Calculating the Length of a Vector from P to Q

The question asks for the length of the vector represented by a directed line segment starting at point P and ending at point Q. This directed line segment defines the vector $\vec{PQ}$.

Given the initial point P and the terminal point Q:

• Initial point P: $(2, -3, 4)$
• Terminal point Q: $(-2, 1, 1)$

To find the components of the vector $\vec{PQ}$, we subtract the coordinates of the initial point from the coordinates of the terminal point:

Let $\vec{PQ} = \langle x, y, z \rangle$.

The components are calculated as follows:

• $x = (\text{Q's x-coordinate}) - (\text{P's x-coordinate}) = -2 - 2 = -4$
• $y = (\text{Q's y-coordinate}) - (\text{P's y-coordinate}) = 1 - (-3) = 1 + 3 = 4$
• $z = (\text{Q's z-coordinate}) - (\text{P's z-coordinate}) = 1 - 4 = -3$

So, the vector $\vec{PQ}$ is $\langle -4, 4, -3 \rangle$ or $-4\mathbf{i} + 4\mathbf{j} - 3\mathbf{k}$.

Finding the Length (Magnitude) of the Vector

The length, or magnitude, of a vector $\mathbf{v} = \langle x, y, z \rangle$ in three dimensions is given by the formula:

$\left| \mathbf{v} \right| = \sqrt{x^2 + y^2 + z^2}$

Using the components of the vector $\vec{PQ} = \langle -4, 4, -3 \rangle$, we can calculate its length:

$\left| \vec{PQ} \right| = \sqrt{(-4)^2 + (4)^2 + (-3)^2}$

Now, we calculate the squares of the components:

• $(-4)^2 = 16$
• $(4)^2 = 16$
• $(-3)^2 = 9$

Substitute these values back into the formula:

$\left| \vec{PQ} \right| = \sqrt{16 + 16 + 9}$

Add the numbers inside the square root:

$16 + 16 + 9 = 32 + 9 = 41$

So, the length of the vector $\vec{PQ}$ is:

$\left| \vec{PQ} \right| = \sqrt{41}$

Comparing with Options

We compare our calculated length with the given options:

• Option 1: $\sqrt{41}$
• Option 2: $\sqrt{32}$
• Option 3: $\sqrt{43}$

Our calculated length $\sqrt{41}$ matches Option 1.

Therefore, the length of the vector represented by the directed line segment with initial point P(2, -3, 4) and terminal point Q(-2, 1, 1) is $\sqrt{41}$.