Question

If \(\mathbf{A}=2 \mathbf{i}-3 \mathbf{j}+\mathbf{k}\) and \(\mathbf{A} \cdot \mathbf{B}=0\), does it follow that \(\mathbf{B}=0:\) (Either prove that it does or give a specific example to show that it doesn't.) Answer the same question if \(\mathrm{A} \times \mathrm{B}=0\). And again answer the same question if \(\mathrm{A} \cdot \mathbf{B}=0\) and \(\mathrm{A} \times \mathbf{B}=0\).

Short answer

No, \(\mathbf{B}\) can be non-zero with only \(\mathbf{A} \cdot \mathbf{B} = 0\) or \(\mathbf{A} \times \mathbf{B} = 0\), but \(\mathbf{B} = 0\) if both conditions hold.

Step-by-step solution

- Definition and Given Information

Given that \(\mathbf{A}=2 \mathbf{i}-3 \mathbf{j}+\mathbf{k}\)and \(\mathbf{A} \cdot \mathbf{B}=0\)and we need to determine if it necessarily means that \(\mathbf{B}=0\).

- Dot Product Condition Analysis

The dot product of two vectors \(\mathbf{A}\) and \(\mathbf{B}\) is zero if and only if the vectors are orthogonal (i.e., the angle between them is 90 degrees). This does not imply that \(\mathbf{B}=0\). To show this with an example, consider another vector \(\mathbf{B} = 3\mathbf{i} + 2\mathbf{j} + z\mathbf{k}\). Then the dot product of \(\mathbf{A}\) and \(\mathbf{B}\) is \(\mathbf{A} \cdot \mathbf{B} = (2\mathbf{i} - 3\mathbf{j} + \mathbf{k}) \cdot (3\mathbf{i} + 2\mathbf{j} + z\mathbf{k}) = 6 - 6 + z = z = 0\). If we choose \(z = 0\), the dot product is zero, although \(\mathbf{B}\) is not zero.

- Cross Product Condition Analysis

Now consider the condition \(\mathbf{A} \times \mathbf{B} = 0\). The cross product of two vectors is zero if and only if the vectors are parallel or one is the zero vector. To see if \(\mathbf{B}\) must be zero, observe that \(\mathbf{B}\) could simply be a scalar multiple of \(\mathbf{A}\), therefore parallel (for instance, \(\mathbf{B} = k \mathbf{A}\), and hence not necessarily zero.

- Both Dot and Cross Product

If both conditions \(\mathbf{A} \cdot \mathbf{B} = 0 \) and \(\mathbf{A} \times \mathbf{B} = 0\) hold, \(\mathbf{A}\) and \(\mathbf{B}\) must be orthogonal and parallel at the same time. The only way this can happen is if \(\mathbf{B} = 0\). If \(\mathbf{B} eq 0\), this would lead to a contradiction.

Key concepts

dot product

The dot product, also known as the scalar product, is an operation that takes two vectors and returns a scalar (a single number). It is denoted as \(\textbf{A} \times \textbf{B}\). To compute the dot product of two vectors, you multiply each corresponding pair of components and sum up these products. For example, if \(\textbf{A} = a_1 \textbf{i} + a_2 \textbf{j} + a_3 \textbf{k}\) and \(\textbf{B} = b_1 \textbf{i} + b_2 \textbf{j} + b_3 \textbf{k}\), then the dot product is calculated as follows:
\(\textbf{A} \times \textbf{B} = a_1 b_1 + a_2 b_2 + a_3 b_3\).
The dot product is particularly useful in finding the angle between two vectors. If the dot product is zero, it implies that the vectors are orthogonal (i.e., the angle between them is 90 degrees). In our given exercise, \(\textbf{A} \times \textbf{B} = 0\) indicates that vectors \(\textbf{A}\) and \(\textbf{B}\) are orthogonal but does not mean \(\textbf{B}\) is the zero vector, as demonstrated by the example provided.

cross product

The cross product, also known as the vector product, is an operation on two vectors in three-dimensional space that results in another vector which is perpendicular to both input vectors. It is denoted as \(\textbf{A} \times \textbf{B}\). To calculate the cross product of two vectors \(\textbf{A}\) and \(\textbf{B}\), you use the determinant of a matrix that includes unit vectors:
\(\textbf{A} \times \textbf{B} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix}\).
This operation results in a new vector that is perpendicular to both \(\textbf{A}\) and \(\textbf{B}\). If the cross product of two vectors is zero, it indicates that the vectors are either parallel or one of the vectors is the zero vector. In our problem, \(\textbf{A} \times \textbf{B} = 0\) shows that \(\textbf{B}\) could be a scalar multiple of \(\textbf{A}\), thereby not necessitating \(\textbf{B}\) to be zero.

orthogonal vectors

Two vectors are orthogonal if their dot product is zero, meaning the vectors are at right angles (90 degrees) to each other. This is a crucial concept in vector mathematics because it indicates that the vectors have no projection on each other.

• Orthogonality is often witnessed in vector spaces and has applications in various realms including computer graphics and physics.
• In the given exercise, the condition \(\textbf{A} \times \textbf{B} = 0\) confirms that \(\textbf{A}\) and \(\textbf{B}\) are orthogonal, but this does not imply that \(\textbf{B}\) is the zero vector.

The given example of \(\textbf{B} = 3 \textbf{i} + 2 \textbf{j} + 0 \textbf{k}\) clarifies that orthogonal vectors need not be zero vectors.

parallel vectors

Two vectors are considered parallel if one is a scalar multiple of the other. This means that they point in the same direction but may have different magnitudes.

• The cross product condition \(\textbf{A} \times \textbf{B} = 0\) signifies that \(\textbf{B}\) can be expressed as \(\textbf{B} = k \textbf{A}\) where \k\ is a scalar.

This implies that \(\textbf{A}\) and \(\textbf{B}\) are parallel. However, this does not mean \(\textbf{B}\) has to be the zero vector. In our exercise, since the cross product is zero, \(\textbf{B}\) can be a parallel vector to \(\textbf{A}\). If both the dot and cross product conditions are zero, then \(\textbf{B}\) must be the zero vector to satisfy both conditions simultaneously.