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3-1.Vectors
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Show that the area of the triangle contained between the vectors $a$ and $b$ is one half of the magnitude of $a \times b .$
Option A
Option B
Option C
Option D
Solution
Consider two vectors $\overrightarrow{ OK }=|\vec{a}| {\text { and }} \overrightarrow{ OM }=|\vec{b}|,$ inclined at an angle $\theta$
In $\Delta$ OMN, we can write the relation:
$\sin \theta=\frac{M N}{O M}=\frac{M N}{|\vec{b}|}$
$MN =|\vec{b}| \sin \theta$
$|\vec{a} \times \vec{a}|=|\vec{a}||\vec{b}| \sin \theta$
$= OK \cdot MN \times \frac{2}{2}$
$=2 \times$ Area of $\Delta OMK$
$\therefore$ Area of $\Delta OMK ^{=}=\frac{1}{2}|\vec{a} \times \vec{b}|$
Standard 11
Physics
