Question

If the angle between the planes $x - 2y + 3z - 5 = 0$ and $x + \alpha y + 2z + 7 = 0$ is $\cos^{-1}\left(\frac{1}{14}\right)$, then the difference between the values of $\alpha$ is

• A. $\frac{12}{11}$
• B. $\frac{62}{55}$
• C. $\frac{31}{11}$
• D. $\frac{8}{5}$

Hint:

The angle between two planes is the angle between their normals; set up the cosine equation and solve the resulting quadratic in alpha.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Normals are \(\vec n_1=(1,-2,3)\) and \(\vec n_2=(1,\alpha,2)\).
\(\cos\theta=\dfrac{|\vec n_1\cdot\vec n_2|}{|\vec n_1||\vec n_2|}\)

Step 2: Set up
\(\vec n_1\cdot\vec n_2=7-2\alpha\), \(|\vec n_1|=\sqrt{14}\), \(|\vec n_2|=\sqrt{5+\alpha^2}\).
\(\dfrac{|7-2\alpha|}{\sqrt{14}\sqrt{5+\alpha^2}}=\dfrac{1}{14}\)

Step 3: Calculation
Squaring: \(196(7-2\alpha)^2=14(5+\alpha^2)\)
\(14(49-28\alpha+4\alpha^2)=5+\alpha^2\)
\(55\alpha^2-392\alpha+681=0\)

Step 4: Difference of roots
\(|\alpha_1-\alpha_2|=\dfrac{\sqrt{392^2-4\cdot55\cdot681}}{55}=\dfrac{\sqrt{3844}}{55}=\dfrac{62}{55}\)
This is the second option.

Final Answer:
The difference between the two values of alpha is 62/55. \[ \boxed{\text{(B) }\dfrac{62}{55}} \]