Problem 12 - Calculus: Early Transcendentals

Question

Find the scalar and vector projections of \langle 1,2,3\rangle onto \(\langle 1,2,0\rangle .\)

Step-by-Step Solution

Answer

The scalar projection is \( \sqrt{5} \), and the vector projection is \( \langle 1, 2, 0 \rangle \).

Step 1: Calculate the Dot Product

To find the projections, we first need the dot product of both vectors. Let \( \mathbf{v} = \langle 1, 2, 3 \rangle \) and \( \mathbf{u} = \langle 1, 2, 0 \rangle \).The dot product \( \mathbf{u} \cdot \mathbf{v} \) is computed as:\[ \mathbf{u} \cdot \mathbf{v} = (1)(1) + (2)(2) + (0)(3) = 1 + 4 + 0 = 5 \]

Step 2: Calculate the Magnitude of Vector u

The magnitude of \( \mathbf{u} \) is needed to find the projections. It is calculated using the formula:\[ \| \mathbf{u} \| = \sqrt{1^2 + 2^2 + 0^2} = \sqrt{1 + 4 + 0} = \sqrt{5} \]

Step 3: Compute the Scalar Projection

The scalar projection of \( \mathbf{v} \) onto \( \mathbf{u} \) is given by:\[ \mathrm{scalar\, projection} = \frac{\mathbf{u} \cdot \mathbf{v}}{\| \mathbf{u} \|} = \frac{5}{\sqrt{5}} = \sqrt{5} \]

Step 4: Find the Vector Projection

The vector projection of \( \mathbf{v} \) onto \( \mathbf{u} \) uses the formula:\[ \mathrm{vector\,projection} = \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u} \]First, find \( \mathbf{u} \cdot \mathbf{u} \):\[ \mathbf{u} \cdot \mathbf{u} = (1)^2 + (2)^2 + (0)^2 = 5 \]Thus, the vector projection is:\[ \mathrm{vector\,projection} = \frac{5}{5} \cdot \langle 1, 2, 0 \rangle = 1 \cdot \langle 1, 2, 0 \rangle = \langle 1, 2, 0 \rangle \]

Key Concepts

Dot Product

The dot product, also known as the scalar product, is a way of multiplying two vectors together to produce a single scalar. It incorporates both the magnitudes of the vectors and the cosine of the angle between them. To find the dot product of two vectors, say \( \mathbf{u} = \langle a, b, c \rangle \) and \( \mathbf{v} = \langle x, y, z \rangle \), use the formula:\[ \mathbf{u} \cdot \mathbf{v} = ax + by + cz \]

• Example: For vectors \( \langle 1, 2, 3 \rangle \) and \( \langle 1, 2, 0 \rangle \), the dot product is calculated as:
• \( (1)(1) + (2)(2) + (3)(0) = 1 + 4 + 0 = 5 \)

Understanding the dot product is crucial because it's the stepping stone to finding vector projections.

Magnitude of a Vector

The magnitude of a vector, often called the length or norm, indicates how long the vector is when visualized in space. The magnitude is found using the Pythagorean theorem, which extends naturally from two dimensions to any number of dimensions. For a vector \( \mathbf{u} = \langle a, b, c \rangle \), the magnitude \( \| \mathbf{u} \| \) is calculated as:\[ \| \mathbf{u} \| = \sqrt{a^2 + b^2 + c^2} \]

• Example: For vector \( \langle 1, 2, 0 \rangle \), the magnitude is \( \sqrt{1^2 + 2^2 + 0^2} = \sqrt{1 + 4 + 0} = \sqrt{5} \).

This measure is vital for normalizing vectors and scaling projections.

Scalar Projection

The scalar projection of one vector onto another gives the length of the shadow or projection of the first vector on the line defined by the second vector. It's given by dividing the dot product of the vectors by the magnitude of the vector onto which you are projecting:\[ \text{Scalar Projection} = \frac{\mathbf{u} \cdot \mathbf{v}}{\| \mathbf{u} \|} \]

• Example: For vectors \( \langle 1, 2, 3 \rangle \) and \( \langle 1, 2, 0 \rangle \), this yields \( \frac{5}{\sqrt{5}} = \sqrt{5} \).

This result tells us how much of the vector \( \mathbf{v} \) actually "lies along" the direction of \( \mathbf{u} \).

Vector Projection Formula

The vector projection is the vector equivalent of the scalar projection, indicating not just how far one vector projects onto another, but also in which direction. It is determined using:\[ \text{Vector Projection} = \left( \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{u} \cdot \mathbf{u}} \right) \mathbf{u} \]

• First, compute \( \mathbf{u} \cdot \mathbf{u} = 5 \).
• Example: For \( \langle 1, 2, 3 \rangle \) onto \( \langle 1, 2, 0 \rangle \), we find this projection to be \( 1 \times \langle 1, 2, 0 \rangle = \langle 1, 2, 0 \rangle \).

This method highlights how much of the vector \( \mathbf{v} \) not just "lies along" \( \mathbf{u} \), but also retains the same direction.