Question

Find vectors parallel to \(\mathbf{v}\) of the given length. $$\mathbf{v}=\overrightarrow{P Q} \text { with } P(1,0,1) \text { and } Q(2,-1,1) ; \text { length }=3$$

Short answer

Find the vector parallel to the vector $\mathbf{v}=\overrightarrow{PQ}$, where P(1, 0, 1) and Q(2, -1, 1), with a length of 3. Solution: The vector parallel to $\mathbf{v}$ with a length of 3 is $\mathbf{w} = \langle 3\sqrt{2}/2, -3\sqrt{2}/2, 0 \rangle$.

Step-by-step solution

Find vector \(\mathbf{v}=\overrightarrow{PQ}\)

Given the coordinates of points P(1, 0, 1) and Q(2, -1, 1), we can find the vector \(\mathbf{v}\) as follows: $$\overrightarrow{PQ} = Q - P = (2 - 1, -1 - 0, 1 - 1) = (1, -1, 0)$$ So, the vector \(\mathbf{v} = \langle 1, -1, 0 \rangle\).

Find the magnitude of \(\mathbf{v}\)

The magnitude of a vector is calculated using the formula: $$||\mathbf{v}|| = \sqrt{x^2 + y^2 + z^2}$$ Now, we can find the magnitude of the vector \(\mathbf{v}\), which is: $$||\mathbf{v}|| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{1 + 1} = \sqrt{2}$$

Find the unit vector of \(\mathbf{v}\)

To find the unit vector of \(\mathbf{v}\), divide the vector by its magnitude: $$\mathbf{u}_v = \frac{\mathbf{v}}{||\mathbf{v}||} = \frac{\langle 1, -1, 0 \rangle}{\sqrt{2}} = \left\langle \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0 \right\rangle$$

Find the vector parallel to \(\mathbf{v}\) with the given length

Given the length of the desired vector is 3, we can find the vector parallel to \(\mathbf{v}\) by multiplying the unit vector by the given length: $$\mathbf{w} = 3\mathbf{u}_v = 3\left\langle \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0 \right\rangle = \left\langle 3\frac{1}{\sqrt{2}}, 3(-\frac{1}{\sqrt{2}}), 3(0) \right\rangle = \langle 3\sqrt{2}/2, -3\sqrt{2}/2, 0 \rangle$$ Thus, the vector parallel to \(\mathbf{v}\) with a length of 3 is \(\mathbf{w} = \langle 3\sqrt{2}/2, -3\sqrt{2}/2, 0 \rangle\).