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For the two vectors in the figure (Figure 1) , find the magnitude of the vector product

– $ \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.

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