Question

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

Short answer

Scalar projection of A on B: \(\frac{-5}{3}\). Scalar projection of B on A: \(-1\). Cosine of the angle: \(-\frac{1}{3}\).

Step-by-step solution

Find the Dot Product of Vectors A and B

Recall the dot product formula: \[ \mathbf{A} \cdot \mathbf{B} = (4i - 3k) \cdot (-2i + 2j - 1k) \] Calculating the dot product: \[ \mathbf{A} \cdot \mathbf{B} = (4)(-2) + (0)(2) + (-3)(-1) = -8 + 0 + 3 = -5 \]

Calculate the Magnitude of Vectors A and B

The magnitude of a vector \( \mathbf{A} = 4i - 3k \) is: \[ |\mathbf{A}| = \sqrt{4^2 + 0^2 + (-3)^2} = \sqrt{16 + 0 + 9} = \sqrt{25} = 5 \] The magnitude of a vector \( \mathbf{B} = -2i + 2j - k \) is: \[ |\mathbf{B}| = \sqrt{(-2)^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \]

Find the Scalar Projection of A on B

The scalar projection of \( \mathbf{A} \) on \( \mathbf{B} \) is given by: \[ \text{Scalar Projection of } \mathbf{A} \text{ on } \mathbf{B} = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{B}|} = \frac{-5}{3} \]

Find the Scalar Projection of B on A

The scalar projection of \( \mathbf{B} \) on \( \mathbf{A} \) is given by: \[ \text{Scalar Projection of } \mathbf{B} \text{ on } \mathbf{A} = \frac{\mathbf{B} \cdot \mathbf{A}}{|\mathbf{A}|} = \frac{-5}{5} = -1 \]

Calculate the Cosine of the Angle Between A and B

The cosine of the angle \( \theta \) between the vectors \( \mathbf{A} \) and \( \mathbf{B} \) is given by: \[ \cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} = \frac{-5}{5 \times 3} = \frac{-5}{15} = -\frac{1}{3} \] Therefore, \[ \cos \theta = -\frac{1}{3} \]

Key concepts

Dot Product

The dot product, or scalar product, of two vectors is a fundamental operation in vector algebra. It essentially measures the extent to which two vectors point in the same direction. The formula for the dot product of vectors \(\textbf{A}\) and \(\textbf{B}\) is given by: \[ \textbf{A} \cdot \textbf{B} = A_x B_x + A_y B_y + A_z B_z \] where \(A_x, A_y, A_z\) are the components of vector \(\textbf{A}\) and \(B_x, B_y, B_z\) are the components of vector \(\textbf{B}\).

In this exercise, the vectors are \(\textbf{A} = 4i - 3k\) and \(\textbf{B} = -2i + 2j - k\). Plugging in these values, we find the dot product as follows: \[ \textbf{A} \cdot \textbf{B} = (4)(-2) + (0)(2) + (-3)(-1) = -8 + 0 + 3 = -5 \] By calculating the dot product, you can understand how aligned two vectors are. A positive dot product indicates that the vectors are pointing in a similar direction, a negative one suggests they are pointing in opposite directions, and a zero value means they are perpendicular.

Vector Magnitude

The magnitude of a vector represents its length and is denoted by \(|\textbf{A}| \) for vector \(\textbf{A}\). It is calculated using the formula: \[ |\textbf{A}| = \sqrt{A_x^2 + A_y^2 + A_z^2} \]

For the vector \(\textbf{A} = 4i - 3k\), the magnitude is: \[ |\textbf{A}| = \sqrt{4^2 + 0^2 + (-3)^2} = \sqrt{16 + 0 + 9} = \sqrt{25} = 5 \] Similarly, for the vector \(\textbf{B} = -2i + 2j - k\), the magnitude is: \[ |\textbf{B}| = \sqrt{(-2)^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \] Understanding the magnitude of vectors is crucial because it helps you quantify the length or distance the vector represents in the space.

Angle Cosine

The cosine of the angle between two vectors can tell us how aligned they are. It is found using the dot product and the magnitudes of the vectors. The formula is: \[ \cos \theta = \frac{\textbf{A} \cdot \textbf{B}}{|\textbf{A}| |\textbf{B}|} \]

For our vectors \(\textbf{A}\) and \(\textbf{B}\), this becomes: \[ \cos \theta = \frac{-5}{5 \times 3} = \frac{-5}{15} = -\frac{1}{3} \] The cosine value helps us understand how vectors behave in relation to one another. A cosine of 1 means they are perfectly aligned, -1 indicates they are directly opposite, and 0 indicates they are perpendicular.

Vector Components

Vectors are often broken down into components along set axes, such as \(i\), \(j\), and \(k\) for 3-dimensional space. These components make it easier to perform mathematical operations like addition, subtraction, and finding projections.

In this exercise, the vectors' components are:

• For \(\textbf{A} = 4i - 3k\), the components are: \(A_x = 4\), \(A_y = 0\), \(A_z = -3\).
• For \(\textbf{B} = -2i + 2j - k\), the components are: \(B_x = -2\), \(B_y = 2\), \(B_z = -1\).

By understanding each vector's components, we can proceed with operations like calculating the dot product and magnitudes efficiently. This breakdown also helps in visualizing the vector's direction and magnitude in space.