Find the scalar and vector projections of $\mathbf{b}$ onto $\mathbf{a}$: $\mathbf{a} = \langle -3, 4 \rangle, \quad \mathbf{b} = \langle 2, 9 \rangle$.

Neetesh Kumar | December 18, 2024

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This is the solution to Math 1C
Assignment: 12.3 Question Number 15
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Step-by-step solution:

Scalar Projection of $\mathbf{b}$ onto $\mathbf{a}$:

The formula for the scalar projection of $\mathbf{b}$ onto $\mathbf{a}$ is:
$\text{scalar projection} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|}.$

Step 1: Compute the dot product $\mathbf{a} \cdot \mathbf{b}$:

The dot product of two vectors $\mathbf{a} = \langle a_1, a_2 \rangle$ and $\mathbf{b} = \langle b_1, b_2 \rangle$ is:
$\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2.$

Substitute $\mathbf{a} = \langle -3, 4 \rangle$ and $\mathbf{b} = \langle 2, 9 \rangle$:
$\mathbf{a} \cdot \mathbf{b} = (-3)(2) + (4)(9).$

Simplify:
$\mathbf{a} \cdot \mathbf{b} = -6 + 36 = 30.$

Step 2: Compute the magnitude of $\mathbf{a}$:

The magnitude of a vector $\mathbf{a} = \langle a_1, a_2 \rangle$ is:
$|\mathbf{a}| = \sqrt{a_1^2 + a_2^2}.$

Substitute $\mathbf{a} = \langle -3, 4 \rangle$:
$|\mathbf{a}| = \sqrt{(-3)^2 + 4^2}.$

Simplify:
$|\mathbf{a}| = \sqrt{9 + 16} = \sqrt{25} = 5.$

Step 3: Find the scalar projection:

Substitute $\mathbf{a} \cdot \mathbf{b} = 30$ and $|\mathbf{a}| = 5$ into the formula:
$\text{scalar projection} = \frac{30}{5}.$

Simplify:
$\text{scalar projection} = 6.$

Vector Projection of $\mathbf{b}$ onto $\mathbf{a}$:

The formula for the vector projection of $\mathbf{b}$ onto $\mathbf{a}$ is:
$\text{vector projection} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|^2} \mathbf{a}.$

Step 1: Compute $|\mathbf{a}|^2$:

From earlier, $|\mathbf{a}| = 5$, so:
$|\mathbf{a}|^2 = 5^2 = 25.$

Step 2: Find the vector projection:

Substitute $\mathbf{a} \cdot \mathbf{b} = 30$, $|\mathbf{a}|^2 = 25$, and $\mathbf{a} = \langle -3, 4 \rangle$:
$\text{vector projection} = \frac{30}{25} \mathbf{a}.$

Simplify the scalar factor:
$\text{vector projection} = \frac{6}{5} \mathbf{a}.$

Step 3: Multiply by $\mathbf{a}$:

$\text{vector projection} = \frac{6}{5} \langle -3, 4 \rangle.$

Distribute the scalar:
$\text{vector projection} = \langle \frac{6}{5}(-3), \frac{6}{5}(4) \rangle.$

Simplify:
$\text{vector projection} = \langle -\frac{18}{5}= -3.6, \frac{24}{5} = 4.8 \rangle.$

Final Answer:

1. Scalar projection of $\mathbf{b}$ onto $\mathbf{a}$:

$\boxed{6}$

2. Vector projection of $\mathbf{b}$ onto $\mathbf{a}$:

$\langle -3.6, 4.8 \rangle$

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