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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the angle between the lines whose direction ratios are $a, b, c$ and $b -c, c -a, a - b.$

$\begin{array}{1 1} 0 \\\theta=45^{\large \circ} \\ \theta=90^{\large \circ} \\ \theta=30^{\large \circ}\end{array} $

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1 Answer

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Toolbox:
  • The angle between two lines is $\cos\theta=\begin{vmatrix}\large\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}\end{vmatrix}$
Step 1:
The direction cosines given are $a,b,c$ are $(b-c),(c-a),(a-b)$
We know the angle between two lines is $\cos\theta=\begin{vmatrix}\large\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}\end{vmatrix}$
On substituting for $a_1,b_1,c_1$ and $a_2,b_2$ and $c_2$
$\Rightarrow \begin{vmatrix}\large\frac{a(b-c)+b(c-a)+c(a-b)}{\sqrt{a^2+b^2+c^2}\sqrt{(b-c)^2+(a-b)^2+(c-a)^2}}\end{vmatrix}$
Step 2:
On expanding we get
$\cos\theta=\begin{vmatrix}\large\frac{ab-ac+bc-ab+ca-bc}{\sqrt{a^2+b^2+c^2}\sqrt{(b-c)^2+(a-b)^2+(c-a)^2}}\end{vmatrix}$
$\Rightarrow 0$
$\theta=90^{\large\circ}$
Hence the angle between the lines is $90^{\large\circ}$
answered Jun 3, 2013 by sreemathi.v
 

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