The Definition of a Vector

Approach:

The coordinates of the directed segment $\vec{AB}$ are the coordinates of the endpoint minus those of the starting point; whether two vectors are equal depends only on their direction and length, and not on where their endpoints happen to be.

Steps:

1. $\vec{AB} = (4 - 1, 6 - 2) = (3, 4)$ and $\vec{CD} = (5 - 2, 7 - 3) = (3, 4)$
2. The two vectors have exactly the same coordinates, the same direction and the same length, so $\vec{AB} = \vec{CD}$, that is, the two vectors are equal
3. $\vec{BA} = (1 - 4, 2 - 6) = (-3, -4) = -\vec{AB}$, so $\vec{BA}$ and $\vec{AB}$ are opposite vectors

Answer: $\vec{AB} = \vec{CD} = (3, 4)$, and the two vectors are equal; $\vec{BA} = (-3, -4) = -\vec{AB}$, and it is the opposite vector of $\vec{AB}$.

Approach:

The magnitude is obtained by taking the square root of the sum of the squares of the coordinates; the unit vector in the same direction is the original vector divided by its own magnitude; the zero vector has length 0 and an arbitrary direction, so its coordinates are all zero.

Steps:

1. $|\vec{c}| = \sqrt{3^2 + 4^2 + 0^2} = \sqrt{25} = 5$
2. The unit vector in the same direction is $\vec{e} = \frac{\vec{c}}{|\vec{c}|} = (\frac{3}{5}, \frac{4}{5}, 0)$, and $|\vec{e}| = \sqrt{\frac{9}{25} + \frac{16}{25}} = 1$
3. The zero vector is $\vec{0} = (0, 0, 0)$, with magnitude 0 and an arbitrary direction
4. $\vec{c} + (-\vec{c}) = (3 - 3, 4 - 4, 0 - 0) = (0, 0, 0) = \vec{0}$, verifying that opposite vectors add up to the zero vector

Answer: $|\vec{c}| = 5$, the unit vector in the same direction is $(\frac{3}{5}, \frac{4}{5}, 0)$, the zero vector is $(0, 0, 0)$, and $\vec{c} + (-\vec{c}) = \vec{0}$.

Approach:

In coordinate representation, the addition, the subtraction and the scalar multiplication of vectors all work component by component; two vectors are collinear if and only if their corresponding components are proportional, that is, one of them is a scalar multiple of the other.

Steps:

1. $\vec{a} + \vec{b} = (2 + (-1), -1 + 2, 3 + 0) = (1, 1, 3)$
2. $\vec{a} - \vec{b} = (2 - (-1), -1 - 2, 3 - 0) = (3, -3, 3)$
3. $2\vec{a} + 3\vec{b} = (4, -2, 6) + (-3, 6, 0) = (1, 4, 6)$
4. If $\vec{a}$ and $\vec{b}$ were collinear there would be a real number $k$ with $\vec{b} = k\vec{a}$; from the first component $-1 = 2k$ we get $k = -\frac{1}{2}$, and from the second component $2 = -k$ we get $k = -2$ — these contradict each other, so they are not collinear

Answer: $\vec{a} + \vec{b} = (1, 1, 3)$, $\vec{a} - \vec{b} = (3, -3, 3)$, $2\vec{a} + 3\vec{b} = (1, 4, 6)$; and $\vec{a}$ and $\vec{b}$ are not collinear.

Approach:

The directed segment $\vec{AB}$ joining two points on the line is a direction vector of that line; dividing the direction vector by its own magnitude gives the unit vector in the same direction.

Steps:

1. A direction vector can be taken as $\vec{AB} = (2 - 1, 3 - 1, 2 - 1) = (1, 2, 1)$
2. $|\vec{AB}| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{6}$
3. The unit vector in the same direction is $\vec{e} = \frac{\vec{AB}}{|\vec{AB}|} = (\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}})$
4. Check: $|\vec{e}| = \sqrt{\frac{1}{6} + \frac{4}{6} + \frac{1}{6}} = 1$, and the direction is the same as that of $\vec{AB}$

Answer: a direction vector can be taken as $\vec{AB} = (1, 2, 1)$, and the unit vector in the same direction is $(\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}})$.