m^2 = 5 + (m-2)^2 \)Expanding \((m-2)^2\):\( 4 + m^2 - 4m \)Substitute back:\( 2 + m^2 = 5 + m^2 - 4m + 4 \)Combine and simplify:\( 2 = 9 - 4m \)\( 4m = 7 \)Thus, \( m = \frac{7}{4} \).Substitute back to find \( n \):\( n = \frac{7}{4} - 2 = \frac{-1}{4} \).
04
Verify Solutions
Substitute \( m = \frac{7}{4} \) and \( n = \frac{-1}{4} \) back into the original equations to verify:For \( 2 - m + n = 0 \):\( 2 - \frac{7}{4} + \frac{-1}{4} = 0 \).Translates to \( \frac{8}{4} - \frac{7}{4} - \frac{1}{4} = 0 \) which is true.For \( 2 + m^2 = 5 + n^2 \):\( 2 + \left(\frac{7}{4}\right)^2 = 5 + \left(\frac{-1}{4}\right)^2 \)Computes to \( 2 + \frac{49}{16} = 5 + \frac{1}{16} \), which simplifies to \( \frac{81}{16} = \frac{81}{16} \), confirming both conditions are satisfied.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding the Dot Product
The dot product, also known as the scalar product, is a crucial concept in vector mathematics, especially when determining if two vectors are perpendicular. Two vectors are perpendicular if their dot product is zero. For vectors \( \mathbf{u} = i + mj + k \) and \( \mathbf{v} = 2i - j + nk \), their dot product is calculated as follows:
- Multiply the corresponding components of the vectors: \( 1 \cdot 2 \) for the \(i\) components, \( m \cdot (-1) \) for the \(j\) components, and \( 1 \cdot n \) for the \(k\) components.
- Add these products together: \( 2 - m + n \).
Exploring Magnitude Equality
The concept of magnitude equality between two vectors means that both vectors have the same length. To find the magnitude of a vector, calculate the square root of the sum of the squares of its components. In our example, the magnitudes of vectors \( \mathbf{u} = i + mj + k \) and \( \mathbf{v} = 2i - j + nk \) are:
- \( |\mathbf{u}| = \sqrt{1^2 + m^2 + 1^2} = \sqrt{2 + m^2} \)
- \( |\mathbf{v}| = \sqrt{2^2 + (-1)^2 + n^2} = \sqrt{5 + n^2} \)
Solving Vector Equations
When tasked with solving equations derived from vector properties, breaking down the process into manageable steps is essential. Here, we have two equations:
- \( 2 - m + n = 0 \), from the dot product condition
- \( 2 + m^2 = 5 + n^2 \), from the magnitude equality condition
- From \( 2 - m + n = 0 \), we find \( n = m - 2 \).
- \( 2 + m^2 = 5 + (m-2)^2 \)
- \( 2 + m^2 = 5 + m^2 - 4m + 4 \)