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3-1.Vectors
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The area of the parallelogram whose sides are represented by the vectors $\hat j + 3\hat k$ and $\hat i + 2\hat j - \hat k$ is
A
$\sqrt {61} $ sq.unit
B
$\sqrt {59} $ sq.unit
C
$\sqrt {49} $ sq.unit
D
$\sqrt {52} $ sq.unit
Solution
(b) $\vec A = \hat j + 3\hat k$,$\vec B = \hat i + 2\hat j – \hat k$
$\vec C = \vec A \times \vec B = \left| {\,\begin{array}{*{20}{c}}{\hat i}&{\hat j}&{\hat k}\\0&1&3\\1&2&{ – 1}\end{array}\,} \right|$$ = – 7\hat i + 3\hat j – \hat k$
Hence area = $|\vec C| = \sqrt {49 + 9 + 1} = \sqrt {59} \,sq\,unit$
Standard 11
Physics
Similar Questions
Vector $A$ is pointing eastwards and vector $B$ northwards. Then, match the following two columns.
| Colum $I$ | Colum $II$ |
| $(A)$ $(A+B)$ | $(p)$ North-east |
| $(B)$ $(A-B)$ | $(q)$ Vertically upwards |
| $(C)$ $(A \times B)$ | $(r)$ Vertically downwards |
| $(D)$ $(A \times B) \times(A \times B)$ | $(s)$ None |
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