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Express the length of a radius vector \(v\) in the Euclidean \(n\) -space (i.e., the distance from the origin to point \(v\) ) by using each of the following: (a) scalars (b) a scalar: product (c) an inner product

Short Answer

Expert verified
The length of vector \( v \) is \( \lVert v \rVert = \sqrt{v_1^2 + v_2^2 + ... + v_n^2} \) using scalars, a scalar product, or an inner product.

Step by step solution

01

Understanding the Radius Vector

The radius vector \( v \) in \( n \)-dimensional Euclidean space is a vector from the origin \( (0,0,...,0) \) to a point \( v = (v_1, v_2, ..., v_n) \). The task is to express the length (or magnitude) of this vector.
02

Length of a Vector using Scalars

The length or magnitude of the vector \( v \) can be calculated using the formula \( \lVert v \rVert = \sqrt{v_1^2 + v_2^2 + ... + v_n^2} \). This involves summing up the squares of its components and then taking the square root of the result.
03

Scalar Product for Length

The scalar product (dot product) of the vector \( v \) with itself is \( v \cdot v = v_1^2 + v_2^2 + ... + v_n^2 \). The length of \( v \) is the square root of this scalar product, i.e., \( \lVert v \rVert = \sqrt{v \cdot v} \).
04

Inner Product for Length

The inner product in \( n \)-dimensional space for vector \( v \) with itself remains \( v \cdot v \) or \( \langle v, v \rangle = v_1^2 + v_2^2 + ... + v_n^2 \). Again, the length is \( \lVert v \rVert = \sqrt{\langle v, v \rangle} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Radius Vector in Euclidean N-Space
When studying Euclidean n-space, a common concept is the **radius vector**. This vector represents the position of a point in space relative to the origin, which is the zero point (0,0,...,0). If we have a point in this space described as \(v = (v_1, v_2, \ldots, v_n)\), the radius vector is the vector drawn from the origin to this point.Understanding the radius vector is crucial because it serves as the building block for determining distances and angles in space. This vector has n components, corresponding to its coordinates within n-dimensional space. Each component \(v_i\) represents the projection of the vector onto the respective axis.
Vector Magnitude or Length
The **magnitude** or length of a vector is a measure of its size or length from the origin to a specific point in space. In the Euclidean n-space, we calculate it using the formula:\[ \lVert v \rVert = \sqrt{v_1^2 + v_2^2 + \ldots + v_n^2} \]This formula effectively captures the distance in space by:
  • Squaring each component of the vector
  • Summing the squares
  • Taking the square root of the total
By performing these steps, one identifies how far a point is from the origin, regardless of the direction. This concept of distance is integral to various applications, from physics to computer graphics.
Scalar Product for Calculating Length
The **scalar product**, also known as the dot product, helps to find the magnitude of a vector in a straightforward way. For vector \(v\), the scalar product with itself is\(v \cdot v = v_1^2 + v_2^2 + \ldots + v_n^2\).Once we obtain this value, the magnitude of the vector is simply the square root:\[ \lVert v \rVert = \sqrt{v \cdot v} \]This method simplifies the process by focusing on the interaction of the vector with itself, providing an efficient means to determine length. The scalar product doesn't just measure length; it's also used when finding angles between vectors and testing important geometric properties.
Inner Product in N-Space
In mathematical terms, the **inner product** is a generalization of the dot product. In Euclidean space, the inner product between a vector and itself is \( \langle v, v \rangle \), which equals \(v_1^2 + v_2^2 + \ldots + v_n^2\).Thus, the inner product also becomes a tool for calculating vector length:\[ \lVert v \rVert = \sqrt{\langle v, v \rangle} \]The inner product offers a systematic way to understand the relationships between vectors through their projections onto each other. This concept extends beyond basic calculations, providing insights into orthogonality, projections, and vector decompositions, which are essential in fields such as machine learning and quantum mechanics.

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Most popular questions from this chapter

Given the following four matrices, test whether any one of them is the inverse of another: \\[ D=\left[\begin{array}{rr} 1 & 12 \\ 0 & 3 \end{array}\right] \quad E=\left[\begin{array}{rr} 1 & 1 \\ 6 & 8 \end{array}\right] \quad F=\left[\begin{array}{rr} 1 & -4 \\ 0 & \frac{1}{3} \end{array}\right] \quad C=\left[\begin{array}{rr} 4 & -\frac{1}{2} \\ -3 & \frac{1}{2} \end{array}\right] \\]

Given \(A=\left[\begin{array}{ll}3 & 6 \\ 2 & 4\end{array}\right], B=\left[\begin{array}{rr}-1 & 7 \\ 8 & 4\end{array}\right],\) and \(C=\left[\begin{array}{ll}3 & 4 \\ 1 & 9\end{array}\right],\) verify that \((a)(A+B)+C=A+(B+C)\) (b) \((A+B)-C=A+(B-C)\)

Prove that \((A+B)(C+D)=A C+A D+B C+B D\)

Rewrite the following in \(\sum\) notation: (a) \(x_{1}\left(x_{1}-1\right)+2 x_{2}\left(x_{2}-1\right)+3 x_{3}\left(x_{3}-1\right)\) (b) \(a_{2}\left(x_{3}+2\right)+a_{3}\left(x_{4}+3\right)+a_{4}\left(x_{5}+4\right)\) (c) \(\frac{1}{x}+\frac{1}{x^{2}}+\dots+\frac{1}{x^{n}} \quad(x \neq 0)\) (d) \(1+\frac{1}{x}+\frac{1}{x^{2}}+\cdots-\frac{1}{x^{n}} \quad(x \neq 0)\)

For \((a)\) through \((d)\) find \(C=A B\) \((a) A=\left[\begin{array}{rr}12 & 14 \\ 20 & 5\end{array}\right] \quad B=\left[\begin{array}{ll}3 & 9 \\ 0 & 2\end{array}\right]\) (b) \(A=\left[\begin{array}{ll}4 & 7 \\ 9 & 1\end{array}\right] \quad B=\left[\begin{array}{lll}3 & 8 & 5 \\ 2 & 6 & 7\end{array}\right]\) (c) \(A=\left[\begin{array}{rr}7 & 11 \\ 2 & 9 \\ 10 & 6\end{array}\right] \quad B=\left[\begin{array}{rrr}12 & 4 & 5 \\ 3 & 6 & 1\end{array}\right]\) \((d) A=\left[\begin{array}{lll}6 & 2 & 5 \\ 7 & 9 & 4\end{array}\right] \quad B=\left[\begin{array}{rr}10 & 1 \\ 11 & 3 \\ 2 & 9\end{array}\right]\) (e) Find (i) \(C=A B,\) and (ii) \(D=B A\), if \(A=\left[\begin{array}{r}-2 \\ 4 \\ 7\end{array}\right] \quad B=\left[\begin{array}{lll}3 & 6 & -2\end{array}\right]\)

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