Define the scalar product of two vectors.
(N/A) The scalar product (or dot product) of two vectors $\vec{A}$ and $\vec{B}$ is defined as the product of the magnitudes of the two vectors and the cosine of the angle $\theta$ between them.
Mathematically,it is expressed as: $\vec{A} \cdot \vec{B} = AB \cos \theta$,where $A$ and $B$ are the magnitudes of vectors $\vec{A}$ and $\vec{B}$ respectively,and $\theta$ is the angle between them $(0 \le \theta \le \pi)$.
The result of the scalar product is a scalar quantity.
Mathematically,it is expressed as: $\vec{A} \cdot \vec{B} = AB \cos \theta$,where $A$ and $B$ are the magnitudes of vectors $\vec{A}$ and $\vec{B}$ respectively,and $\theta$ is the angle between them $(0 \le \theta \le \pi)$.
The result of the scalar product is a scalar quantity.
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Three vectors $\vec{A}, \vec{B}$ and $\vec{C}$ are such that $\vec{A} \cdot \vec{B} = 0$ and $\vec{A} \cdot \vec{C} = 0$. Then $\vec{A}$ is parallel to:
Given two vectors $\vec{A} = 3\hat{i} + \hat{j}$ and $\vec{B} = \hat{j} + 2\hat{k}$. Find the unit vector perpendicular to both $\vec{A}$ and $\vec{B}$.
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View SolutionExplain the geometrical interpretation of the scalar product of two vectors.
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View SolutionIf $|\vec{A}| = 2$ and $|\vec{B}| = 4$,then match the relation in Column-$I$ with the angle $\theta$ between $\vec{A}$ and $\vec{B}$ in Column-$II$.
Column-$I$ Column-$II$ $(a) |\vec{A} \times \vec{B}| = 0$ $(i) \theta = 30^{\circ}$ $(b) |\vec{A} \times \vec{B}| = 8$ $(ii) \theta = 45^{\circ}$ $(c) |\vec{A} \times \vec{B}| = 4$ $(iii) \theta = 90^{\circ}$ $(d) |\vec{A} \times \vec{B}| = 4\sqrt{2}$ $(iv) \theta = 0^{\circ}$
| Column-$I$ | Column-$II$ |
| $(a) |\vec{A} \times \vec{B}| = 0$ | $(i) \theta = 30^{\circ}$ |
| $(b) |\vec{A} \times \vec{B}| = 8$ | $(ii) \theta = 45^{\circ}$ |
| $(c) |\vec{A} \times \vec{B}| = 4$ | $(iii) \theta = 90^{\circ}$ |
| $(d) |\vec{A} \times \vec{B}| = 4\sqrt{2}$ | $(iv) \theta = 0^{\circ}$ |
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View SolutionFind the component of vector $\vec{P} = 2 \hat{i} + 3 \hat{j}$ along the direction of vector $\vec{Q} = \hat{i} + \hat{j}$.
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