Q. What is the angle between the line 6x = 4y = 3z and the plane 3x + 2y - 3z = 4 ?

3627 233 Three Dimensional Geometry Report Error

A

0

23%

B

$π /6$

23%

C

$π /3$

27%

D

$π /2$

27%

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 $(\frac{1}{6}, \frac{1}{4}, \frac{1}{3})$ and equation of the plane is 3x + 2y -3z - 4 = 0
If $θ$ be the angle between line and plane, then direction ratios of the normal to this plane is (3, 2, - 3)
$sin θ = ∣ ∣ \frac{a _{1} a _{2} + b _{1} b _{2} + c _{1} c _{2}}{a _{1}^{2} + b _{1}^{2} + c _{1}^{2} a _{2}^{2} + b _{2}^{2} + c _{2}^{2}} ∣ ∣$
$= ∣ ∣ \frac{\frac{1}{6} \times 3 + \frac{1}{4} \times 2 + \frac{1}{3} ( - 3 )}{\frac{1}{36} + \frac{1}{16} + \frac{1}{9} 9 + 4 + 9} ∣ ∣$
$= ∣ ∣ \frac{1 - 1}{\frac{1}{36} + \frac{1}{16} + \frac{1}{9} 22} ∣ ∣ = 0$
$\Rightarrow θ = 0^{\circ}$