Define the scalar product and obtain the magnitude of a vector from it. Mention the direction of scalar product.
(N/A) The scalar product (or dot product) of two vectors $\vec{A}$ and $\vec{B}$ is defined as the product of their magnitudes and the cosine of the angle $\theta$ between them: $\vec{A} \cdot \vec{B} = AB \cos \theta$.
To obtain the magnitude of a vector $\vec{A}$ from the scalar product,we take the dot product of the vector with itself: $\vec{A} \cdot \vec{A} = A A \cos(0^\circ) = A^2$. Thus,the magnitude $A = \sqrt{\vec{A} \cdot \vec{A}}$.
The scalar product is a scalar quantity,meaning it has magnitude but no direction. Therefore,it does not have a direction.
To obtain the magnitude of a vector $\vec{A}$ from the scalar product,we take the dot product of the vector with itself: $\vec{A} \cdot \vec{A} = A A \cos(0^\circ) = A^2$. Thus,the magnitude $A = \sqrt{\vec{A} \cdot \vec{A}}$.
The scalar product is a scalar quantity,meaning it has magnitude but no direction. Therefore,it does not have a direction.
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If $\overrightarrow{A}$ and $\overrightarrow{B}$ are two vectors,then which of the following are correct?
$(a) \ (\overrightarrow{A} \times \overrightarrow{B}) \perp \overrightarrow{A}$
$(b) \ (\overrightarrow{A} \times \overrightarrow{B}) \perp \overrightarrow{B}$
$(c) \ (\overrightarrow{A} \times \overrightarrow{B}) \perp (\overrightarrow{A} + \overrightarrow{B})$
$(d) \ (\overrightarrow{A} \times \overrightarrow{B}) \perp (\overrightarrow{A} - \overrightarrow{B})$
$(e) \ (\overrightarrow{A} \times \overrightarrow{B}) \perp (\overrightarrow{A} \cdot \overrightarrow{B})$
$(a) \ (\overrightarrow{A} \times \overrightarrow{B}) \perp \overrightarrow{A}$
$(b) \ (\overrightarrow{A} \times \overrightarrow{B}) \perp \overrightarrow{B}$
$(c) \ (\overrightarrow{A} \times \overrightarrow{B}) \perp (\overrightarrow{A} + \overrightarrow{B})$
$(d) \ (\overrightarrow{A} \times \overrightarrow{B}) \perp (\overrightarrow{A} - \overrightarrow{B})$
$(e) \ (\overrightarrow{A} \times \overrightarrow{B}) \perp (\overrightarrow{A} \cdot \overrightarrow{B})$
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