Question

Let \(\quad \mathbf{a}=\langle\sqrt{3} / 3, \sqrt{3} / 3, \sqrt{3} / 3\rangle, \mathbf{b}=\langle 1,-1,0\rangle, \quad\) and \(\mathbf{c}=\langle-2,-2,1\rangle .\) Find the angle between each pair of vectors.

Short answer

The angles are 90° between \(\mathbf{a}\) and \(\mathbf{b}\), 125.26° between \(\mathbf{a}\) and \(\mathbf{c}\), and 90° between \(\mathbf{b}\) and \(\mathbf{c}\).

Step-by-step solution

Determine dot product between vectors a and b

Calculate the dot product of given vectors \( \mathbf{a} = \langle \sqrt{3}/3, \sqrt{3}/3, \sqrt{3}/3 \rangle \) and \( \mathbf{b} = \langle 1, -1, 0 \rangle \). Use the formula for dot product: \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \). So, we have:\[\mathbf{a} \cdot \mathbf{b} = (\sqrt{3}/3)\times1 + (\sqrt{3}/3)\times(-1) + (\sqrt{3}/3)\times0 = \frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3} + 0 = 0\]

Calculate magnitudes of vectors a and b

Find the magnitude of vectors \( \mathbf{a} \) and \( \mathbf{b} \) using \( \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2} \).For \( \mathbf{a} \):\[\|\mathbf{a}\| = \sqrt{(\sqrt{3}/3)^2 + (\sqrt{3}/3)^2 + (\sqrt{3}/3)^2} = \sqrt{3\times(\sqrt{3}/3)^2} = 1\]For \( \mathbf{b} \):\[\|\mathbf{b}\| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{2}\]

Compute angle between a and b

Use the dot product to find the angle between vectors. Since \( \mathbf{a} \cdot \mathbf{b} = 0 \), the angle \( \theta \) between them can be found using:\[\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} = \frac{0}{1 \times \sqrt{2}} = 0\]Thus, \( \theta = 90^\circ \).

Determine dot product between vectors a and c

Calculate the dot product of vectors \( \mathbf{a} = \langle \sqrt{3}/3, \sqrt{3}/3, \sqrt{3}/3 \rangle \) and \( \mathbf{c} = \langle -2, -2, 1 \rangle \):\[\mathbf{a} \cdot \mathbf{c} = (\sqrt{3}/3)\times(-2) + (\sqrt{3}/3)\times(-2) + (\sqrt{3}/3)\times1 = -\frac{2\sqrt{3}}{3} - \frac{2\sqrt{3}}{3} + \frac{\sqrt{3}}{3} = -\frac{3\sqrt{3}}{3} = -\sqrt{3}\]

Calculate magnitude of vector c

Find the magnitude of vector \( \mathbf{c} \):\[\|\mathbf{c}\| = \sqrt{(-2)^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = 3\]

Compute angle between a and c

Find the angle between vectors \( \mathbf{a} \) and \( \mathbf{c} \):\[\cos \theta = \frac{\mathbf{a} \cdot \mathbf{c}}{\|\mathbf{a}\| \|\mathbf{c}\|} = \frac{-\sqrt{3}}{1 \times 3} = -\frac{\sqrt{3}}{3}\]From \( \cos^{-1}(-\frac{\sqrt{3}}{3}) \), we can approximate \( \theta \approx 125.26^\circ \).

Determine dot product between vectors b and c

Calculate the dot product of vectors \( \mathbf{b} = \langle 1, -1, 0 \rangle \) and \( \mathbf{c} = \langle -2, -2, 1 \rangle \):\[\mathbf{b} \cdot \mathbf{c} = 1\times(-2) + (-1)\times(-2) + 0\times1 = -2 + 2 + 0 = 0\]

Compute angle between b and c

Since \( \mathbf{b} \cdot \mathbf{c} = 0 \), use the relation:\[\cos \theta = \frac{\mathbf{b} \cdot \mathbf{c}}{\|\mathbf{b}\| \|\mathbf{c}\|} = \frac{0}{\sqrt{2} \times 3} = 0\]Therefore, the angle \( \theta \) is \( 90^\circ \).

Key concepts

Dot Product

The dot product is a foundational operation in vector mathematics. It allows for the calculation of the angle between two vectors, which can be very useful in physics and engineering. The dot product of two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \) is computed as: \[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \].The result is a scalar, not a vector, and it incorporates information about both the magnitudes of the vectors and the cosine of the angle between them.The dot product can tell us several things:

• If \( \mathbf{a} \cdot \mathbf{b} = 0 \), then the vectors \( \mathbf{a} \) and \( \mathbf{b} \) are orthogonal (i.e., they form a 90-degree angle).
• If \( \mathbf{a} \cdot \mathbf{b} > 0 \), the vectors point in more than 90 degrees with respect to each other.
• If \( \mathbf{a} \cdot \mathbf{b} < 0 \), they point in more than 90 degrees.

In the context of our exercise, calculating the dot product between different pairs of vectors was crucial in determining their angular relationships.

Magnitude of Vectors

The magnitude of a vector is analogous to the length of the vector. It provides a measure of how far away the end of the vector is from its origin point. For a vector \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \), the magnitude is given by:\[ \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2} \].This formula is derived from the Pythagorean theorem and applies in n-dimensions.Knowing the magnitude of vectors \( \mathbf{a} \), \( \mathbf{b} \), and \( \mathbf{c} \) helps in normalizing the vectors. Additionally, we use the magnitude to compute the cosine of the angle between two vectors by dividing the dot product by the product of their magnitudes. This method is effective in finding angles between vectors in multi-dimensional space.In our exercise, determining the magnitudes allowed us to apply the cosine formula:\[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \],which is essential in converting the scalar result of a dot product into an angle measurement.

Unit Vectors

Unit vectors are incredibly important because they are vectors with a magnitude of exactly one, and they indicate direction in space. A unit vector \( \mathbf{u} \) corresponding to a vector \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \) is determined by dividing each component of the vector by its magnitude:\[ \mathbf{u} = \frac{\mathbf{v}}{\|\mathbf{v}\|} = \langle \frac{v_1}{\|\mathbf{v}\|}, \frac{v_2}{\|\mathbf{v}\|}, \frac{v_3}{\|\mathbf{v}\|} \rangle \].Unit vectors are often used to simplify vector calculations and to ensure vectors have consistent lengths. They are particularly helpful in physics where direction is vital, such as in determining the direction of travel or force.In vector problems like our exercise, transforming vectors into their unit form helps make calculations more intuitive and manageable. It simplifies finding angles since unit vectors, by definition, account only for direction, making their dot product directly equal to the cosine of the angle they form with another vector.