Chapter 3: Problem 7
Let \(\mathbf{v}=(-2,3,0,6) .\) Find all scalars \(k\) such that \(\|k \mathbf{v}\|=5\).
Short Answer
Expert verified
The scalars \( k \) are \( \frac{5}{7} \) and \( -\frac{5}{7} \).
Step by step solution
01
Understand the Vector
The vector \( \mathbf{v} \) given is \((-2, 3, 0, 6)\). It represents the coordinates in a four-dimensional space. We will need to find the norm of this vector for the computations later.
02
Calculate the Norm of \( \mathbf{v} \)
The norm or magnitude of a vector \( \mathbf{v} = (a, b, c, d) \) is calculated using the formula \( \|\mathbf{v}\| = \sqrt{a^2 + b^2 + c^2 + d^2} \). So, for \( \mathbf{v} = (-2, 3, 0, 6) \), the norm is \( \|\mathbf{v}\| = \sqrt{(-2)^2 + 3^2 + 0^2 + 6^2} \). This simplifies to \( \|\mathbf{v}\| = \sqrt{4 + 9 + 0 + 36} = \sqrt{49} = 7 \).
03
Understand the Condition \( \|k\mathbf{v}\| = 5 \)
The expression \( k\mathbf{v} \) denotes the scalar multiplication of \( k \) with each component of \( \mathbf{v} \). The condition states that this new vector should have a norm of 5. This means \( \|k\mathbf{v}\| = 5 \).
04
Calculate the Norm of \( k\mathbf{v} \)
The norm of \( k\mathbf{v} \) is given by \( |k| \times \|\mathbf{v}\| \). From Step 2, \( \|\mathbf{v}\| = 7 \). Thus, \( |k| \times 7 = 5 \).
05
Solve for \( k \)
To solve for \( k \), we need \( |k| \times 7 = 5 \). Dividing both sides by 7, we get \( |k| = \frac{5}{7} \). Therefore, \( k = \frac{5}{7} \) or \( k = -\frac{5}{7} \) since \(|k|\) accounts for both the positive and negative values of \( k \).
06
Final Check
Ensure both calculated values of \( k \), \( \frac{5}{7} \) and \( -\frac{5}{7} \), maintain the condition \( \|k\mathbf{v}\| = 5 \). Both indeed provide \( 5 \) when their absolute values are multiplied by \( 7 \). Thus, both are correct.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Scalar Multiplication
Scalar multiplication in the context of vectors involves multiplying each component of a vector by a scalar value. Imagine you have a vector \( \mathbf{v} = (a, b, c, d) \) and a scalar \( k \). To perform scalar multiplication, you apply \( k \) to each element: \( k\mathbf{v} = (k \cdot a, k \cdot b, k \cdot c, k \cdot d) \).
This operation effectively "scales" the vector. If \( k > 1 \), the vector is stretched, and if \( 0 < k < 1 \), it is compressed. A negative \( k \) not only changes the size but also reverses the direction of the vector.
This operation effectively "scales" the vector. If \( k > 1 \), the vector is stretched, and if \( 0 < k < 1 \), it is compressed. A negative \( k \) not only changes the size but also reverses the direction of the vector.
- If \( k = 0 \), the vector becomes the zero vector \( (0,0,0,0) \), having no direction or magnitude.
- Scalar multiplication preserves the original direction when \( k \) is positive.
- With negative \( k \), the direction is flipped by 180 degrees.
Four-Dimensional Space
In four-dimensional space, vectors have four components, which sounds quite complex but is an extension of the more familiar two-dimensional and three-dimensional spaces. The vector \( \mathbf{v} = (x_1, x_2, x_3, x_4) \) resides in a space defined by four axes. This concept is crucial in higher mathematics and physics.
We often think of 4D space in terms of vector components, such as \( (-2, 3, 0, 6) \). Each number represents a measure along a different dimension. It's not something easily visualized, like 2D or 3D objects, yet it is a mathematical representation useful in various analyses.
We often think of 4D space in terms of vector components, such as \( (-2, 3, 0, 6) \). Each number represents a measure along a different dimension. It's not something easily visualized, like 2D or 3D objects, yet it is a mathematical representation useful in various analyses.
- Four-dimensional vectors are used in computer graphics, particularly in transformations involving time or additional physical aspects.
- They can be instrumental in solving high-dimensional mathematical problems.
Magnitude of a Vector
The magnitude or norm of a vector reflects its length. For a 4D vector like \( \mathbf{v} = (a, b, c, d) \), it is determined using the formula \( \|\mathbf{v}\| = \sqrt{a^2 + b^2 + c^2 + d^2} \). This is akin to finding the length of a line in a simpler, lower-dimensional space.
Magnitude serves as a key concept in understanding vector size, irrespective of direction. For instance, with \( \mathbf{v} = (-2, 3, 0, 6) \), it is calculated as \( \|\mathbf{v}\| = \sqrt{4 + 9 + 0 + 36} = \sqrt{49} = 7 \).
Magnitude serves as a key concept in understanding vector size, irrespective of direction. For instance, with \( \mathbf{v} = (-2, 3, 0, 6) \), it is calculated as \( \|\mathbf{v}\| = \sqrt{4 + 9 + 0 + 36} = \sqrt{49} = 7 \).
- The magnitude tells us how far the vector extends from the intersection of its component axes, rendering it essential for vector operations.
- In expressions like \( \|k\mathbf{v}\| = 5 \), it is a crucial determinant in assessing the effect of scalar multiplication on the vector's length.