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 :
Solution
$\left| {\overrightarrow a } \right| = 2,\left| {\overrightarrow b } \right| = 5$<br><br>$$\left| {\overrightarrow a \times \overrightarrow b } \right| = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\sin \theta = \pm 8$$<br><br>$\sin \theta = \pm \,{4 \over 5}$<br><br>$\therefore$ $$\overrightarrow a .\,\overrightarrow b = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta $$<br><br>$= 10.\left( { \pm \,{3 \over 5}} \right) = \pm 6$<br><br>$\left| {\overrightarrow a .\,\overrightarrow b } \right| = 6$
About this question
Subject: Mathematics · Chapter: Vector Algebra · Topic: Types of Vectors
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