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Chapter 12: Problem 69

Find vectors parallel to \(\mathbf{v}\) of the given length. $$\mathbf{v}=\overrightarrow{P Q} \text { with } P(3,4,0) \text { and } Q(2,3,1) ; \text { length }=3$$

Short Answer

Expert verified
Question: Find a vector parallel to the vector connecting points P(3, 4, 0) and Q(2, 3, 1) with a length of 3. Answer: The parallel vector with a length of 3 is given by: (-鈭�3, -鈭�3, 鈭�3).

Step by step solution

01

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

We can find the components of the vector \(\mathbf{v}\) by subtracting the coordinates of point P from the coordinates of point Q: $$\mathbf{v} = \overrightarrow{P Q} = Q - P = (2-3, 3-4, 1-0)$$ Then we find the components of the vector as:$$\mathbf{v} = (-1, -1, 1)$$
02

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

Next, we need to compute the magnitude (length) of vector \(\mathbf{v}\). The magnitude of a vector can be calculated using the formula: $$|\mathbf{v}| = \sqrt{(-1)^2 + (-1)^2 + 1^2}$$. So the magnitude of \(\mathbf{v}\) is:$$|\mathbf{v}| = \sqrt{3}$$
03

Find the unit vector parallel to \(\mathbf{v}\)

To find the unit vector parallel to \(\mathbf{v}\), we divide each component of \(\mathbf{v}\) by its magnitude. This gives us the unit vector \(\mathbf{u}\):$$\mathbf{u} = \frac{\mathbf{v}}{|\mathbf{v}|} = \frac{1}{\sqrt{3}}(-1, -1, 1) = \left(-\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$$
04

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

Since we want to find a vector parallel to \(\mathbf{v}\) with a length of 3, we can multiply the unit vector \(\mathbf{u}\) by the desired length. This will give us a new vector \(\mathbf{w}\) with the same direction as \(\mathbf{v}\) but with a length of 3: $$\mathbf{w} = 3\mathbf{u} = 3\left(-\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) = (-\sqrt{3}, -\sqrt{3}, \sqrt{3})$$ So, the parallel vector to \(\mathbf{v}\) with a length of 3 is given by: $$\mathbf{w} = (-\sqrt{3}, -\sqrt{3}, \sqrt{3})$$.

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