Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The normal vectors to the given planes are $\overrightarrow{n_1}=\hat{i}-2\hat{j}+3\hat{k}$ and $\overrightarrow{n_2}=\hat{i}+\alpha \hat{j}+2\hat{k}$.

The angle $\theta$ between the planes is given by:

$\cos \theta =\frac{|\overrightarrow{n_1}\cdot \overrightarrow{n_2}|}{|\overrightarrow{n_1}||\overrightarrow{n_2}|}$

Given $\theta ={\cos }^{-1}\left(\frac{1}{14}\right)$, we have $\cos \theta =\frac{1}{14}$.

Substituting the values:

$\frac{1}{14}=\frac{|(1)(1)+(-2)(\alpha )+(3)(2)|}{\sqrt{1^2+(-2{)}^2+3^2}\sqrt{1^2+{\alpha }^2+2^2}}$

$\frac{1}{14}=\frac{|7-2\alpha |}{\sqrt{14}\sqrt{{\alpha }^2+5}}$

Squaring both sides:

$\frac{1}{196}=\frac{(7-2\alpha {)}^2}{14({\alpha }^2+5)}$

$\frac{1}{14}=\frac{49-28\alpha +4{\alpha }^2}{{\alpha }^2+5}$

${\alpha }^2+5=14(49-28\alpha +4{\alpha }^2)$

${\alpha }^2+5=686-392\alpha +56{\alpha }^2$

$55{\alpha }^2-392\alpha +681=0$

Let the roots of this quadratic equation be ${\alpha }_1$ and ${\alpha }_2$. The difference between the values of $\alpha$ is given by $\frac{\sqrt{D}}{a}$ where $D=b^2-4ac$.

$D=(-392{)}^2-4(55)(681)$

$D=153664-149820=3844$

$\sqrt{D}=\sqrt{3844}=62$

Therefore, the difference between the values of $\alpha$ is:

$|{\alpha }_1-{\alpha }_2|=\frac{62}{55}$

Answer: $\frac{62}{55}$