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Magazine Chapter 9: Problem 12
Find \(t\) such that \(\mathbf{a}=t \mathbf{i}-\mathbf{j}\) and \(\mathbf{b}=\mathbf{i}+\) j make an angle of \(\frac{3}{4} \pi\) radians.
Short Answer
Expert verified
The possible values of \( t \) are 0 or -1.
Step by step solution
01
Write the dot product formula for vectors
The dot product of two vectors \( \mathbf{a} = t \mathbf{i} - \mathbf{j} \) and \( \mathbf{b} = \mathbf{i} + \mathbf{j} \) is given by the formula:\[ \mathbf{a} \cdot \mathbf{b} = (t \cdot 1) + ((-1) \cdot 1) = t - 1 \]
02
Calculate the magnitudes of the vectors
The magnitude of the vector \( \mathbf{a} \) is:\[ \| \mathbf{a} \| = \sqrt{t^2 + (-1)^2} = \sqrt{t^2 + 1} \]The magnitude of the vector \( \mathbf{b} \) is:\[ \| \mathbf{b} \| = \sqrt{1^2 + 1^2} = \sqrt{2} \]
03
Use the angle formula with dot product
The dot product is related to the angle \( \theta \) between the two vectors by:\[ \mathbf{a} \cdot \mathbf{b} = \| \mathbf{a} \| \cdot \| \mathbf{b} \| \cdot \cos \theta \]Substituting for \( \theta = \frac{3}{4} \pi \), we have:\[ t - 1 = \sqrt{t^2 + 1} \cdot \sqrt{2} \cdot \cos \left( \frac{3}{4} \pi \right) \]
04
Calculate \( \cos \left( \frac{3}{4} \pi \right) \)
The angle \( \frac{3}{4} \pi \) is in the second quadrant, where cosine is negative. Calculate:\[ \cos \left( \frac{3}{4} \pi \right) = -\frac{\sqrt{2}}{2} \]
05
Substitute and simplify the equation
Substitute \( \cos \left( \frac{3}{4} \pi \right) = -\frac{\sqrt{2}}{2} \) into the equation:\[ t - 1 = \sqrt{t^2 + 1} \cdot \sqrt{2} \cdot \left( -\frac{\sqrt{2}}{2} \right) \]This simplifies to:\[ t - 1 = - (t^2 + 1) \]
06
Solve for \( t \)
Rearrange the equation and solve for \( t \):\[ t - 1 = -t^2 - 1 \]\[ t + t^2 = 0 \]\[ t(t + 1) = 0 \]This gives us the solutions \( t = 0 \) or \( t = -1 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Dot Product
The dot product, also known as the scalar product, is a way to multiply two vectors that results in a scalar. It involves multiplying corresponding components of the vectors and summing the results. For instance, for two-dimensional vectors \( \mathbf{a} = t \mathbf{i} - \mathbf{j} \) and \( \mathbf{b} = \mathbf{i} + \mathbf{j} \), the dot product is calculated as:
- Multiply the \( x \)-components: \( t \times 1 = t \)
- Multiply the \( y \)-components: \( (-1) \times 1 = -1 \)
- Add them together: \( t - 1 \)
Magnitude of Vectors
The magnitude of a vector, also known as its length or norm, is found using the Pythagorean theorem. For a vector \( \mathbf{a} = t \mathbf{i} - \mathbf{j} \), it can be thought of as the hypotenuse of a right triangle formed by its components.
- Calculate the square of each component: \( t^2 \) and \((-1)^2 = 1\).
- Add these squares together: \( t^2 + 1 \).
- Take the square root to find the magnitude: \( \| \mathbf{a} \| = \sqrt{t^2 + 1} \).
Trigonometric Functions
Trigonometric functions like sine and cosine help in relating angles to sides of a right triangle. Specifically, the cosine of an angle in mathematics is a strong tool used to find projections of vectors and, consequently, angles.For example, when calculating the cosine of an angle such as \( \frac{3}{4} \pi \), you are dealing with angles in radians, being about 135 degrees, located in the second quadrant of the unit circle.
- Cosine is negative in the second quadrant.
- Use the identity: \( \cos \left( \frac{3}{4} \pi \right) = -\frac{\sqrt{2}}{2} \).
Angle Between Vectors
The angle between two vectors can be found using their dot product and magnitudes. The formula to do this is:\[\mathbf{a} \cdot \mathbf{b} = \| \mathbf{a} \| \cdot \| \mathbf{b} \| \cdot \cos \theta\]For vectors \( \mathbf{a} \) and \( \mathbf{b} \), substitute the dot product and magnitude values into the formula. Here, the angle \( \theta \) is provided as \( \frac{3}{4} \pi \).
- Substitute magnitudes: \( t - 1 = \sqrt{t^2 + 1} \cdot \sqrt{2} \cdot \cos \left( \frac{3}{4} \pi \right) \).
- Use the cosine: \( -\frac{\sqrt{2}}{2} \).
- Simplify to find \( t \): this results in an equation that upon solving gives \( t = 0 \) or \( t = -1 \).
