Question

Find the angle between the position vectors to the points \((3,-4,0)\) and \((-2,1,0)\) and find the direction cosines of a vector perpendicular to both.

Short answer

The angle is approximately 116.57 degrees. The direction cosines of a perpendicular vector (0,0,-5) are (0,0,-1).

Step-by-step solution

- Find Dot Product

The dot product of vectors \(\textbf{a} = (3, -4, 0)\) and \(\textbf{b} = (-2, 1, 0)\) is calculated using the formula: \(\textbf{a} \bullet \textbf{b} = a_1b_1 + a_2b_2 + a_3b_3\). Substituting the values, \(\textbf{a} \bullet \textbf{b} = 3(-2) + (-4)(1) + 0(0) = -6 - 4 = -10\)

- Calculate Magnitudes

The magnitude of vector \(\textbf{a}\) is found using \(\big| \textbf{a} \big| = \sqrt{a_1^2 + a_2^2 + a_3^2}\). Thus, \(\big| \textbf{a} \big| = \sqrt{3^2 + (-4)^2 + 0^2} = \sqrt{9 + 16} = \sqrt{25} = 5\). The magnitude of \(\textbf{b}\) is \(\big| \textbf{b} \big| = \sqrt{(-2)^2 + 1^2 + 0^2} = \sqrt{4 + 1} = \sqrt{5}\).

- Calculate Cosine of the Angle

The cosine of the angle \(\theta\) between the vectors is given by \(\text{cos} \theta = \frac{\textbf{a} \bullet \textbf{b}}{\big| \textbf{a} \big| \big| \textbf{b} \big|}\). Substituting the values, \(\text{cos} \theta = \frac{-10}{5 \sqrt{5}} = \frac{-10}{5 \sqrt{5}} = \frac{-2}{\big| \sqrt{5} \big|} = - \frac{2}{\sqrt{5}} = - \frac{2\sqrt{5}}{5} \).

- Calculate the Angle

The angle \( \theta \) is given by \(\theta = \arccos \left( - \frac{2 \sqrt{5}}{5} \right)\). Calculating this gives an angle of approximately 116.57 degrees.

- Find a Perpendicular Vector

A vector perpendicular to both \(\textbf{a}\) and \(\textbf{b}\) can be found using the cross product \( \textbf{a} \times \textbf{b} \). Compute \(\textbf{a} \times \textbf{b}\) with: \(\textbf{a} = (3, -4, 0)\) and \(\textbf{b} = (-2, 1, 0)\). The cross product \( \textbf{a} \times \textbf{b}\) is calculated as follows: \(\textbf{a} \times \textbf{b} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \ 3 & -4 & 0 \ -2 & 1 & 0 \ \end{vmatrix}\) which simplifies to: \(\textbf{i}(0) - \textbf{j}(0) + \textbf{k}(3 \times 1 - (-4) \times (-2)) = (0 \textbf{i} - 0 \textbf{j} + \textbf{k}(3 - 8)) = (0, 0, -5)\). Thus, one perpendicular vector is \( \textbf{k}(0, 0, -5) = (0, 0, -5) \).

- Determine Direction Cosines

The direction cosines are the cosines of the angles between the vector and the coordinate axes. For a vector \( \textbf{v} = (v_1, v_2, v_3) \), the direction cosines are \( \text{cos} \alpha = \frac{v_1}{\big| \textbf{v} \big|} \, \text{cos} \beta = \frac{v_2}{\big| \textbf{v} \big|} \, \text{cos} \gamma = \frac{v_3}{\big| \textbf{v} \big|} \). For \( \textbf{v} = (0, 0, -5) \), the magnitude is \( \big| \textbf{v} \big| = 5 \). So, the direction cosines are: \( \text{cos} \alpha = \frac{0}{5} = 0 \, \text{cos} \beta = \frac{0}{5} = 0 \, and \text{cos} \gamma = \frac{-5}{5} = -1 \).

