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Compute the dot product of the vectors \(\mathbf{u}\) and \(\mathbf{v},\) and find the angle between the vectors. $$\mathbf{u}=\langle-10,0,4\rangle \text { and } \mathbf{v}=\langle 1,2,3\rangle$$

Short Answer

Expert verified
Answer: The angle between vectors \(\mathbf{u}\) and \(\mathbf{v}\) is approximately \(94.5^\circ\).

Step by step solution

01

Find the dot product of \(\mathbf{u}\) and \(\mathbf{v}\)

To find the dot product of \(\mathbf{u}\) and \(\mathbf{v},\) we multiply the corresponding components of the vectors and sum them up: $$\mathbf{u} \cdot \mathbf{v} = (-10)(1) + (0)(2) + (4)(3)$$ Calculating the dot product: $$\mathbf{u} \cdot \mathbf{v} = -10 + 0 + 12 = 2$$ So, the dot product of \(\mathbf{u}\) and \(\mathbf{v}\) is \(2\).
02

Find the magnitudes of \(\mathbf{u}\) and \(\mathbf{v}\)

To find the magnitudes of \(\mathbf{u}\) and \(\mathbf{v},\) we use the formula: $$||\mathbf{u}|| = \sqrt{(-10)^2 + 0^2 + 4^2} = \sqrt{100 + 0 + 16} = \sqrt{116}$$ $$||\mathbf{v}|| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$$ So, the magnitudes of \(\mathbf{u}\) and \(\mathbf{v}\) are \(\sqrt{116}\) and \(\sqrt{14}\), respectively.
03

Find the angle between \(\mathbf{u}\) and \(\mathbf{v}\) using the cosine formula

We'll use the cosine formula for the dot product, which relates the angle \(\theta\) between two vectors to their dot product and magnitudes: $$\cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| \hspace{0.2cm} ||\mathbf{v}||}$$ Now, substituting the values found in Steps 1 and 2: $$\cos\theta = \frac{2}{\sqrt{116} \hspace{0.2cm} \sqrt{14}}$$ We now have to find the inverse cosine (arccos) of this value to find the angle \(\theta\): $$\theta = \arccos\left(\frac{2}{\sqrt{116} \hspace{0.2cm} \sqrt{14}}\right)$$ So, the angle between the vectors \(\mathbf{u}\) and \(\mathbf{v}\) is approximately \(94.5^\circ\) when calculated using a calculator.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Magnitudes
When discussing vectors, the magnitude is like the vector's length, starting from the origin of the coordinate system.
It's a measure of how long the vector is, regardless of its direction. To find the magnitude, you use a formula involving the square root of the sum of the squares of its components.
For a vector \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \), the magnitude is computed as follows:
  • First, square each component of the vector.
  • Next, add all these squared values together.
  • Then, take the square root of this sum to find the magnitude.
For the vector \( \mathbf{u} = \langle -10, 0, 4 \rangle \), the magnitude is found with \( ||\mathbf{u}|| = \sqrt{(-10)^2 + 0^2 + 4^2} = \sqrt{116} \).
This follows the steps outlined above, demonstrating how to break down the calculation into simpler parts.
Understanding vector magnitudes is crucial since it's a stepping stone to more complex concepts, such as the dot product and finding angles between vectors.
Angle Between Vectors
The angle between two vectors is an important concept in vector mathematics.
It tells us how much one vector needs to turn to align with the other. This angle has countless applications, from physics to computer graphics.
To find this angle, you can't just use a protractor between the vectors because they're abstract, not physical lines. Instead, we use mathematical formulas.
  • First, calculate the dot product of the vectors.
  • Second, determine the magnitudes of both vectors.
With these values ready, you can use the cosine formula for vectors: \( \cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| \hspace{0.2cm} ||\mathbf{v}||} \). This formula essentially divides the dot product by the product of the magnitudes.
The angle \( \theta \) can then be identified by taking the arccosine (inverse cosine) of this expression.
The result gives you \( \theta = \arccos\left( \frac{2}{\sqrt{116} \times \sqrt{14}} \right) \), showing that the vectors form an angle of approximately \( 94.5^\circ \). Understanding how angles between vectors are calculated helps in visualizing vector orientation in space.
Cosine Formula
The cosine formula is an elegant tool that connects the dot product of two vectors with the cosine of the angle between them.
It's a bridge between algebraic operations and geometric interpretation. Practically, it provides a way to express how vectors align in space.
  • The cosine of the angle \( \theta \) is calculated using the formula: \( \cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| \hspace{0.2cm} ||\mathbf{v}||} \).
  • Here, \( \mathbf{u} \cdot \mathbf{v} \) is the dot product of the vectors, which measures how much the vectors extend in the same direction.
  • \( ||\mathbf{u}|| \) and \( ||\mathbf{v}|| \) are their magnitudes. They help normalize the vector comparison.
The beauty of the cosine formula lies in its ability to quantify the direct relationship between two vectors as a single value between -1 and 1.
If the cosine is 1 or -1, the vectors are perfectly aligned or opposite, respectively.
A cosine of 0 indicates perpendicular vectors. This concise and effective formula is fundamental in a variety of fields, serving as a linchpin for understanding the spatial relationship between vectors.

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In contrast to the proof in Exercise \(83,\) we now use coordinates and position vectors to prove the same result. Without loss of generality, let \(P\left(x_{1}, y_{1}, 0\right)\) and \(Q\left(x_{2}, y_{2}, 0\right)\) be two points in the \(x y\) -plane, and let \(R\left(x_{3}, y_{3}, z_{3}\right)\) be a third point such that \(P, Q,\) and \(R\) do not lie on a line. Consider \(\Delta P Q R\) a. Let \(M_{1}\) be the midpoint of the side \(P Q .\) Find the coordinates of \(M_{1}\) and the components of the vector \(R M_{1}\) b. Find the vector \(\overrightarrow{O Z}_{1}\) from the origin to the point \(Z_{1}\) two-thirds of the way along \(\hat{R} \vec{M}_{1}\) c. Repeat the calculation of part (b) with the midpoint \(M_{2}\) of \(R Q\) and the vector \(P M_{2}\) to obtain the vector \(\overline{O Z}_{2}\) d. Repeat the calculation of part (b) with the midpoint \(M_{3}\) of \(P R\) and the vector \(Q M_{3}\) to obtain the vector \(\overline{O Z}_{3}\) e. Conclude that the medians of \(\Delta P Q R\) intersect at a point. Give the coordinates of the point. f. With \(P(2,4,0), Q(4,1,0),\) and \(R(6,3,4),\) find the point at which the medians of \(\Delta P Q R\) intersect.

Equations of planes Find an equation of the following planes. The plane containing the \(x\) -axis and the point \(P_{0}(1,2,3)\)

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