Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

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, $2,-3$ ) $\begin{array}{rl}\sin \theta & =\left|\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}\right|\\ & =\left|\frac{\frac{1}{6}\times 3+\frac{1}{4}\times 2+\frac{1}{3}(-3)}{\sqrt{\frac{1}{36}+\frac{1}{16}+\frac{1}{9}}\sqrt{9+4+9}}\right|=0\\ \Rightarrow \theta & =0^{\circ }\end{array}$