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Q.
What is the angle between the line 6x = 4y = 3z and the plane 3x + 2y - 3z = 4 ?
Three Dimensional Geometry
Solution:
The equation of the given line is 6x = 4y = 3z which is written in symmetric form as $\frac{x - 0}{1 /6} = \frac{y -0}{1/4} = \frac{z - 0}{ 1/3}$
Direction ratios of this line are $\left(\frac{1}{6} , \frac{1}{4} , \frac{1}{3}\right)$ and equation of the plane is 3x + 2y -3z - 4 = 0
If $\theta$ be the angle between line and plane, then direction ratios of the normal to this plane is (3, 2, - 3)
$\sin\theta = \left| \frac{a_{1}a_{2} +b_{1}b_{2} +c_{1}c_{2}}{\sqrt{a_{1}^{2} + b_{1}^{2} +c_{1}^{2}} \sqrt{a_{2}^{2} +b_{2}^{2} +c_{2}^{2}}}\right| $
$= \left|\frac{\frac{1}{6}\times3 + \frac{1}{4} \times2+ \frac{1}{3}\left(-3\right)}{\sqrt{\frac{1}{36} + \frac{1}{16} + \frac{1}{9}}\sqrt{9+4+9}} \right| $
$= \left|\frac{1-1}{\sqrt{\frac{1}{36} + \frac{1}{16} + \frac{1}{9}} \sqrt{22}}\right| = 0$
$ \Rightarrow \theta= 0^{\circ}$