Problem 75 - Precalculus

Question

The projection of v onto \(\mathbf{u}\) is defined by proju \(\mathbf{v}=\left(\frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}|^{2}}\right) \mathbf{u}\) This vector is depicted below. Heuristically, this is the "shadow" of v on u. (GRAPH CAN'T COPY) a. Compute proju \(2 \mathbf{u}\) b. What is proju \(c \mathbf{u}\) for any \(c>0 ?\)

Step-by-Step Solution

Answer

a. \( \text{proj}_u (2 \mathbf{u}) = 2 \mathbf{u} \); b. For any \( c > 0 \), \( \text{proj}_u (c \mathbf{u}) = c \mathbf{u} \).

Step 1: Understand the Formula

The projection of a vector \( \mathbf{v} \) onto another vector \( \mathbf{u} \) is given by the formula: \[ \text{proj}_u \mathbf{v} = \left( \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}|^2} \right) \mathbf{u} \]. This formula gives the component of \( \mathbf{v} \) that lies in the direction of \( \mathbf{u} \).

Step 2: Apply the Formula for 2u

To find \( \text{proj}_u (2 \mathbf{u}) \), substitute \( 2 \mathbf{u} \) into the formula for \( \mathbf{v} \). \[ \text{proj}_u (2 \mathbf{u}) = \left( \frac{\mathbf{u} \cdot (2 \mathbf{u})}{|\mathbf{u}|^2} \right) \mathbf{u} = \left( \frac{2 (\mathbf{u} \cdot \mathbf{u})}{|\mathbf{u}|^2} \right) \mathbf{u} \]Since \( \mathbf{u} \cdot \mathbf{u} = |\mathbf{u}|^2 \), this simplifies to: \( \text{proj}_u (2 \mathbf{u}) = 2 \mathbf{u} \).

Step 3: Generalize for cu

For any constant \( c > 0 \), substitute \( c \mathbf{u} \) into the formula:\[ \text{proj}_u (c \mathbf{u}) = \left( \frac{\mathbf{u} \cdot (c \mathbf{u})}{|\mathbf{u}|^2} \right) \mathbf{u} = \left( \frac{c (\mathbf{u} \cdot \mathbf{u})}{|\mathbf{u}|^2} \right) \mathbf{u} \]Using the identity \( \mathbf{u} \cdot \mathbf{u} = |\mathbf{u}|^2 \), this simplifies to: \( \text{proj}_u (c \mathbf{u}) = c \mathbf{u} \).

Key Concepts

Dot Product

When we talk about the dot product, we are looking at a special operation that we can do with vectors. Imagine two vectors, \( \mathbf{a} \) and \( \mathbf{b} \). The dot product is a way to multiply these vectors, resulting in a scalar value. It's written as \( \mathbf{a} \cdot \mathbf{b} \). The formula to calculate it is: \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \) for vectors in a three-dimensional space.
- This product is handy because it helps determine if two vectors are perpendicular. If their dot product is zero, they are at a right angle!- It also shows how much of one vector goes in the direction of another.
For projection, the dot product \( \mathbf{u} \cdot \mathbf{v} \) tells us how much of \( \mathbf{v} \) is "in" \( \mathbf{u} \). In the context of vector projection, it's the top part of the fraction in the projection formula. This part is crucial when breaking down how much of vector \( \mathbf{v} \) goes along vector \( \mathbf{u} \). Understanding this component gives depth to how we turn vector operations into real-world applications, like finding the "shadow" of \( \mathbf{v} \) on \( \mathbf{u} \).

Magnitude of a Vector

The concept of the magnitude of a vector is like thinking of the length of a line. For a vector \( \mathbf{u} \), its magnitude, written as \(|\mathbf{u}|\), represents how long or large this vector is. The magnitude can be found using the formula: \(|\mathbf{u}| = \sqrt{u_1^2 + u_2^2 + u_3^2}\) for a three-dimensional vector.
When projecting vectors, knowing the magnitude helps normalize the vector, which means making it a unit vector with a length of one. This process helps simplify calculations when the direction is more relevant than the size.
- The magnitude appears in the denominator of the projection formula.- It plays a role in reducing distortion when comparing different vectors.For projection, using \(|\mathbf{u}|^2\) in the formula \( \text{proj}_u \mathbf{v} = \left( \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}|^2} \right) \mathbf{u} \), helps in figuring out the size of the component of \( \mathbf{v} \) that lies along \( \mathbf{u} \). It is like determining how much it "stretches" over \( \mathbf{u} \).

Scalar Multiplication

Scalar multiplication involves taking a vector and multiplying it by a real number. This changes the size of the vector without altering its direction. If you have a vector \( \mathbf{u} \) and a scalar \( c \), the scalar multiplication is \( c \mathbf{u} = (cu_1, cu_2, cu_3) \). So:- It stretches or shrinks the vector depending on whether the scalar \( c \) is greater or less than one.- If \( c \) is negative, it also reverses the vector’s direction.
In context, the formula for the projection of \( c \mathbf{u} \) helps show this principle. By simplifying \( \text{proj}_u (c \mathbf{u}) \) to \( c \mathbf{u} \), it illustrates how multiplying a vector by a scalar affects its projection.
When using projection formulas:- The vector’s direction stays the same, but the component's length along another vector might change.- The flexibility and transformation power of scalar multiplication make it a valuable tool in vector calculus.