Question

a. Find two unit vectors parallel to \(\mathbf{v}=6 \mathbf{i}-8 \mathbf{j}\) b. Find \(b\) if \(\mathbf{v}=\left\langle\frac{1}{3}, b\right\rangle\) is a unit vector. c. Find all values of \(a\) such that \(w=a \mathbf{i}-\frac{a}{3} \mathbf{j}\) is a unit vector.

Short answer

Question: Find the possible values of \(a\) for which the vector \(w = a \mathbf{i} - \frac{a}{3} \mathbf{j}\) is a unit vector. Answer: \(a = \pm\sqrt{\frac{9}{10}}\)

Step-by-step solution

a. Find two unit vectors parallel to \(\mathbf{v}=6 \mathbf{i}-8 \mathbf{j}\)

First, calculate the magnitude of the vector \(\mathbf{v}\): \(\|\mathbf{v}\| = \sqrt{(6)^2 + (-8)^2} = \sqrt{100} = 10\) Now, dividing the components of \(\mathbf{v}\) by its magnitude, we find the unit vectors: \(\frac{1}{\|\mathbf{v}\|}\mathbf{v} = \frac{1}{10}(6\mathbf{i} - 8\mathbf{j}) = \frac{3}{5}\mathbf{i} - \frac{4}{5}\mathbf{j}\) This gives us a unit vector in the direction of \(\mathbf{v}\). Since unit vectors can have two directions, our final answer is: \(\pm(\frac{3}{5}\mathbf{i} - \frac{4}{5}\mathbf{j})\)

b. Find \(b\) if \(\mathbf{v}=\left\langle\frac{1}{3}, b\right\rangle\) is a unit vector

A unit vector has a magnitude of 1. So we will find the magnitude of \(\mathbf{v}\) and set it equal to 1: \(\|\mathbf{v}\| = \sqrt{(\frac{1}{3})^2 + b^2} = 1\) Squaring both sides and solving for \(b\) gives us: \((\frac{1}{3})^2 + b^2 = 1 \Rightarrow b^2 = 1 - (\frac{1}{3})^2 = \frac{8}{9} \Rightarrow b = \pm\sqrt{\frac{8}{9}} = \pm\frac{2\sqrt{2}}{3}\) So, \(b = \pm\frac{2\sqrt{2}}{3}\).

c. Find all values of \(a\) such that \(w=a \mathbf{i}-\frac{a}{3} \mathbf{j}\) is a unit vector

To find all values of \(a\) that make this vector a unit vector, we need the magnitude of \(w\) to be equal to 1: \(\|w\| = \sqrt{(a)^2 + (-\frac{a}{3})^2} = 1\) Squaring both sides and solving for \(a\) gives us: \(a^2 + \frac{a^2}{9} = 1 \Rightarrow \frac{10}{9}a^2 = 1 \Rightarrow a^2 = \frac{9}{10} \Rightarrow a = \pm\sqrt{\frac{9}{10}}\) So, \(a = \pm\sqrt{\frac{9}{10}}\).

Key concepts

Unit Vectors

Unit vectors are vectors that have a magnitude of exactly 1. They are crucial in vector calculus because they establish direction without influencing magnitude. Imagine them as the direction indicators on a map.To find a unit vector in the same direction as a given vector, you need to divide the vector by its own magnitude. For instance, if you have a vector \( \mathbf{v} = 6\mathbf{i} - 8\mathbf{j} \), you first calculate its magnitude.

• Calculate the magnitude: \( \|\mathbf{v}\| = \sqrt{6^2 + (-8)^2} = 10\).
• Divide the vector by its magnitude: \( \frac{1}{10}(6\mathbf{i} - 8\mathbf{j}) = \frac{3}{5}\mathbf{i} - \frac{4}{5}\mathbf{j} \).

Thus, \( \frac{3}{5}\mathbf{i} - \frac{4}{5}\mathbf{j} \) is a unit vector in the direction of \( \mathbf{v} \). Since unit vectors can have opposite directions, the two unit vectors parallel to \( \mathbf{v} \) are \( \pm \left( \frac{3}{5}\mathbf{i} - \frac{4}{5}\mathbf{j} \right) \). Remember, using unit vectors often simplifies complex vector problems by providing a simple, scaled reference.

Vector Magnitude

Understanding vector magnitude is essential when dealing with vectors, as it tells us the length or size of the vector.The magnitude of a vector \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} \) is determined by the formula:\[ \|\mathbf{v}\| = \sqrt{a^2 + b^2} \]This formula is derived from the Pythagorean theorem because in a two-dimensional plane, vectors themselves create a right triangle with horizontal and vertical components.Let's consider a vector \( \mathbf{v} = \left\langle \frac{1}{3}, b \right\rangle \). To find the condition when this vector is a unit vector, we ensure its magnitude equals 1.

• Apply the formula: \( \sqrt{\left(\frac{1}{3}\right)^2 + b^2} = 1\).
• Solve for \( b \): \( b^2 = \frac{8}{9} \) and \( b = \pm \frac{2\sqrt{2}}{3} \).

These calculations confirm that the specified vector becomes a unit vector when \( b \) is either of the calculated values.Understanding how to work with magnitude is a step towards mastering vector operations.

Parallel Vectors

Vectors are parallel if they have the same or exact opposite direction. Working with parallel vectors involves understanding the proportional relationship between components.For a vector \( \mathbf{w} = a\mathbf{i} - \frac{a}{3}\mathbf{j} \), such that it is also a unit vector, we precisely determine the values of \( a \) that maintain this condition.The process involves setting the magnitude of \( \mathbf{w} \) equal to 1:\[ \|\mathbf{w}\| = \sqrt{a^2 + \left(-\frac{a}{3}\right)^2} = 1 \]From this equation, you solve for \( a \):

• Re-organize the equation: \( a^2 + \frac{a^2}{9} = 1 \).
• Find \( \frac{10}{9}a^2 = 1 \) which simplifies to \( a^2 = \frac{9}{10} \).
• Thus, \( a = \pm \sqrt{\frac{9}{10}} \).

These computations identify all possible \( a \) values, ensuring that \( \mathbf{w} \) remains both parallel to any vector of the form \( a\mathbf{i} - \frac{a}{3}\mathbf{j} \), and a unit vector. Such exercises deepen your understanding of how vector directions and magnitudes interplay to create parallel vectors.