Question

What is unit vector along \(\hat{i}+\hat{\jmath}\) ? (A) \([(1+\hat{j}) / 2]\) (C) \([(\hat{1}+\hat{j}) / \sqrt{3}]\) (D) \([(\hat{1}-\hat{j}) / \sqrt{2}]\)

Short answer

The correct answer is \(\frac{1}{\sqrt{2}}(\hat{i}+\hat{\jmath})\). None of the provided options match this answer.

Step-by-step solution

Write down the given vector in component form

We are given the vector \(\hat{i}+\hat{\jmath}\) which in component form can be written as: \[ \vec{v} = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \]

Calculate the magnitude of the vector

To find the unit vector along \(\vec{v}\), we first need to find the magnitude of \(\vec{v}\), which is the square root of the sum of the squares of the components: \[ |\vec{v}| = \sqrt{(1^2)+(1^2)} = \sqrt{2} \]

Normalize the vector

Now that we have the magnitude, we can normalize the vector by dividing each component by the magnitude to get the unit vector: \[ \hat{u} = \frac{\vec{v}}{|\vec{v}|} = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix} \] The unit vector in i-j representation is: \[ \hat{u} = \frac{1}{\sqrt{2}}(\hat{i}+\hat{\jmath}) \] Comparing with the given options, we find that the correct answer is not among the options. There must be a mistake in the options provided.

Key concepts

Vector Normalization

Vector normalization is a process used to convert a vector into a unit vector, which is a vector with a magnitude of 1. This is done by scaling the original vector by the reciprocal of its magnitude. Normalizing a vector retains its direction but changes its length. This process is crucial in many areas such as physics, computer graphics, and machine learning.

To normalize a vector \( \vec{v} \), you divide each component of \( \vec{v} \) by \( |\vec{v}| \), the magnitude of the vector. This is represented by the formula:

• \( \hat{u} = \frac{\vec{v}}{|\vec{v}|} \)

By following this process, \( \hat{u} \) will be a unit vector, pointing in the same direction as \( \vec{v} \) but with a magnitude of 1. Successfully normalizing a vector can be checked by ensuring its magnitude is exactly 1.

Magnitude of a Vector

The magnitude of a vector represents its length in a vector space and is a critical concept when working with vectors. To find the magnitude of a 2-dimensional vector \( \vec{v} = \begin{bmatrix} x \ y \end{bmatrix} \), you apply the formula:

\( |\vec{v}| = \sqrt{x^2 + y^2} \)

This formula is derived from the Pythagorean theorem and calculates the diagonal length of the rectangle whose sides are aligned with the components of the vector. This diagonal represents the vector itself. For the vector \( \begin{bmatrix} 1 \ 1 \end{bmatrix} \),
the magnitude is calculated as:

• \( |\vec{v}| = \sqrt{1^2 + 1^2} = \sqrt{2} \)

Understanding the magnitude of a vector is essential for applications such as vector addition or normalization, ensuring you're working accurately within vector spaces.

Vector Components

Vector components are the individual parts of a vector that define its direction and magnitude in the respective axes of a coordinate system. For example, in a 2D plane, a vector \( \vec{v} \) can be expressed in terms of its components along the x and y axes as \( \begin{bmatrix} x \ y \end{bmatrix} \). These components essentially serve as the instructions for how far and in which direction you move along each axis from the origin.

In the exercise's context, the vector \( \hat{i} + \hat{\jmath} \) was expressed in component form as \( \begin{bmatrix} 1 \ 1 \end{bmatrix} \). Here:

• \( x = 1 \) (along the x-axis, or \( \hat{i} \))
• \( y = 1 \) (along the y-axis, or \( \hat{\jmath} \))

These values direct you to move one unit along the x-axis and one unit along the y-axis. Understanding how to break down vectors into components is useful when performing operations like addition, scaling, or finding the resultant magnitude, as these components are foundational in calculating these vector properties.