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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the angle between the lines $r=(3\hat i+2\hat j+6\hat k) +\lambda(2\hat i+\hat j+2\hat k) \;and\; r=(2\hat i-5\hat k)+\mu(6\hat i+3\hat j+2\hat k)$

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  • Angle between two lines is given by,$ cos^{-1}\bigg( \large\frac{\overrightarrow b_1.\overrightarrow b_2}{|\overrightarrow b_1| |\overrightarrow b_2|}\bigg)$ where the two lines are $\overrightarrow r=\overrightarrow a_1+\lambda\overrightarrow b_1\:and\:\overrightarrow r=\overrightarrow a_2+\mu\overrightarrow b_2$
The vector equation of the given lines are
$\overrightarrow r=(3\hat i+2\hat j+6\hat k)+\lambda(2\hat i+\hat j+2\hat k)$ and
$\overrightarrow r=(2\hat i-5\hat k)+\mu (6\hat i+3\hat j+2\hat k)$
Here $ \overrightarrow b_1 = 2\hat i + \hat j + 2\hat k$ and
$ \overrightarrow b_2 = 6\hat i + 3\hat j + 2\hat k$
$|\overrightarrow b_1|=\sqrt{2^2+1^2+2^2}$
$\qquad =\sqrt {4+4+1}$
$\qquad =3$
$|\overrightarrow b_2|=\sqrt{6^2+3^2+2^2}$
$\qquad =\sqrt {36+9+4}$
$\qquad =7$
Angle between two lines is $\theta = cos^{-1}\bigg( \large\frac{\overrightarrow b_1.\overrightarrow b_2}{|\overrightarrow b_1| |\overrightarrow b_2|}\bigg)$
Therefore $\cos \theta=\large\frac{(2\hat i+\hat j+ 2\hat k).(6 \hat i+3 \hat j+2 \hat k)}{3 \times 7}$
Multiplying by dot product rule
$\cos \theta=\large\frac{12+3+4}{21}$
$\cos \theta=\large\frac{19}{21}$
Hence $\theta = cos^{-1} \bigg( \large\frac{19}{21} \bigg)$
answered Jun 10, 2013 by meena.p
 

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