‹ Back to the paper
Prove the following.
If A, B, C are three non-collinear points with position vectors a⃗, b⃗, c⃗, respectively, then show…
If $A, B, C$ are three non-collinear points with position vectors $\vec{a}, \vec{b}, \vec{c}$, respectively, then show that the length of the perpendicular from $C$ on $AB$ is $\frac{|(\vec{a} \times \vec{b}) + (\vec{b} \times \vec{c}) + (\vec{c} \times \vec{a})|}{|\vec{b} - \vec{a}|}$.
To show: $\frac{|(\vec{a} \times \vec{b}) + (\vec{b} \times \vec{c}) + (\vec{c} \times \vec{a})|}{|\vec{b} - \vec{a}|}$
Answer
Answer
Hence the length of the perpendicular from $C$ on $AB$ is $\frac{|(\vec a\times\vec b)+(\vec b\times\vec c)+(\vec c\times\vec a)|}{|\vec b-\vec a|}$.
- Let $CM$ be the perpendicular from $C$ on $AB$, of length $p$. Then $\overrightarrow{AB} = \vec b - \vec a$ and $\overrightarrow{AC} = \vec c - \vec a$
- Area of $\triangle ABC = \frac12|\overrightarrow{AB}\times\overrightarrow{AC}|$, by the cross-product area formula
- Also area of $\triangle ABC = \frac12\times AB\times p = \frac12|\vec b-\vec a|\,p$, by area $=\frac12$ base $\times$ height
- Equating: $p = \frac{|(\vec b-\vec a)\times(\vec c-\vec a)|}{|\vec b-\vec a|}$
- Expanding: $(\vec b-\vec a)\times(\vec c-\vec a) = \vec b\times\vec c - \vec b\times\vec a - \vec a\times\vec c + \vec a\times\vec a$, by the distributive law
- Using $\vec a\times\vec a = \vec 0$, $-\vec b\times\vec a = \vec a\times\vec b$ and $-\vec a\times\vec c = \vec c\times\vec a$: $(\vec b-\vec a)\times(\vec c-\vec a) = \vec a\times\vec b + \vec b\times\vec c + \vec c\times\vec a$
- Hence $p = \frac{|(\vec a\times\vec b)+(\vec b\times\vec c)+(\vec c\times\vec a)|}{|\vec b-\vec a|}$