The area of the triangle formed by $2\hat i + \hat j - \hat k$ and $\hat i + \hat j + \hat k$ is
The area of the triangle formed by $2\hat i + \hat j - \hat k$ and $\hat i + \hat j + \hat k$ is
- A
$3$ sq.unit
- B
$2\sqrt 3 $ sq. unit
- C
$2\sqrt {14} $ sq. unit
- D
$\frac{{\sqrt {14} }}{2}$ sq. unit
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The diagonals of a parallelogram are $2\,\hat i$ and $2\hat j.$What is the area of the parallelogram.........$units$
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If $\theta$ is the angle between two vectors $A$ and $B$, then match the following two columns.
colum $I$
colum $II$
$(A)$ $A \cdot B =| A \times B |$
$(p)$ $\theta=90^{\circ}$
$(B)$ $A \cdot B = B ^2$
$(q)$ $\theta=0^{\circ}$ or $180^{\circ}$
$(C)$ $|A+B|=|A-B|$
$(r)$ $A=B$
$(D)$ $|A \times B|=A B$
$(s)$ None
| colum $I$ | colum $II$ |
| $(A)$ $A \cdot B =| A \times B |$ | $(p)$ $\theta=90^{\circ}$ |
| $(B)$ $A \cdot B = B ^2$ | $(q)$ $\theta=0^{\circ}$ or $180^{\circ}$ |
| $(C)$ $|A+B|=|A-B|$ | $(r)$ $A=B$ |
| $(D)$ $|A \times B|=A B$ | $(s)$ None |
The vector sum of two forces is perpendicular to their vector differences. In that case, the forces
- [AIPMT 2003]