Problem 7 - Thomas Calculus

Question

Let \(\mathbf{u}=\langle 3,-2\rangle\) and \(\mathbf{v}=\langle- 2,5\rangle .\) Find the (a) component form and (b) magnitude (length) of the vector. \(\frac{3}{5} \mathbf{u}+\frac{4}{5} \mathbf{v}\)

Step-by-Step Solution

Answer

The component form is \(\langle \frac{1}{5}, \frac{14}{5} \rangle\) and the magnitude is \(\frac{\sqrt{197}}{5}\).

Step 1: Scale Vector u

Calculate \( \frac{3}{5} \mathbf{u} \) by multiplying each component of \( \mathbf{u} = \langle 3, -2 \rangle \) by \( \frac{3}{5} \).\[\frac{3}{5} \mathbf{u} = \left\langle \frac{3}{5} \times 3, \frac{3}{5} \times (-2) \right\rangle = \left\langle \frac{9}{5}, -\frac{6}{5} \right\rangle\]

Step 2: Scale Vector v

Calculate \( \frac{4}{5} \mathbf{v} \) by multiplying each component of \( \mathbf{v} = \langle -2, 5 \rangle \) by \( \frac{4}{5} \).\[\frac{4}{5} \mathbf{v} = \left\langle \frac{4}{5} \times (-2), \frac{4}{5} \times 5 \right\rangle = \left\langle -\frac{8}{5}, 4 \right\rangle\]

Step 3: Add Scaled Vectors

Add the scaled vectors from Step 1 and Step 2 to find their resultant component form.\[\frac{3}{5} \mathbf{u} + \frac{4}{5} \mathbf{v} = \left\langle \frac{9}{5}, -\frac{6}{5} \right\rangle + \left\langle -\frac{8}{5}, 4 \right\rangle = \left\langle \frac{9}{5} - \frac{8}{5}, -\frac{6}{5} + 4 \right\rangle = \left\langle \frac{1}{5}, \frac{14}{5} \right\rangle\]

Step 4: Calculate Magnitude of Resultant Vector

Find the magnitude of the resultant vector \( \langle \frac{1}{5}, \frac{14}{5} \rangle \) using the formula for the magnitude of a vector \( \langle a, b \rangle \), which is \( \sqrt{a^2 + b^2} \).\[\text{Magnitude} = \sqrt{\left(\frac{1}{5}\right)^2 + \left(\frac{14}{5}\right)^2} = \sqrt{\frac{1}{25} + \frac{196}{25}} = \sqrt{\frac{197}{25}} = \frac{\sqrt{197}}{5}\]

Key Concepts

Scaling Vectors

Scaling a vector involves multiplying each of its components by a scalar. In the context of the exercise, the vectors \( \mathbf{u} \) and \( \mathbf{v} \) were scaled by different factors. This operation helps modify the magnitude of vectors without altering their direction, unless the scalar is negative. This kind of transformation is particularly useful when you need to adjust a vector's length proportionally in vector algebra.

For \( \mathbf{u} = \langle 3, -2 \rangle \), scaling it by \( \frac{3}{5} \) gives:

• First component: \( 3 \times \frac{3}{5} = \frac{9}{5} \)
• Second component: \( (-2) \times \frac{3}{5} = -\frac{6}{5} \)

For \( \mathbf{v} = \langle -2, 5 \rangle \), scaling it by \( \frac{4}{5} \) results in:

• First component: \( (-2) \times \frac{4}{5} = -\frac{8}{5} \)
• Second component: \( 5 \times \frac{4}{5} = 4 \)

Addition of Vectors

Adding vectors is about combining their components to form a new resultant vector. This operation can be visualized as placing the tail of one vector at the head of another. The sum of two scaled vectors from the exercise is found by adding each corresponding component separately.

To add the scaled vectors from Step 1 and Step 2:

• Add the first components: \( \frac{9}{5} + (-\frac{8}{5}) = \frac{1}{5} \)
• Add the second components: \( -\frac{6}{5} + 4 = \frac{14}{5} \)

These additions result in a new vector \( \left\langle \frac{1}{5}, \frac{14}{5} \right\rangle \). Such vector addition is fundamental to understanding how multiple forces or directions combine in physics and engineering.

Vector Magnitude

The magnitude of a vector represents its length, essentially indicating how far it extends. To compute the magnitude of a vector \( \langle a, b \rangle \), use the Pythagorean theorem extended into two dimensions. The formula \( \sqrt{a^2 + b^2} \) gives the length of the vector.

In the exercise, the resultant vector \( \langle \frac{1}{5}, \frac{14}{5} \rangle \) has a magnitude that is calculated as follows:

• Square each component: \( \left( \frac{1}{5} \right)^2 = \frac{1}{25} \) and \( \left( \frac{14}{5} \right)^2 = \frac{196}{25} \)
• Add these squares: \( \frac{1}{25} + \frac{196}{25} = \frac{197}{25} \)
• Take the square root: \( \sqrt{\frac{197}{25}} = \frac{\sqrt{197}}{5} \)

Understanding vector magnitude is crucial in fields like physics for determining quantities like speed and force.