Chapter 7: Problem 32
Find a vector \(b\) that is parallel to the given vector and has the indicated magnitude. \(\mathbf{a}=\frac{1}{2} \mathbf{i}-\frac{1}{2} \mathbf{j},\|\mathbf{b}\|=3\)
Short Answer
Expert verified
\( \mathbf{b} = \frac{3\sqrt{2}}{4} \mathbf{i} - \frac{3\sqrt{2}}{4} \mathbf{j} \)
Step by step solution
01
Find the Magnitude of Vector a
The first step is to calculate the magnitude of vector \( \mathbf{a} = \frac{1}{2} \mathbf{i} - \frac{1}{2} \mathbf{j} \). Use the formula for the magnitude of a vector: \( \| \mathbf{a} \| = \sqrt{ \left( \frac{1}{2} \right)^2 + \left( -\frac{1}{2} \right)^2 } = \sqrt{ \frac{1}{4} + \frac{1}{4} } = \sqrt{ \frac{1}{2} } = \frac{1}{\sqrt{2}} \).
02
Find the Unit Vector in the Direction of a
To find a unit vector \( \mathbf{u} \) in the direction of \( \mathbf{a} \), divide \( \mathbf{a} \) by its magnitude: \( \mathbf{u} = \frac{\mathbf{a}}{\| \mathbf{a} \|} = \frac{1}{\sqrt{2}} \left( \frac{1}{2} \mathbf{i} - \frac{1}{2} \mathbf{j} \right) = \left( \frac{1}{\sqrt{8}} \right) \mathbf{i} - \left( \frac{1}{\sqrt{8}} \right) \mathbf{j} \).
03
Scale the Unit Vector to the Desired Magnitude
Now, multiply the unit vector \( \mathbf{u} \) by the desired magnitude of \( 3 \) to find \( \mathbf{b} \): \( \mathbf{b} = 3 \left( \frac{1}{\sqrt{8}} \right) \mathbf{i} - 3 \left( \frac{1}{\sqrt{8}} \right) \mathbf{j} = \frac{3}{\sqrt{8}} \mathbf{i} - \frac{3}{\sqrt{8}} \mathbf{j} \).
04
Simplify the Expression for b
Simplify \( \frac{3}{\sqrt{8}} \) by rationalizing the denominator: \( \frac{3}{\sqrt{8}} = \frac{3 \times \sqrt{8}}{8} = \frac{3\sqrt{8}}{8} = \frac{3\sqrt{2}\sqrt{4}}{8} = \frac{3\times 2\sqrt{2}}{8} = \frac{3\sqrt{2}}{4} \). Thus, vector \( \mathbf{b} = \frac{3\sqrt{2}}{4} \mathbf{i} - \frac{3\sqrt{2}}{4} \mathbf{j} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Magnitude
Understanding the magnitude of a vector is crucial in vector calculus. The magnitude, which is often referred to as the "length" or "norm" of a vector, is a representation of the vector's size or extent. For a vector represented in Cartesian coordinates as \( \mathbf{a} = x \mathbf{i} + y \mathbf{j} \), its magnitude can be calculated using the formula:
- \( \| \mathbf{a} \| = \sqrt{x^2 + y^2} \)
Unit Vector
A unit vector is a vector of length 1 that indicates direction. It is very useful in physics and engineering for representing direction without regard to magnitude. To convert any vector into a unit vector, you need to divide the vector by its magnitude.
- For vector \( \mathbf{a} \), the unit vector \( \mathbf{u} \) is given by: \( \mathbf{u} = \frac{\mathbf{a}}{\| \mathbf{a} \|} \)
- \( \mathbf{u} = \left( \frac{1}{\sqrt{8}} \right) \mathbf{i} - \left( \frac{1}{\sqrt{8}} \right) \mathbf{j} \)
Parallel Vectors
Vectors that have the same or exact opposite direction are considered parallel. Two vectors are parallel if one is a scalar multiple of the other. This means if you multiply one vector by a constant, you should be able to get the other.
- For example, if \( \mathbf{a} = k \mathbf{b} \), where \( k \) is any scalar, they are parallel.
Vector Scaling
Scaling a vector involves changing its magnitude without altering its direction. This operation is crucial when you wish to stretch or shrink a vector by a certain factor. To scale a unit vector to a specific magnitude, simply multiply it by the desired magnitude.
- If \( \mathbf{u} \) is a unit vector, to scale it to a magnitude \( k \), multiply: \( \mathbf{v} = k \mathbf{u} \)
- \( \mathbf{b} = 3\left( \frac{1}{\sqrt{8}} \right) \mathbf{i} - 3\left( \frac{1}{\sqrt{8}} \right) \mathbf{j} \)