Match Column-$I$ with Column-$II$.
Column-$I$ Column-$II$ $(1)$ Resultant of two vectors of equal magnitude $(a)$ Along the bisector of the angle between them $(2)$ Direction of $\overrightarrow A \times \overrightarrow B$ $(b)$ Coplanar $(c)$ Perpendicular to the plane containing $\overrightarrow A$ and $\overrightarrow B$
| Column-$I$ | Column-$II$ |
|---|---|
| $(1)$ Resultant of two vectors of equal magnitude | $(a)$ Along the bisector of the angle between them |
| $(2)$ Direction of $\overrightarrow A \times \overrightarrow B$ | $(b)$ Coplanar |
| $(c)$ Perpendicular to the plane containing $\overrightarrow A$ and $\overrightarrow B$ |
- A$(1-a), (2-b)$
- B$(1-a), (2-c)$
- C$(1-c), (2-b)$
- D$(1-b), (2-a)$
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Similar Questions
If $\theta$ is the angle between two vectors $\vec{A}$ and $\vec{B}$,then match the following two columns.
Column $I$ Column $II$ $(A)$ $\vec{A} \cdot \vec{B} = |\vec{A} \times \vec{B}|$ $(p)$ $\theta = 45^{\circ}$ or $135^{\circ}$ $(B)$ $\vec{A} \cdot \vec{B} = B^2$ $(q)$ $\theta = 0^{\circ}$ $(C)$ $|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|$ $(r)$ $\vec{A} = \vec{B}$ $(D)$ $|\vec{A} \times \vec{B}| = AB$ $(s)$ $\theta = 90^{\circ}$
| Column $I$ | Column $II$ |
| $(A)$ $\vec{A} \cdot \vec{B} = |\vec{A} \times \vec{B}|$ | $(p)$ $\theta = 45^{\circ}$ or $135^{\circ}$ |
| $(B)$ $\vec{A} \cdot \vec{B} = B^2$ | $(q)$ $\theta = 0^{\circ}$ |
| $(C)$ $|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|$ | $(r)$ $\vec{A} = \vec{B}$ |
| $(D)$ $|\vec{A} \times \vec{B}| = AB$ | $(s)$ $\theta = 90^{\circ}$ |
Medium
View Solution$A$ vector $\vec{A}$ having magnitude $6$ units is added to vector $\vec{B}$,which is along the $x$-axis. The resultant of $\vec{A}$ and $\vec{B}$ is along the $y$-axis. If the magnitude of the resultant of $\vec{A}$ and $\vec{B}$ is three times that of $\vec{B}$,then the magnitude of $\vec{B}$ is:
Given $\vec{A} + \vec{B} + \vec{C} = \vec{0}$. The magnitudes of two vectors are equal,and the magnitude of the third vector is $\sqrt{2}$ times that of the other two. What are the angles between the vectors?
Difficult
View SolutionTwo particles are located at an equal distance from the origin. The position vectors of these are represented by $\overrightarrow{A} = 2\hat{i} + 3n\hat{j} + 2\hat{k}$ and $\overrightarrow{B} = 2\hat{i} - 2\hat{j} + 4p\hat{k}$,respectively. If both vectors are at a right angle to each other,the value of $n^{-1}$ is . . . . . . .
Two vectors $A$ and $B$ have equal magnitude $x$. The angle between them is $60^{\circ}$. Match the following two columns:
Column $I$ Column $II$ $(A) |A+B|$ $(p) \frac{\sqrt{3}}{2} x^2$ $(B) |A-B|$ $(q) x$ $(C) A \cdot B$ $(r) \sqrt{3} x$ $(D) |A \times B|$ $(s) \frac{x^2}{2}$
| Column $I$ | Column $II$ |
|---|---|
| $(A) |A+B|$ | $(p) \frac{\sqrt{3}}{2} x^2$ |
| $(B) |A-B|$ | $(q) x$ |
| $(C) A \cdot B$ | $(r) \sqrt{3} x$ |
| $(D) |A \times B|$ | $(s) \frac{x^2}{2}$ |
Medium
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