|
|
|
STEP II 1992 question 9 solution
From The Student RoomTSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 1992 question 9 solution
Solution by generalebriety. | 
|
|
|
|
|
STEP II 1992 question 9 solution
From The Student RoomTSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 1992 question 9 solution
Solution by generalebriety. |
|
![\displaystyle (1-\lambda -\mu )\mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c} = \mathbf{a} - \lambda\mathbf{a} - \mu\mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c} \\
\mathbf{r} = \mathbf{a} + \lambda (\mathbf{b} - \mathbf{a}) + \mu (\mathbf{c} - \mathbf{a}) \longleftarrow \text{Plane through }\mathbf{a}. \\
\text{Let } \mathbf{n} = (\mathbf{b} -\mathbf{a})\cross (\mathbf{c} -\mathbf{a}) = \mathbf{b}\times\mathbf{c} - \mathbf{a}\times\mathbf{c} - \mathbf{b}\times\mathbf{a} = \mathbf{a}\times\mathbf{b} + \mathbf{b}\times\mathbf{c} + \mathbf{c}\times\mathbf{a}. \\
\text{Then }\mathbf{r}\cdot\mathbf{n} = \mathbf{a}\cdot\mathbf{n} . \\ \\
[(1-\lambda -\mu )\mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c}]\cdot(\mathbf{a}\times\mathbf{b} + \mathbf{b}\times\mathbf{c} + \mathbf{c}\times\mathbf{a}) = \mathbf{a}\cdot (\mathbf{a}\times\mathbf{b} + \mathbf{b}\times\mathbf{c} + \mathbf{c}\times\mathbf{a})\\
(1-\lambda -\mu )\mathbf{a}\cdot\mathbf{b}\times\mathbf{c} + \lambda\mathbf{b}\cdot\mathbf{c}\times\mathbf{a} + \mu\mathbf{c}\cdot\mathbf{a}\times\mathbf{b} = \mathbf{a}\cdot\mathbf{b}\times\mathbf{c} \text{ and first result follows}. \\
\bigg[\mathbf{r}\cdot\mathbf{n} =\bigg] \mathbf{a}\cdot\mathbf{n} = \mathbf{b}\cdot\mathbf{n} = \mathbf{c}\cdot\mathbf{n} \\
\mathbf{a}\cdot(\mathbf{a}\times\mathbf{b} + \mathbf{b}\times\mathbf{c} + \mathbf{c}\times\mathbf{a}) = \mathbf{b}\cdot(\mathbf{a}\times\mathbf{b} + \mathbf{b}\times\mathbf{c} + \mathbf{c}\times\mathbf{a}) = \mathbf{c}\cdot(\mathbf{a}\times\mathbf{b} + \mathbf{b}\times\mathbf{c} + \mathbf{c}\times\mathbf{a} ). \\
\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}) = \mathbf{b}\cdot(\mathbf{c}\times\mathbf{a}) = \mathbf{c}\cdot(\mathbf{a}\times\mathbf{b}) \text{ as the dot product of two perpendicular vectors is }\mathbf{0}. \\
A, B, C, O \text{ all in same plane }\Rightarrow \mathbf{a}\cdot (\mathbf{b}\times\mathbf{c}) = \mathbf{b}\cdot (\mathbf{c}\times\mathbf{a}) = \mathbf{c}\cdot (\mathbf{a}\times\mathbf{b}) = \mathbf{0}.](http://thestudentroom.co.uk/../latexrender/pictures/340b026d0717a62f7537a01c64815a42.png "\displaystyle (1-\lambda -\mu )\mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c} = \mathbf{a} - \lambda\mathbf{a} - \mu\mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c} \\
\mathbf{r} = \mathbf{a} + \lambda (\mathbf{b} - \mathbf{a}) + \mu (\mathbf{c} - \mathbf{a}) \longleftarrow \text{Plane through }\mathbf{a}. \\
\text{Let } \mathbf{n} = (\mathbf{b} -\mathbf{a})\cross (\mathbf{c} -\mathbf{a}) = \mathbf{b}\times\mathbf{c} - \mathbf{a}\times\mathbf{c} - \mathbf{b}\times\mathbf{a} = \mathbf{a}\times\mathbf{b} + \mathbf{b}\times\mathbf{c} + \mathbf{c}\times\mathbf{a}. \\
\text{Then }\mathbf{r}\cdot\mathbf{n} = \mathbf{a}\cdot\mathbf{n} . \\ \\
[(1-\lambda -\mu )\mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c}]\cdot(\mathbf{a}\times\mathbf{b} + \mathbf{b}\times\mathbf{c} + \mathbf{c}\times\mathbf{a}) = \mathbf{a}\cdot (\mathbf{a}\times\mathbf{b} + \mathbf{b}\times\mathbf{c} + \mathbf{c}\times\mathbf{a})\\
(1-\lambda -\mu )\mathbf{a}\cdot\mathbf{b}\times\mathbf{c} + \lambda\mathbf{b}\cdot\mathbf{c}\times\mathbf{a} + \mu\mathbf{c}\cdot\mathbf{a}\times\mathbf{b} = \mathbf{a}\cdot\mathbf{b}\times\mathbf{c} \text{ and first result follows}. \\
\bigg[\mathbf{r}\cdot\mathbf{n} =\bigg] \mathbf{a}\cdot\mathbf{n} = \mathbf{b}\cdot\mathbf{n} = \mathbf{c}\cdot\mathbf{n} \\
\mathbf{a}\cdot(\mathbf{a}\times\mathbf{b} + \mathbf{b}\times\mathbf{c} + \mathbf{c}\times\mathbf{a}) = \mathbf{b}\cdot(\mathbf{a}\times\mathbf{b} + \mathbf{b}\times\mathbf{c} + \mathbf{c}\times\mathbf{a}) = \mathbf{c}\cdot(\mathbf{a}\times\mathbf{b} + \mathbf{b}\times\mathbf{c} + \mathbf{c}\times\mathbf{a} ). \\
\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}) = \mathbf{b}\cdot(\mathbf{c}\times\mathbf{a}) = \mathbf{c}\cdot(\mathbf{a}\times\mathbf{b}) \text{ as the dot product of two perpendicular vectors is }\mathbf{0}. \\
A, B, C, O \text{ all in same plane }\Rightarrow \mathbf{a}\cdot (\mathbf{b}\times\mathbf{c}) = \mathbf{b}\cdot (\mathbf{c}\times\mathbf{a}) = \mathbf{c}\cdot (\mathbf{a}\times\mathbf{b}) = \mathbf{0}.")
