Problem 42 - Algebra and Trigonometry

Question

Let $$\mathbf{u}=-\mathbf{i}+\mathbf{j}, \quad \mathbf{v}=3 \mathbf{i}-2 \mathbf{j}, \quad \text { and } \quad \mathbf{w}=-5 \mathbf{j}$$ Find each specified scalar or vector. $$ \operatorname{proj}_{\mathbf{u}}(\mathbf{v}-\mathbf{w}) $$

Step-by-Step Solution

Answer

The projection of vector \(\mathbf{v}-\mathbf{w}\) onto vector \(\mathbf{u}\) is \(\mathbf{0}\), a zero vector.

Step 1: Subtract Vectors

Subtract the vectors \(\mathbf{v}\) and \(\mathbf{w}\) to get a new vector. This can be done by subtracting the respective components of the vectors: \((3\mathbf{i}-2\mathbf{j})-(-5\mathbf{j}) = 3\mathbf{i} + 3 \mathbf{j}\). So, \(\mathbf{v}-\mathbf{w} = 3\mathbf{i} + 3 \mathbf{j}\).

Step 2: Compute Dot Product of U and (V-W)

Compute the dot product of vectors \(\mathbf{u}\) and \(\mathbf{v}-\mathbf{w}\). This is given by \(\mathbf{u} \cdot (\mathbf{v}-\mathbf{w}) = (-\mathbf{i}+\mathbf{j}) \cdot (3\mathbf{i}+3\mathbf{j}) = -3 + 3 = 0\).

Step 3: Compute Dot Product of U with Itself

Compute the dot product of vector \(\mathbf{u}\) with itself. This is given by \(\mathbf{u} \cdot \mathbf{u} = (-\mathbf{i}+\mathbf{j}) \cdot (-\mathbf{i}+\mathbf{j}) = 1 + 1 = 2\).

Step 4: Calculate The Projection of (V-W) onto U

Use the formula of projection to calculate the projection of \(\mathbf{v}-\mathbf{w}\) onto \(\mathbf{u}\). This is given by: \(\operatorname{proj}_{\mathbf{u}}(\mathbf{v}-\mathbf{w}) = \frac{{\mathbf{u} \cdot (\mathbf{v}-\mathbf{w})}}{{\mathbf{u} \cdot \mathbf{u}}} \mathbf{u} = \frac{0}{2} (-\mathbf{i}+\mathbf{j}) = \mathbf{0}\). Hence, the projection is a zero vector.

Key Concepts

Vector Subtraction

Vector subtraction is an essential concept in vector algebra, allowing us to find the difference between two vectors. The process involves subtracting the corresponding components of each vector. For instance, if we have vectors \( \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} \) and \( \mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} \), the subtraction \( \mathbf{a} - \mathbf{b} \) is performed component-wise:

• Subtract the \( \mathbf{i} \)-components: \( a_1 - b_1 \)
• Subtract the \( \mathbf{j} \)-components: \( a_2 - b_2 \)

This results in a new vector with components \( (a_1 - b_1)\mathbf{i} + (a_2 - b_2)\mathbf{j} \). In practical terms, this operation helps determine how much one vector changes in relation to another.
In the exercise, vector \( \mathbf{v} \) is \( 3\mathbf{i} - 2\mathbf{j} \) and vector \( \mathbf{w} \) is \( -5\mathbf{j} \). By using vector subtraction, we find that \( \mathbf{v} - \mathbf{w} = 3\mathbf{i} + (3)\mathbf{j} \).

Dot Product

The dot product, also known as the scalar product, is a crucial operation in vector mathematics. It combines two vectors to produce a scalar value, reflecting the extent to which the vectors align. To compute the dot product of two vectors \( \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} \) and \( \mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} \), use the formula:\[\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2\]The result is a single number summarizing the interaction between the vectors.
In the exercise, the dot product \( \mathbf{u} \cdot (\mathbf{v} - \mathbf{w}) \) yields zero. This implies that the vector \( \mathbf{u} \) and \( \mathbf{v} - \mathbf{w} \) are perpendicular. Such insights help identify orthogonal dimensions and interactions in vector spaces.

Scalar Projection

Scalar projection is another fascinating concept that reveals the magnitude of one vector's component along the direction of another vector. The scalar projection of vector \( \mathbf{a} \) onto vector \( \mathbf{b} \) is calculated using:\[\text{proj}_{\mathbf{b}}\mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{b}\|}\]However, if only a directional component is needed, the formula simplifies to:\[\frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}}\]This provides a scalar that, when multiplied by the unit vector of \( \mathbf{b} \), yields the actual vector projection.
In the given problem, the scalar projection \( \frac{\mathbf{u} \cdot (\mathbf{v}-\mathbf{w})}{\mathbf{u} \cdot \mathbf{u}} \) turns out to be zero because the dot product is zero, showing no overlap in the direction of \( \mathbf{u} \).

Zero Vector

A zero vector, denoted as \( \mathbf{0} \), has all of its components equal to zero. It is crucial in vector algebra because it represents no movement or no displacement from the origin. For any vector \( \mathbf{a} \), adding or subtracting the zero vector leaves \( \mathbf{a} \) unchanged:

• \( \mathbf{a} + \mathbf{0} = \mathbf{a} \)
• \( \mathbf{a} - \mathbf{0} = \mathbf{a} \)

In projections, a zero vector result indicates no component of one vector along the direction of another. This is often seen when vectors are perpendicular.
In the exercise's final step, the vector projection was calculated as a zero vector, meaning there is no component of \( \mathbf{v}-\mathbf{w} \) along \( \mathbf{u} \). This result emphasizes their perpendicular relationship and absence of interaction in that direction.