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Question

Show that the area of the triangle contained between the vectors $$\textbf a$$ and $$\textbf b$$ is one half of the magnitude of $$\textbf{a} \times \textbf{b}$$.


Solution

Consider two vectors OK = vector $$|a| $$and OM = vector$$ |b|$$, inclined at an angle $$\theta$$ as shown in the following figure.
In $$\triangle OMN,$$ we can write the relation:
$$sin\theta=\dfrac{MN}{OM}=\dfrac{MN}{\left|\vec{b}\right|}$$
$$\implies MN=\left|\vec{b}\right| sin\theta$$

$$\left|\vec{a}\times \vec{b}\right|=\left|\vec{a}\right| \left| \vec{b}\right| sin\theta$$
             $$=OK\times MN\\ =2\times \dfrac{1}{2}\times OK\times MN$$
             $$=2\times $$ Area of $$\triangle OMK$$

$$\implies $$ Area of $$\triangle$$ $$OMK=\dfrac{1}{2}\times \left| \vec{a}\times \vec{b}\right|$$

476743_458297_ans_11d97b3448654b1886aae6147d7247ce.png

Physics
NCERT
Standard XI

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