– $ \overrightarrow A \space \times \overrightarrow B $
– Determine the vector product’s direction $ \overrightarrow A \space \times \overrightarrow B$.
– Calculate the scalar product when the angle is $ 60 { \circ} $ and the vector magnitude is $ 5 and 4 $.
– Calculate the scalar product when the angle is $ 60 { \circ} $ and the vector magnitude is $ 5 \space and \space 5 $.
The main purpose of this guide is to find the direction and magnitude of the vector product.
This question uses the concept of magnitude and direction of vector product. A vector product has both magnitude and direction. Mathematically, the vector product is represented as:
\[A \space \times \space B \space = \space ||A || \space || B || \space sin \theta n \]
Expert Answer
We first have to find the direction and magnitude of the vector product.
a) \[A \space \times \space B \space = \space (2.80[cos60 \hat x \space + \space sin60 \hat y]) \space \times \space (1.90[cos60 \hat x \space + \space sin60 \hat y]) \]
By simplifying, we get:
\[= \space -2.80 \space \times \space 1.90cos60sin60 \hat z \space – \space 2.80 \space \times \space 1.90cos60sin60 \hat z \]
\[= \space -2 \space \times \space 2.80 \space \times 1.90cos60sin60 \hat z \]
Thus:
\[A \space \times \space B \space = \space – 4.61 \space cm^2 \space \hat z \]
Now the magnitude is:
\[=\space 4.61 \space cm^2 \space \hat z \]
b) Now we have to calculate the direction for the vector product.
The vector product is pointed in the negative direction of the z-axis.
c) Now, we have to find the scalar product.
\[(\overrightarrow A \space . \space \overrightarrow B \space = \space AB \space cos \theta) \]
By putting values, we get:
\[= \space 20 \space cos 60 \]
\[= \space – \space 19.04 \]
d) We have to find the scalar product.
\[(\overrightarrow A \space . \space \overrightarrow B \space = \space AB \space cos \theta) \]
By putting values, we get:
\[= \space 25 \space cos 60 \]
\[= \space – \space 23.81 \]
Numerical Answer
The magnitude of the cross product is $ 4.61 \space cm^2 \space \hat z$.
The direction is along the z-axis.
The scalar product is $ – \space 19.04 $.
The scalar product is $ – \space 23.81 $.
Example
Calculate the scalar product when the angle is $ 30 { \circ} $, $ 90 { \circ} $ and the vector magnitude is $ 5 and 5 $.
First, we have to calculate the scalar product for the angle of $ 30 $ degrees.
We know that:
\[(\overrightarrow A \space . \space \overrightarrow B \space = \space AB \space cos \theta) \]
By putting values, we get:
\[= \space 25 \space cos 30 \]
\[= \space 3.85 \]
Now we have to calculate the scalar product for the angle of 90 degrees.
We know that:
\[(\overrightarrow A \space . \space \overrightarrow B \space = \space AB \space cos \theta) \]
By putting values, we get:
\[= \space 25 \space cos 90 \]
\[= \space 25 \space \times \space 0 \]
\[= \space 0 \]
Thus the scalar product between two vectors is equal to $ 0 $ when the angle is $ 90 $ degrees.