Problem 32 - APEX Calculus

Question

Vectors \(\vec{u}\) and \(\vec{v}\) are given. Write \(\vec{u}\) as the sum of two vectors, one of which is parallel to \(\vec{v}\) and one of which is perpendicular to \(\vec{v}\). Note: these are the same pairs of vectors as found in Exercises 21-26. \(\vec{u}=\langle 3,-1,2\rangle, \vec{v}=\langle 2,2,1\rangle\)

Step-by-Step Solution

Answer

\( \vec{u}_{\text{parallel}} = \langle \frac{4}{3}, \frac{4}{3}, \frac{2}{3} \rangle \); \( \vec{u}_{\text{perpendicular}} = \langle \frac{5}{3}, -\frac{7}{3}, \frac{4}{3} \rangle \).

Step 1: Compute the Dot Product

To find the component of \( \vec{u} \) parallel to \( \vec{v} \), we first need the dot product \( \vec{u} \cdot \vec{v} \). Calculate it using the formula: \[ \vec{u} \cdot \vec{v} = u_1 v_1 + u_2 v_2 + u_3 v_3 \]For \( \vec{u} = \langle 3, -1, 2 \rangle \) and \( \vec{v} = \langle 2, 2, 1 \rangle \):\[ \vec{u} \cdot \vec{v} = 3 \times 2 + (-1) \times 2 + 2 \times 1 = 6 - 2 + 2 = 6 \]

Step 2: Find the Magnitude Squared of \(\vec{v}\)

Calculate the square of the magnitude of \( \vec{v} \) using the formula:\[ \| \vec{v} \|^2 = v_1^2 + v_2^2 + v_3^2 \]For \( \vec{v} = \langle 2, 2, 1 \rangle \):\[ \| \vec{v} \|^2 = 2^2 + 2^2 + 1^2 = 4 + 4 + 1 = 9 \]

Step 3: Find the Parallel Component

The component of \( \vec{u} \) parallel to \( \vec{v} \) is calculated by:\[ \vec{u}_{\text{parallel}} = \frac{\vec{u} \cdot \vec{v}}{\| \vec{v} \|^2} \vec{v} \]Substitute \( \vec{u} \cdot \vec{v} = 6 \) and \( \| \vec{v} \|^2 = 9 \):\[ \vec{u}_{\text{parallel}} = \frac{6}{9} \vec{v} = \frac{2}{3} \langle 2, 2, 1 \rangle = \langle \frac{4}{3}, \frac{4}{3}, \frac{2}{3} \rangle \]

Step 4: Find the Perpendicular Component

The component of \( \vec{u} \) perpendicular to \( \vec{v} \) is given by:\[ \vec{u}_{\text{perpendicular}} = \vec{u} - \vec{u}_{\text{parallel}} \]Substitute \( \vec{u} = \langle 3, -1, 2 \rangle \) and \( \vec{u}_{\text{parallel}} = \langle \frac{4}{3}, \frac{4}{3}, \frac{2}{3} \rangle \):\[ \vec{u}_{\text{perpendicular}} = \langle 3, -1, 2 \rangle - \langle \frac{4}{3}, \frac{4}{3}, \frac{2}{3} \rangle = \langle 3 - \frac{4}{3}, -1 - \frac{4}{3}, 2 - \frac{2}{3} \rangle \]Simplify the components:\[ \vec{u}_{\text{perpendicular}} = \langle \frac{5}{3}, -\frac{7}{3}, \frac{4}{3} \rangle \]

Key Concepts

Dot Product

The dot product is a way to multiply two vectors, resulting in a scalar (a single number). To find the dot product between vectors \( \vec{u} = \langle u_1, u_2, u_3 \rangle \) and \( \vec{v} = \langle v_1, v_2, v_3 \rangle \), you'll use the formula: \[\vec{u} \cdot \vec{v} = u_1 v_1 + u_2 v_2 + u_3 v_3\] This calculation helps determine how much one vector goes in the direction of another.

• If the dot product is positive, the vectors point roughly in the same direction.
• If it's zero, there's no directional influence; they're perpendicular.
• If negative, the vectors point in opposite directions.

Here, we find the dot product of \( \vec{u} = \langle 3, -1, 2 \rangle \) and \( \vec{v} = \langle 2, 2, 1 \rangle \) to be 6, implying a certain directional alignment.

Parallel Vectors

Two vectors are parallel if one is a scalar multiple of the other. To express this mathematically, for vectors \( \vec{u} \) and \( \vec{v} \), they are parallel if there is a number \( k \) such that \( \vec{u} = k\vec{v} \). Parallel vectors point in exactly the same or opposite directions. When decomposing a vector into parallel components, we utilize the formula: \[\vec{u}_{\text{parallel}} = \frac{\vec{u} \cdot \vec{v}}{\| \vec{v} \|^2} \vec{v}\] This formula finds the component of \( \vec{u} \) that runs in the same direction as \( \vec{v} \). In our case, substituting in the values gives us \( \vec{u}_{\text{parallel}} = \langle \frac{4}{3}, \frac{4}{3}, \frac{2}{3} \rangle \), confirming it lies along the direction defined by \( \vec{v} \).

Perpendicular Vectors

Perpendicular vectors are at a 90-degree angle to each other, meaning they have no component in each other's direction. If the dot product of two vectors is zero, they are perpendicular. The component of vector \( \vec{u} \) that is perpendicular to vector \( \vec{v} \) can be found using the relation: \[\vec{u}_{\text{perpendicular}} = \vec{u} - \vec{u}_{\text{parallel}}\] This calculation gives us the part of \( \vec{u} \) that does not align with \( \vec{v} \). By isolating this component, we get the vector \( \vec{u}_{\text{perpendicular}} = \langle \frac{5}{3}, -\frac{7}{3}, \frac{4}{3} \rangle \), indicating it does not follow the path of \( \vec{v} \) and stands at right angles with the vector's direction.

Magnitude of a Vector

The magnitude of a vector is essentially its length. Compute it using the Pythagorean theorem analogy, which in vector terms is: \[\| \vec{v} \| = \sqrt{v_1^2 + v_2^2 + v_3^2}\] This gives a real sense of how long, or how much 'energy' a vector contains. For our vector \( \vec{v} = \langle 2, 2, 1 \rangle \), first find the magnitude squared, \( \| \vec{v} \|^2 = 9 \).

• Magnitude provides a single, positive number indicating size.
• It can help in normalizing vectors or finding unit vectors.

Thus, knowing the magnitude is key to identifying how much vector direction needs scaling to achieve a unit vector or determine its impact in specific contexts.