Problem 64 - Algebra and Trigonometry

Question

In Exercises \(61-64\), find the magninude ly to the nearest hundredth, and the direction angle \(\vec{\theta},\) to the nearest tenth of a degree, for each given vector \(\mathbf{v}\) $$\mathbf{v}=(7 \mathbf{i}-3 \mathbf{j})-(10 \mathbf{i}-3 \mathbf{j})$$

Step-by-Step Solution

Answer

The magnitude of the vector \(\mathbf{v}\) is 3, and its direction angle is \(180^{\circ}\).

Step 1: Simplify the Vector Equation

Subtracting one vector from another involves subtracting their corresponding components. The resulting vector \(\mathbf{v}\) will therefore be \((7 \mathbf{i}-3 \mathbf{j}) - (10 \mathbf{i}-3 \mathbf{j}) = (7-10) \mathbf{i} + (-3+3) \mathbf{j} = -3\mathbf{i} + 0\mathbf{j}\).

Step 2: Calculate the Magnitude of the Vector

The magnitude of a vector \(\mathbf{v}\) with components \(a \mathbf{i} + b \mathbf{j}\) is given by \(\|\mathbf{v}\| = \sqrt{a^{2}+b^{2}}\). Applying this to the vector \(\mathbf{v} = -3\mathbf{i} + 0\mathbf{j}\), we get \(\|\mathbf{v}\| = \sqrt{(-3)^{2}+(0)^{2}} = \sqrt{9} = 3\).

Step 3: Calculate the Direction Angle of the Vector

The direction angle \(\theta\) of a vector with components \(a \mathbf{i} + b \mathbf{j}\) can be found using the formula \(\theta = \arctan(\frac{b}{a})\). For the vector \(\mathbf{v} = -3\mathbf{i} + 0\mathbf{j}\), \(\theta = \arctan(\frac{0}{-3}) = 0^{\circ}\). However, because the vector points in the negative x-direction, we add 180^{\circ} to the result, getting \(\theta = 180^{\circ}\).

Key Concepts

Vector Magnitude

The magnitude of a vector is a measure of its length or size. To compute the magnitude of a vector with components, it utilizes a formula similar to the Pythagorean theorem.
An essential concept here is that this calculation gives us a scalar quantity, meaning it tells us "how long" the vector is but does not give any directional information.

For any vector \(\mathbf{v} = a\mathbf{i} + b\mathbf{j}\),the formula for its magnitude is given by:\[\|\mathbf{v}\| = \sqrt{a^2 + b^2}\] This formula comes from the realization that a vector forms the hypotenuse of a right triangle, with its components along the x and y axes serving as the other two sides.
In our original exercise, for the vector \(\mathbf{v} = -3\mathbf{i} + 0\mathbf{j}\),substituting the components into the magnitude formula gives:\[\|\mathbf{v}\| = \sqrt{(-3)^2 + 0^2} = \sqrt{9} = 3\]This simple calculation indicates that the vector has a length of 3 units.
Remember, the magnitude is always a non-negative number, even if the components themselves are negative.

Direction Angle

The direction angle of a vector tells us in which direction the vector points relative to the positive x-axis.
This is particularly useful in applications where direction needs to be understood alongside magnitude, such as navigation or physics problems.

For a vector\(\mathbf{v} = a\mathbf{i} + b\mathbf{j}\),the direction angle \(\theta\) can be determined using the arctangent function:\[\theta = \tan^{-1}\left(\frac{b}{a}\right)\]This formula essentially measures the angle of the slope created by the vector's components.
In the given problem, the vector\(\mathbf{v} = -3\mathbf{i} + 0\mathbf{j}\)results in:\[\theta = \arctan\left(\frac{0}{-3}\right) = \arctan(0)\]For vectors aligning perfectly with an axis, the angle can be zero or a full straight line, depending on the vector's direction.
In this case, because the vector points in the negative x-direction, the angle is actually\(180^{\circ}\),since it's effectively pointing backward from the positive x-axis.
This final adjustment accounts for vectors in the different quadrants of the coordinate system.

Vector Subtraction

Vector subtraction is a fundamental operation used to determine the "difference" between two vectors in terms of both direction and magnitude.
This involves simply subtracting corresponding components from another to yield a new vector.
Here's a simple breakdown of how vector subtraction works:

• For two vectors, \(\mathbf{u} = u_1\mathbf{i} + u_2\mathbf{j}\) and \(\mathbf{v} = v_1\mathbf{i} + v_2\mathbf{j}\), subtraction results in:\[\mathbf{u} - \mathbf{v} = (u_1 - v_1)\mathbf{i} + (u_2 - v_2)\mathbf{j}\]
• This operation adjusts the vector's position and can change its magnitude and direction.
• It essentially represents the geometric distance from the endpoint of one vector to the endpoint of another.

In the exercise problem, the subtraction of \(\mathbf{v} = (7\mathbf{i} - 3\mathbf{j}) - (10\mathbf{i} - 3\mathbf{j})\)was simplified to:\[\mathbf{v} = (7 - 10)\mathbf{i} + (-3 + 3)\mathbf{j} = -3\mathbf{i} + 0\mathbf{j}\]You see the process focuses on matching and subtracting each component, leading to a new vector that can be analyzed for magnitude and direction.
This concept also helps visualize changes in physical space, as subtraction provides a direct line from one point to another.