JEE MAIN - Mathematics (2021 - 25th July Evening Shift - No. 11)
If $$\left| {\overrightarrow a } \right| = 2,\left| {\overrightarrow b } \right| = 5$$ and $$\left| {\overrightarrow a \times \overrightarrow b } \right|$$ = 8, then $$\left| {\overrightarrow a .\,\overrightarrow b } \right|$$ is equal to :
6
4
3
5
Explanation
$$\left| {\overrightarrow a } \right| = 2,\left| {\overrightarrow b } \right| = 5$$
$$\left| {\overrightarrow a \times \overrightarrow b } \right| = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\sin \theta = \pm 8$$
$$\sin \theta = \pm \,{4 \over 5}$$
$$\therefore$$ $$\overrightarrow a .\,\overrightarrow b = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta $$
$$ = 10.\left( { \pm \,{3 \over 5}} \right) = \pm 6$$
$$\left| {\overrightarrow a .\,\overrightarrow b } \right| = 6$$
$$\left| {\overrightarrow a \times \overrightarrow b } \right| = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\sin \theta = \pm 8$$
$$\sin \theta = \pm \,{4 \over 5}$$
$$\therefore$$ $$\overrightarrow a .\,\overrightarrow b = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta $$
$$ = 10.\left( { \pm \,{3 \over 5}} \right) = \pm 6$$
$$\left| {\overrightarrow a .\,\overrightarrow b } \right| = 6$$
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