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Let \(\mathbf{a}, \mathbf{b}, \mathbf{c}\)  be vectors such that \(\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} = \begin{pmatrix} -3 \\ -12 \\ 4 \end{pmatrix} .\)
Find the area of the triangle whose vertices are \(\mathbf{a}, \mathbf{b}\) and \(\mathbf{c}.\)

 Apr 29, 2020

Best Answer 

 #1
avatar+24983 
+3

Let \(\mathbf{a}\), \(\mathbf{b}\), \(\mathbf{c} \) be vectors such that \(\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} = \begin{pmatrix} -3 \\ -12 \\ 4 \end{pmatrix} \).
Find the area of the triangle whose vertices are \(\mathbf{a}\)\(\mathbf{b}\) and \(\mathbf{c} \).

 

\(\text{Let the area of the triangle $=A$ }\)

 

\(\begin{array}{|rcll|} \hline 2A &=& |~ (\mathbf{a}-\mathbf{b})\times (\mathbf{a}-\mathbf{c}) ~| \\ 2A &=& |~ \mathbf{a}\times\mathbf{a}+\mathbf{a}\times(\mathbf{-c})+(\mathbf{-b})\times\mathbf{a}+(\mathbf{-b})\times(\mathbf{-c}) ~| \\ && \boxed{ \mathbf{a}\times\mathbf{a} = 0 \\ \mathbf{a}\times(\mathbf{-c}) = \mathbf{c}\times\mathbf{a} \\ (\mathbf{-b})\times\mathbf{a} = \mathbf{a}\times\mathbf{b} \\ (\mathbf{-b})\times(\mathbf{-c}) = \mathbf{b}\times\mathbf{c} } \\ 2A &=& |~ 0+\mathbf{c}\times\mathbf{a}+\mathbf{a}\times\mathbf{b} +\mathbf{b}\times\mathbf{c} ~| \\ 2A &=& |~ \mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} ~| \\ 2A &=& |~ \begin{pmatrix} -3 \\ -12 \\ 4 \end{pmatrix} ~| \\ 2A &=& \sqrt{(-3)^2+(-12)^2+(4)^2} \\ 2A &=& \sqrt{9+144+16} \\ 2A &=& \sqrt{169} \\ 2A &=& 13 \\ A &=& \dfrac{13}{2} \\ \mathbf{A} &=& \mathbf{6.5} \\ \hline \end{array}\)

 

laugh

 Apr 30, 2020
 #1
avatar+24983 
+3
Best Answer

Let \(\mathbf{a}\), \(\mathbf{b}\), \(\mathbf{c} \) be vectors such that \(\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} = \begin{pmatrix} -3 \\ -12 \\ 4 \end{pmatrix} \).
Find the area of the triangle whose vertices are \(\mathbf{a}\)\(\mathbf{b}\) and \(\mathbf{c} \).

 

\(\text{Let the area of the triangle $=A$ }\)

 

\(\begin{array}{|rcll|} \hline 2A &=& |~ (\mathbf{a}-\mathbf{b})\times (\mathbf{a}-\mathbf{c}) ~| \\ 2A &=& |~ \mathbf{a}\times\mathbf{a}+\mathbf{a}\times(\mathbf{-c})+(\mathbf{-b})\times\mathbf{a}+(\mathbf{-b})\times(\mathbf{-c}) ~| \\ && \boxed{ \mathbf{a}\times\mathbf{a} = 0 \\ \mathbf{a}\times(\mathbf{-c}) = \mathbf{c}\times\mathbf{a} \\ (\mathbf{-b})\times\mathbf{a} = \mathbf{a}\times\mathbf{b} \\ (\mathbf{-b})\times(\mathbf{-c}) = \mathbf{b}\times\mathbf{c} } \\ 2A &=& |~ 0+\mathbf{c}\times\mathbf{a}+\mathbf{a}\times\mathbf{b} +\mathbf{b}\times\mathbf{c} ~| \\ 2A &=& |~ \mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} ~| \\ 2A &=& |~ \begin{pmatrix} -3 \\ -12 \\ 4 \end{pmatrix} ~| \\ 2A &=& \sqrt{(-3)^2+(-12)^2+(4)^2} \\ 2A &=& \sqrt{9+144+16} \\ 2A &=& \sqrt{169} \\ 2A &=& 13 \\ A &=& \dfrac{13}{2} \\ \mathbf{A} &=& \mathbf{6.5} \\ \hline \end{array}\)

 

laugh

heureka Apr 30, 2020

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