Question

\(\mathrm{A}=+\mathrm{i} \wedge+\mathrm{j} \wedge-2 \mathrm{k} \wedge\) and \(\mathrm{B} \overrightarrow{\mathrm{i}} \wedge-\mathrm{j} \wedge+\mathrm{k} \wedge\) Find the unit vector in direction of \(\mathrm{A} \rightarrow \times \mathrm{B}^{\rightarrow}\) (A) \([1 / \sqrt{(23)}](-\mathrm{i} \wedge-5 \mathrm{j} \wedge-2 \mathrm{k} \wedge)\) (B) \([1 / \sqrt{(35)]}(-\mathrm{i} \wedge-5 \mathrm{j} \wedge-3 \mathrm{k} \wedge)\) (C) \([1 / \sqrt{(29})](-i \wedge-5 j \wedge-3 k \wedge)\) (D) \([1 / \sqrt{(35)]}(-\mathrm{i} \wedge-5 j \wedge-3 \mathrm{k} \wedge)\)

Short answer

\(\hat{C} = [\frac{1}{\sqrt{14}}](- \mathrm{i} - 3 \mathrm{j} - 2 \mathrm{k})\)

Step-by-step solution

Calculate the cross product of A and B

To find the cross product of A and B, we will use the following formula for cross product of two vectors: \(A \times B = (A_yB_z - A_zB_y) \mathrm{i} + (A_zB_x - A_xB_z) \mathrm{j} + (A_xB_y - A_yB_x) \mathrm{k}\) Given A = (1, 1, -2) and B = (1, -1, 1), we have \(A_x = 1, A_y = 1, A_z = -2\), and \(B_x = 1, B_y = -1, B_z = 1\). Now, substituting the values in the formula, we get: \(A \times B = ((1)(1) - (-2)(-1)) \mathrm{i} + ((-2)(1) - (1)(1)) \mathrm{j} + ((1)(-1) - (1)(1)) \mathrm{k}\)

Simplify the cross product expression

Now, we will simplify the cross product expression: \(A \times B = (1 - 2) \mathrm{i} + (-2 - 1) \mathrm{j} + (-1 - 1) \mathrm{k} = - \mathrm{i} - 3 \mathrm{j} - 2 \mathrm{k}\)

Calculate the magnitude of the cross product

To calculate the magnitude (length) of the cross product, we use the formula: \(|A \times B| = \sqrt{(\mathrm{-1})^2 + (\mathrm{-3})^2 + (\mathrm{-2})^2}\) \(|A \times B| = \sqrt{1 + 9 + 4}\) \(|A \times B| = \sqrt{14}\)

Calculate the unit vector in the direction of A x B

To find the unit vector, we divide each component of the cross product by its magnitude: \(\hat{C} = \frac{(- \mathrm{i} - 3 \mathrm{j} - 2 \mathrm{k})}{\sqrt{14}}\) Now, we can rewrite it as: \(\hat{C} = [\frac{1}{\sqrt{14}}](- \mathrm{i} - 3 \mathrm{j} - 2 \mathrm{k})\) Comparing this with the given options, it doesn't match any of the choices, which indicates that there may be an error in the given options. However, our calculated unit vector \(\hat{C}\) is correct.

Key concepts

Unit Vector

A unit vector is a vector that has a magnitude of 1. It is commonly used to indicate direction. When you have a vector in any direction, you can turn it into a unit vector by dividing it by its own magnitude. This process retains the direction of the original vector while converting its size to 1.

In physics and engineering, unit vectors are important because they help to simplify vector calculations and represent direction without scale. They are usually denoted with a "hat" symbol, for example, \( \hat{i}, \hat{j}, \hat{k} \), which correspond to the unit vectors along the x, y, and z axes respectively.

Magnitude of Vector

The magnitude of a vector measures its length in space. It is always a non-negative quantity and is calculated using components of the vector along each axis. For a vector \(\overrightarrow{V} = a \mathrm{i} + b \mathrm{j} + c \mathrm{k} \, \), the magnitude \(|\overrightarrow{V}|\) is determined by the formula:
\[ |\overrightarrow{V}| = \sqrt{a^2 + b^2 + c^2} \]

• The components \(a, b, c\) represent the vector's projections onto the x, y, and z axes.
• This formula is derived from the Pythagorean Theorem, applied in three-dimensional space.

Understanding the magnitude is crucial because it gives you the "size" of the vector, separate from its direction. Magnitude helps define how long the vector is in whatever context it is used.

Unit Vector Calculation

To calculate a unit vector from a given vector, you first need to know the magnitude of the vector. The process involves dividing each component of the vector by its magnitude.

Let's say we have a vector \(\overrightarrow{V} = a \mathrm{i} + b \mathrm{j} + c \mathrm{k}\). The unit vector \(\hat{u}\) in the direction of \(\overrightarrow{V}\) can be found using:
\[\hat{u} = \frac{1}{|\overrightarrow{V}|} (a \mathrm{i} + b \mathrm{j} + c \mathrm{k})\]

• Calculate the magnitude \(|\overrightarrow{V}|\) using the formula \(\sqrt{a^2 + b^2 + c^2}\).
• Divide each component \(a, b, c\) by the magnitude.

By doing this, you ensure that the new vector maintains the same direction as \(\overrightarrow{V}\) but has a length of 1.

Vector i, j, k Components

Vectors in three-dimensional space are often expressed in terms of the standard basis vectors \(\mathrm{i}, \mathrm{j}, \mathrm{k} \), which align with the x, y, and z axes. These components give a vector a specific direction and magnitude in space.

• \(\mathrm{i}\) is the unit vector along the x-axis, \(\mathrm{j}\) is along the y-axis, and \(\mathrm{k}\) is along the z-axis.
• Any vector \(\overrightarrow{V} = a \mathrm{i} + b \mathrm{j} + c \mathrm{k} \) is a combination of these unit vectors, with \(a, b,\) and \(c\) being coefficients that determine the vector's direction and length.

Understanding these components helps you break down the vector into smaller, more manageable parts, making calculations easier. By adjusting the magnitudes of \(a, b,\) and \(c\), you can precisely represent any vector in three-dimensional space.