2021 RI Promo Q3
2021 RI Promo Q3
6 marks
Promo
The points \(A\), \(B\) and \(C\) C have position vectors \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\) respectively.
- Show that the area of triangle \(ABC\) is \(\frac{1}{2}\left| \mathbf{a}\times \mathbf{b}+\mathbf{b}\times \mathbf{c}+\mathbf{c}\times \mathbf{a} \right|\). Hence show that the shortest distance from \(B\) to \(AC\) is \(\frac{\left| \mathbf{a}\times \mathbf{b}+\mathbf{b}\times \mathbf{c}+\mathbf{c}\times \mathbf{a} \right|}{\left| \mathbf{c}-\mathbf{a} \right|}\).[4]
- Given that \(\mathbf{a}\) and \(\mathbf{b}\) are non-zero vectors such that \(\left| \mathbf{a}-\mathbf{b} \right|=\left| \mathbf{a}+\mathbf{b} \right|\), find the value of \(\mathbf{a}\cdot \mathbf{b}\).[2]
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Answer:(a) \(\operatorname{Area}(\triangle ABC)=\dfrac12\left|\mathbf a\times\mathbf b+\mathbf b\times\mathbf c+\mathbf c\times\mathbf a\right|\), shortest distance \(=\dfrac{\left|\mathbf a\times\mathbf b+\mathbf b\times\mathbf c+\mathbf c\times\mathbf a\right|}{|\mathbf c-\mathbf a|}\) (b) \(\mathbf a\mathbin{\cdot}\mathbf b=0\)
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