For the following problem, find a particular equation of the plane described.

Perpendicular to $\vec{n}=-\vec{i}+3 \vec{j}-2 \vec{k}$, containing the point (4,7,5)

In the previous problem, the vector in the x-direction is 4 units long. Vectors $\vec{i}$ and $\vec{j}$ are unit vectors in the x- and y-directions, respectively. A unit vector is a vector that is 1 unit long. So the x-component of $\vec{p}$ is the vector $4 \vec{i}$, a product of a scalar and a vector. A scalar is a quantity that has only magnitude, not direction. What is the y-component of $\vec{p}$ in the figure shown before?

For the following problem, find the direction cosines and direction angles for the position vector to the given point.

(5,7,-1)

For $\vec{a}$ and $\vec{b}$ in the previous part, find the position vector to the point 70% of the way from the head of $\vec{a}$ to the head of $\vec{b}$.

If $\vec{a}$ and $\vec{b}$ in the previous part are position vectors, find the displacement vector from the head of $\vec{a}$ to the head of $\vec{b}$.

Find a unit vector in the direction of $\vec{a}$ in the previousp art.

Refer to vectors $\vec{a}$ and $\vec{b}$ in the figure mentioned.

Find $\vec{a}-\vec{b}$ numerically from the coordinates of $\vec{a}$ and $\vec{b}$, and show that the answer agrees with your drawings in parts $a$ and $b$.

Find $|\vec{a}|$ for $\vec{a}$ in the previous part.

If $\vec{a}=6 \vec{i}-5 \vec{j}+2 \vec{k}$ and $\vec{b}=3 \vec{i}+4 \vec{j}-7 \vec{k}$, find $3 \vec{a}-2 \vec{b}$.

Refer to vectors $\vec{a}$ and $\vec{b}$ in the figure mentioned.

Draw a displacement vector from the head of $\vec{b}$ to the head of $\vec{a}$. Explain why this vector is equivalent to $\vec{a}-\vec{b}$ from part a. Explain verbally how you can tell whether $\vec{a}-\vec{b}$ goes from $\vec{a}$ to $\vec{b}$ or from $\vec{b}$ to $\vec{a}$.

Draw $\vec{v}=5 \vec{i}+9 \vec{j}+4 \vec{k}$ as a position vector. Show the "box" that makes it look three dimensional. Indicate the unit vectors $\vec{i}, \vec{j}$, and $\vec{k}$ on the drawing.

Update your journal with what you have learned since the last entry. Include the definition, computational technique, and uses of the dot product, along with its two other names.

Refer to vectors $\vec{a}$ and $\vec{b}$ in the figure mentioned.

On another copy of the figure, draw $-\vec{b}$, the opposite of vector $\vec{b}$, as a position vector (starting at the origin). Then translate $-\vec{b}$ so that its tail is at the head of $\vec{a}$. Using the definition of vector addition, draw vector $\vec{a}+(-\vec{b})$. Explain why $\vec{a}-\vec{b}$ is equivalent to $\vec{a}+(-\vec{b})$.

For the following problem, use the definition of dot product to find $\vec{a} \cdot \vec{b}$, where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$ when they are placed tail-to-tail.

$|\vec{a}|=29,|\vec{b}|=50$, and $\theta=127^{\circ}$