Key concepts

Dot Product

The dot product, also called the scalar product, is a way to multiply two vectors that results in a scalar (a single number). For two vectors \(\textbf{a} = (a_1, a_2, a_3)\) and \(\textbf{b} = (b_1, b_2, b_3)\), the dot product is calculated as:
\(\textbf{a} \bullet \textbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3\).
This formula helps in various applications like finding the angle between vectors and in projections.
For example, if \(\textbf{a} = (3, -4, 0)\) and \(\textbf{b} = (-2, 1, 0)\), the dot product is:
\(\textbf{a} \bullet \textbf{b} = 3(-2) + (-4)(1) + 0(0) = -6 - 4 = -10\).

Vector Magnitude

The magnitude (or length) of a vector is a measure of how long the vector is. It's calculated using the formula:
\(\big| \textbf{a} \big| = \sqrt{a_1^2 + a_2^2 + a_3^2}\).
This helps in normalizing vectors and in various geometric calculations.
For instance, the magnitude of vector \(\textbf{a} = (3, -4, 0)\) is found as:
\(\big| \textbf{a} \big| = \sqrt{3^2 + (-4)^2 + 0^2} = \sqrt{9 + 16} = \sqrt{25} = 5\).
The magnitude of \(\textbf{b} = (-2, 1, 0)\) is:
\(\big| \textbf{b} \big| = \sqrt{-2^2 + 1^2 + 0^2} = \sqrt{4 + 1} = \sqrt{5}\).

Trigonometric Functions

Trigonometric functions like cosine (cos) are crucial in vector calculations, especially when finding angles. The cosine of the angle between two vectors \(\textbf{a}\) and \(\textbf{b}\) can be found using the dot product:
\(\text{cos} \theta = \frac{\textbf{a} \bullet \textbf{b}}{\big| \textbf{a} \big| \big| \textbf{b} \big|}\).
This shows the relationship between the vectors and helps calculate the angle directly.
For example, from the given vectors, we find:
\(\text{cos} \theta = \frac{-10}{5 \sqrt{5}} = - \frac{2\sqrt{5}}{5}\), which results in an angle:
\(\theta = \arccos \left( - \frac{2 \sqrt{5}}{5} \right)\) approximately equals to 116.57 degrees.

Cross Product

The cross product of two vectors results in another vector that is perpendicular to both. It's useful in finding such perpendicular vectors and determining torque in physics.
For two vectors \(\textbf{a} = (a_1, a_2, a_3)\) and \(\textbf{b} = (b_1, b_2, b_3)\), the cross product is:
\(\textbf{a} \times \textbf{b} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \ 3 & -4 & 0 \ -2 & 1 & 0 \end{vmatrix}\).
Solving this, we find:
\(\textbf{k}(3*1 - (-4)*(-2)) = \textbf{k}(3 - 8) = (0, 0, -5)\). So, a perpendicular vector is \( \textbf{k}(0, 0, -5) = (0, 0, -5)\).

Direction Cosines

Direction cosines are the cosines of angles that a vector makes with the coordinate axes. They tell us about the vector's orientation in space.
For a vector \( \textbf{v} = (v_1, v_2, v_3)\), these cosines are found as:
\(\text{cos} \alpha = \frac{v_1}{\big| \textbf{v} \big|} \ \text{cos} \beta = \frac{v_2}{\big| \textbf{v} \big|} \ \text{cos} \gamma = \frac{v_3}{\big| \textbf{v} \big|}\).
For the vector \( \textbf{v} = (0, 0, -5)\), the magnitude is \(\big| \textbf{v} \big| = 5\).
Thus, the direction cosines are:
\(\text{cos} \alpha = \frac{0}{5} = 0 \ \text{cos} \beta = \frac{0}{5} = 0 \ \text{cos} \gamma = \frac{-5}{5} = -1\).