Question

If \(\mathbf{A}=4 \mathbf{i}-3 \mathbf{k}\) and \(\mathbf{B}=-2 \mathbf{i}+2 \mathbf{j}-\mathbf{k},\) find the scalar projection of \(\mathbf{A}\) on \(\mathbf{B},\) the scalar projection of \(\mathbf{B}\) on \(\mathbf{A},\) and the cosine of the angle between \(\mathbf{A}\) and \(\mathbf{B}\).

Short answer

Scalar projection of \(\textbf{A}\) on \(\textbf{B}\) is \(-5/3\), scalar projection of \(\textbf{B}\) on \(\textbf{A}\) is \(-1\), and \(\text{cos}(\theta) = -1/3\).

Step-by-step solution

- Calculate Dot Product

Find the dot product of vectors \(\textbf{A}\) and \(\textbf{B}\): \[ \textbf{A} \bullet \textbf{B} = (4\textbf{i} - 3\textbf{k}) \bullet (-2\textbf{i} + 2\textbf{j} - \textbf{k}) = (4)(-2) + (0)(2) + (-3)(-1) = -8 + 3 = -5 \]

- Calculate Magnitudes

Find the magnitudes of \(\textbf{A}\) and \(\textbf{B}\): \[ ||\textbf{A}|| = \sqrt{(4)^2 + (0)^2 + (-3)^2} = \sqrt{16 + 0 + 9} = \sqrt{25} = 5 \] \[ ||\textbf{B}|| = \sqrt{(-2)^2 + (2)^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \]

- Scalar Projection of A on B

Use the scalar projection formula: \[\text{Scalar projection of } \textbf{A} \text{ on } \textbf{B} = \frac{\textbf{A} \bullet \textbf{B}}{||\textbf{B}||} = \frac{-5}{3} = -\frac{5}{3} \]

- Scalar Projection of B on A

Use the scalar projection formula: \[\text{Scalar projection of } \textbf{B} \text{ on } \textbf{A} = \frac{\textbf{B} \bullet \textbf{A}}{||\textbf{A}||} = \frac{-5}{5} = -1 \]

- Cosine of the Angle

Use the dot product and magnitudes to find the cosine of the angle \(\theta\) between \(\textbf{A}\) and \(\textbf{B}\): \[ \text{cos}(\theta) = \frac{\textbf{A} \bullet \textbf{B}}{||\textbf{A}|| \times ||\textbf{B}||} = \frac{-5}{5 \times 3} = \frac{-5}{15} = -\frac{1}{3} \]

Key concepts

dot product

The dot product, also known as the scalar product, is a fundamental operation in vector calculus. It combines two vectors into a single number (a scalar). The result helps in understanding how much one vector goes in the direction of another. For two vectors \(\textbf{A}\) and \(\textbf{B}\), given their components, the dot product is calculated as:
\[ \textbf{A} \bullet \textbf{B} = A_xB_x + A_yB_y + A_zB_z \] Here, \(\textbf{A} = 4 \textbf{i} - 3 \textbf{k}\) and \(\textbf{B} = -2 \textbf{i} + 2 \textbf{j} - \textbf{k}\). Applying the dot product formula, we get:
\[4 \times (-2) + 0 \times 2 + (-3) \times (-1) = -8 + 0 + 3 = -5 \] By this, \(\textbf{A} \bullet \textbf{B} = -5\). The negative result suggests that \(\textbf{A}\) and \(\textbf{B}\) have more directionality against each other than with each other, indicating an angle greater than 90 degrees between them.

vector magnitudes

The magnitude of a vector gives the length or size of the vector. It is computed using the Pythagorean theorem in a 3D space. For a vector \(\textbf{A} = 4 \textbf{i} - 3 \textbf{k}\), the magnitude, noted as \(||\textbf{A}||\), can be found by:
\[ ||\textbf{A}|| = \text{\textbf{A's magnitude}} = \sqrt{(4)^2 + (0)^2 + (-3)^2} = \sqrt{16 + 0 + 9} = \sqrt{25} = 5 \] Similarly, the magnitude of vector \(\textbf{B} = -2 \textbf{i} + 2 \textbf{j} - \textbf{k}\) is:
\[ ||\textbf{B}|| = \text{\textbf{B's magnitude}} = \sqrt{(-2)^2 + (2)^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \] Understanding magnitudes is crucial for quantifying the length of vectors, which aids in various calculations like projections and angles.

scalar projection

Scalar projection measures how much one vector extends in the direction of another vector. For vector \(\textbf{A}\) on \(\textbf{B}\), the scalar projection is calculated as:
\[ \text{\textbf{Scalar projection of } \textbf{A} \text{ on } \textbf{B}} = \frac{\textbf{A} \bullet \textbf{B}}{||\textbf{B}||} \] From the previous sections, we have \(\textbf{A} \bullet \textbf{B} = -5\) and \(||\textbf{B}|| = 3\). So,
\[ \text{\textbf{Scalar projection}} = \frac{-5}{3} = -\frac{5}{3} \] Similarly, the scalar projection of \(\textbf{B}\) on \(\textbf{A}\) is:
\[ \text{\textbf{Scalar projection of } \textbf{B} \text{ on } \textbf{A}} = \frac{\textbf{B} \bullet \textbf{A}}{||\textbf{A}||} = \frac{-5}{5} = -1 \] These results show how much of each vector is pointing in the direction of the other, and a negative value indicates opposite direction.

cosine of angle between vectors

The cosine of the angle between two vectors helps to determine the orientation of one vector with respect to another. It can be calculated using the dot product and the magnitudes of the vectors:
\[ \text{cos}(\theta) = \frac{\textbf{A} \bullet \textbf{B}}{||\textbf{A}|| \times ||\textbf{B}||} \] With \(\textbf{A} \bullet \textbf{B} = -5\), \(||\textbf{A}|| = 5\), and \(||\textbf{B}|| = 3\), we get:
\[ \text{cos}(\theta) = \frac{-5}{5 \times 3} = \frac{-5}{15} = -\frac{1}{3} \] This result means that the angle \(\theta\) is \(\text{cos}^{-1}(-\frac{1}{3})\), which is greater than 90 degrees. This verifies that \(\textbf{A}\) and \(\textbf{B}\) are oriented more against each other rather than in the same direction.