Worksheet titled "Dot and Cross Product Worksheet" featuring four problems on vector operations, including finding angles between vectors, projections, vector equations of lines, and points of intersection between a plane and a line.

Let's solve each problem step-by-step.

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1) Find the angle between the vectors

The angle $\theta$ between two vectors $\vec{v}$ and $\vec{w}$ is given by:

$$
\cos \theta = \frac{\vec{v} \cdot \vec{w}}{\|\vec{v}\| \|\vec{w}\|}
$$

#### (a) $\vec{v} = (2, 3), \quad \vec{w} = (-3, 4)$

- Dot product:
$$
\vec{v} \cdot \vec{w} = 2(-3) + 3(4) = -6 + 12 = 6
$$

- Magnitudes:
$$
\|\vec{v}\| = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}
$$
$$
\|\vec{w}\| = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5
$$

- So:
$$
\cos \theta = \frac{6}{\sqrt{13} \cdot 5} = \frac{6}{5\sqrt{13}}
$$
$$
\theta = \cos^{-1}\left(\frac{6}{5\sqrt{13}}\right)
$$

Let’s compute numerically:
$$
\frac{6}{5\sqrt{13}} \approx \frac{6}{5 \cdot 3.6056} \approx \frac{6}{18.028} \approx 0.3328
$$
$$
\theta \approx \cos^{-1}(0.3328) \approx 70.6^\circ
$$

✔ Answer (a): $\theta \approx 70.6^\circ$

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#### (b) $\vec{v} = (1, -1, 3), \quad \vec{w} = (2, 2, -1)$

- Dot product:
$$
\vec{v} \cdot \vec{w} = 1(2) + (-1)(2) + 3(-1) = 2 - 2 - 3 = -3
$$

- Magnitudes:
$$
\|\vec{v}\| = \sqrt{1^2 + (-1)^2 + 3^2} = \sqrt{1 + 1 + 9} = \sqrt{11}
$$
$$
\|\vec{w}\| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3
$$

- So:
$$
\cos \theta = \frac{-3}{\sqrt{11} \cdot 3} = \frac{-1}{\sqrt{11}}
$$
$$
\theta = \cos^{-1}\left(\frac{-1}{\sqrt{11}}\right)
$$

Numerically:
$$
\frac{-1}{\sqrt{11}} \approx \frac{-1}{3.3166} \approx -0.3015
$$
$$
\theta \approx \cos^{-1}(-0.3015) \approx 107.5^\circ
$$

✔ Answer (b): $\theta \approx 107.5^\circ$

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2) Find the projection of $\vec{v}$ onto $\vec{w}$

The vector projection of $\vec{v}$ onto $\vec{w}$ is:

$$
\text{proj}_{\vec{w}} \vec{v} = \frac{\vec{v} \cdot \vec{w}}{\|\vec{w}\|^2} \vec{w}
$$

#### (a) $\vec{v} = (2, 3), \vec{w} = (-3, 4)$

We already computed:
- $\vec{v} \cdot \vec{w} = 6$
- $\|\vec{w}\|^2 = (-3)^2 + 4^2 = 9 + 16 = 25$

So:
$$
\text{proj}_{\vec{w}} \vec{v} = \frac{6}{25} (-3, 4) = \left(-\frac{18}{25}, \frac{24}{25}\right)
$$

✔ Answer (a): $\left(-\frac{18}{25}, \frac{24}{25}\right)$

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#### (b) $\vec{v} = (1, -1, 3), \vec{w} = (2, 2, -1)$

We already have:
- $\vec{v} \cdot \vec{w} = -3$
- $\|\vec{w}\|^2 = 2^2 + 2^2 + (-1)^2 = 4 + 4 + 1 = 9$

So:
$$
\text{proj}_{\vec{w}} \vec{v} = \frac{-3}{9} (2, 2, -1) = -\frac{1}{3} (2, 2, -1) = \left(-\frac{2}{3}, -\frac{2}{3}, \frac{1}{3}\right)
$$

✔ Answer (b): $\left(-\frac{2}{3}, -\frac{2}{3}, \frac{1}{3}\right)$

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3) Find a vector equation for the line in $\mathbb{R}^3$ containing the points $(3, -5, 4)$ and $(2, -1, 2)$

A vector equation of a line is:
$$
\vec{r}(t) = \vec{r}_0 + t \vec{d}
$$
where $\vec{r}_0$ is a point on the line and $\vec{d}$ is the direction vector.

Let:
- $\vec{r}_0 = (3, -5, 4)$
- Direction vector: $\vec{d} = (2 - 3, -1 - (-5), 2 - 4) = (-1, 4, -2)$

So:
$$
\vec{r}(t) = (3, -5, 4) + t(-1, 4, -2)
$$

✔ Answer: $\vec{r}(t) = (3, -5, 4) + t(-1, 4, -2)$

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4) Find all points of intersection, if any, of the plane $2x - 3y + 4z + 15 = 0$ and the line $\vec{p}(t) = (2, -3, 0) + t(4, 2, 3)$

We substitute the parametric equations of the line into the plane equation.

From the line:
$$
x = 2 + 4t,\quad y = -3 + 2t,\quad z = 0 + 3t = 3t
$$

Substitute into the plane:
$$
2x - 3y + 4z + 15 = 0
$$
$$
2(2 + 4t) - 3(-3 + 2t) + 4(3t) + 15 = 0
$$

Compute:
$$
4 + 8t + 9 - 6t + 12t + 15 = 0
$$
$$
(4 + 9 + 15) + (8t - 6t + 12t) = 0
$$
$$
28 + 14t = 0
\Rightarrow 14t = -28 \Rightarrow t = -2
$$

Now plug $t = -2$ back into the line to get the point:
$$
x = 2 + 4(-2) = 2 - 8 = -6 \\
y = -3 + 2(-2) = -3 - 4 = -7 \\
z = 3(-2) = -6
$$

So the point of intersection is $(-6, -7, -6)$

✔ Answer: The line intersects the plane at the point $(-6, -7, -6)$

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✔ Final Answers Summary:

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1) Angle between vectors:

(a) $\theta \approx 70.6^\circ$
(b) $\theta \approx 107.5^\circ$

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2) Projection of $\vec{v}$ onto $\vec{w}$:

(a) $\left(-\frac{18}{25}, \frac{24}{25}\right)$
(b) $\left(-\frac{2}{3}, -\frac{2}{3}, \frac{1}{3}\right)$

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3) Vector equation of the line:
$$
\vec{r}(t) = (3, -5, 4) + t(-1, 4, -2)
$$

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4) Point of intersection:
$$
(-6, -7, -6)
$$

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