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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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If a, b & c are the lengths of the sides of a triangle, using vector method, show that its area is $\sqrt {s(s-a)(s-b)(s-c)}$ where $2s=a+b+c$

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Toolbox:
  • Area of $\Delta ABC=\frac{1}{2}|\overrightarrow {BC}\times\overrightarrow {BA}|$
  • $\large|\overrightarrow a\times\overrightarrow c|=|\overrightarrow a||\overrightarrow b|sin\theta$
  • $\large|\overrightarrow a.\overrightarrow c|=|\overrightarrow a||\overrightarrow b|cos\theta$
  • $\large|\overrightarrow a+\overrightarrow c|^2=|\overrightarrow a|^2+\overrightarrow c|^2+2\overrightarrow a.\overrightarrow c$
  • $\large2s=a+b+c$
  • $\large\:x^2-y^2=(x+y)(x-y)$
  • $\large(x+y)^2=x^2+y^2+2xy$
  • $\large(x-y)^2=x^2+y^2-2xy$
Let ABC be $\Delta$ and $\large\overrightarrow {AB}=\overrightarrow c$, $\large\overrightarrow {BC}=\overrightarrow a$ and $\large\overrightarrow {CA}=\overrightarrow b$
$\Rightarrow \:\overrightarrow a+\overrightarrow b+\overrightarrow c=\overrightarrow 0$
We know that Area of $\Delta ABC=\frac{1}{2}|\overrightarrow {BC}\times\overrightarrow {BA}|$ $=\frac{1}{2}|\overrightarrow a\times\overrightarrow c|$
Squaring both the sides we get
$4\Delta^2=\large|\overrightarrow a\times\overrightarrow c|^2$
We know that $|\overrightarrow a\times\overrightarrow c|=|\overrightarrow a||\overrightarrow c|sin\theta$ where $\theta$ is angle between $\overrightarrow a\:and\:\overrightarrow c$
$4\Delta^2=|\overrightarrow a|^2|\overrightarrow c|^2sin^2(\pi-B)$
We know that $sin^2(\pi-B)=1-cos^2(\pi-B)$ and given that$ |\overrightarrow a|=a,\:|\overrightarrow b|=b\:and\:|\overrightarrow c|=c$
$\Rightarrow 4\Delta^2=a^2c^2(1-cos^2(\pi-B$))
$=\Rightarrow\:4\Delta^2=a^2c^2-a^2c^2cos^2(\pi-B)$
But we know that $|\overrightarrow a.\overrightarrow c|=|\overrightarrow a||\overrightarrow c|cos\theta$
$i.e.,(\overrightarrow a.\overrightarrow c)^2=a^2c^2cos^2(\pi-B)$
$\Rightarrow \:4\Delta^2=a^2c^2-(\overrightarrow a.\overrightarrow c)^2$
Multiplying both the sides by 4 we get
$\Rightarrow \:16\Delta^2=4a^2c^2-(2\overrightarrow a.\overrightarrow c)^2$
We know that $|\overrightarrow a+\overrightarrow c|^2=|\overrightarrow a|^2+\overrightarrow c|^2+2\overrightarrow a.\overrightarrow c$
$\Rightarrow\: 2\overrightarrow a.\overrightarrow c=|\overrightarrow a+\overrightarrow c|^2-|\overrightarrow a|^2-|\overrightarrow c|^2|$
Substituting the value of $ 2\overrightarrow a.\overrightarrow b$
$\Rightarrow \:16\Delta^2=4a^2c^2-(|\overrightarrow a+\overrightarrow c|^2-|\overrightarrow a|^2-|\overrightarrow c|^2|)^2$
But we know that $\overrightarrow a+\overrightarrow b+\overrightarrow c=\overrightarrow 0$
or $|\overrightarrow a+\overrightarrow c|=|-\overrightarrow b|=|\overrightarrow b|$
Substituting the value of $|\overrightarrow a+\overrightarrow c|$ we get
$\Rightarrow\:16\Delta^2=4a^2c^2-(|\overrightarrow b|^2-|\overrightarrow a|^2-|\overrightarrow c|^2|)^2$
$\Rightarrow\:16\Delta^2=(2ac)^2-(b^2-a^2-c^2)^2$
Using the formula $\large\:x^2-y^2=(x+y)(x-y)$ we get
$\Rightarrow\:16\Delta^2=(2ac+b^2-a^2-c^2)(2ac-b^2+a^2+c^2)$
$=(b^2-(a^2+c^2-2ac))((a^2+c^2+2ac)-b^2)$
But we know that $a^2+c^2-2ac=(a-c)^2$ and $a^2+c^2+2ac=(a+c)^2$
$\Rightarrow\:16\Delta^2=(b^2-(a-c)^2)((a+c)^2-b^2)$
Again using the formula $\large\:x^2-y^2=(x+y)(x-y)$ we get
$(b^2-(a-c)^2)=(b+a-c)(b-a+c) $ and
$( (a+c)^2-b^2)=(a+c+b)(a+c-b)$
$\Rightarrow\:16\Delta^2=(b+a-c)(b-a+c)(a+c+b)(a+c-b)$
Given $2s=a+b+c$ $\Rightarrow\:(a+b-c)=2s-2c$
$(a+c-b)=2s-2b$ and $(b+c-a)=2s-2a$
$=(2s-2c)(2s-2a)(2s(-2b)(2s)$
$\Rightarrow\:16\Delta^2=16s(s-a)(s-b)(s-c).$
$\Rightarrow \: Area\:of\: triangle\:\Delta=\sqrt{s(s-a)(s-b)(s-c)}.$
Hence proved.

 

answered Apr 6, 2013 by rvidyagovindarajan_1
edited Apr 6, 2013 by rvidyagovindarajan_1
 

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