Question

Under what condition A.B is not equal to zero and A×B is equal to zero when A and B are two non zero vectors?

Answer 1

Answer on Question #67156 – Math – Linear Algebra

Question

Under what condition A.B is not equal to zero and $A \times B$ is equal to zero when A and B are two non-zero vectors?

Solution

A. B is not equal to zero and $A \times B$ is equal to zero when vectors A and B are collinear.

Let's show it.

Let $\mathbf{B}$ be $\lambda \mathbf{A}$, that is,

$$\mathbf {B} = \left(B _ {x}, B _ {y}, B _ {z}\right) = \left(\lambda A _ {x}, \lambda A _ {y}, \lambda A _ {z}\right),$$

then

$$\mathbf {A} \cdot \mathbf {B} = A _ {x} B _ {x} + A _ {y} B _ {y} + A _ {z} B _ {z} = \lambda \left(A _ {x} A _ {x} + A _ {y} A _ {y} + A _ {z} A _ {z}\right) = \lambda \left(A _ {x} ^ {2} + A _ {y} ^ {2} + A _ {z} ^ {2}\right).$$

If $\mathbf{A}$ is a non-zero vector, then $\mathbf{A} \cdot \mathbf{B} = \lambda \left(A_x^2 + A_y^2 + A_z^2\right) > 0$.

Next, we find $\mathbf{A} \times \mathbf{B}$:

$$\begin{array}{l} \mathbf {A} \times \mathbf {B} = \left| \begin{array}{c c c} \mathbf {i} & \mathbf {j} & \mathbf {k} \\ A _ {x} & A _ {y} & A _ {z} \\ B _ {x} & B _ {y} & B _ {z} \end{array} \right| = \left| \begin{array}{c c c} \mathbf {i} & \mathbf {j} & \mathbf {k} \\ A _ {x} & A _ {y} & A _ {z} \\ \lambda A _ {x} & \lambda A _ {y} & \lambda A _ {z} \end{array} \right| \\ = \mathbf {i} \left| \begin{array}{l l} A _ {y} & A _ {z} \\ \lambda A _ {y} & \lambda A _ {z} \end{array} \right| - \mathbf {j} \left| \begin{array}{l l} A _ {x} & A _ {z} \\ \lambda A _ {x} & \lambda A _ {z} \end{array} \right| + \mathbf {k} \left| \begin{array}{l l} A _ {x} & A _ {y} \\ \lambda A _ {x} & \lambda A _ {y} \end{array} \right| = \\ = \mathbf {i} \left(\lambda A _ {y} A _ {z} - \lambda A _ {y} A _ {z}\right) - \mathbf {j} \left(\lambda A _ {x} A _ {z} - \lambda A _ {x} A _ {z}\right) + \mathbf {k} \left(\lambda A _ {x} A _ {y} - \lambda A _ {x} A _ {y}\right) = 0 \\ \end{array}$$

Answer: if vectors $\mathbf{A}$ and $\mathbf{B}$ are collinear, non-zero vectors, then $\mathbf{A} \cdot \mathbf{B} \neq 0$ and $\mathbf{A} \times \mathbf{B} = 0$.

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