Question

Let \[ \vec a=4\hat i-\hat j+\alpha\hat k \] and \[ \vec b=\hat i+\alpha\hat j-4\hat k \] be two vectors. If $\alpha_1,\alpha_2$ ($\alpha_1<\alpha_2$) are two different values of $\alpha$ such that \[ (\vec a,\vec b)=\cos^{-1}\left(-\frac{2}{7}\right), \] then \[ \alpha_1+2\alpha_2= \]

• A. $15$
• B. $24$
• C. $33$
• D. $52$

Hint:

When the magnitudes are equal, the cosine formula becomes much simpler.

Answer 1

The Correct Option is C

Step 1: Concept
Use the dot product formula involving the angle between vectors.

Step 2: Meaning
We have \[ \vec a\cdot\vec b = 4(1)+(-1)(\alpha)+\alpha(-4) = 4-5\alpha. \] Also, \[ |\vec a| = |\vec b| = \sqrt{\alpha^2+17}. \]

Step 3: Analysis
Since \[ \cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|} = -\frac{2}{7}, \] we get \[ \frac{4-5\alpha}{\alpha^2+17} = -\frac{2}{7}. \] Cross-multiplying, \[ 28-35\alpha = -2\alpha^2-34. \] Hence \[ 2\alpha^2-35\alpha+62=0. \] Factoring, \[ (2\alpha-31)(\alpha-2)=0. \] Thus \[ \alpha_1=2, \qquad \alpha_2=\frac{31}{2}. \] Therefore, \[ \alpha_1+2\alpha_2 = 2+31 = 33. \]

Step 4: Conclusion
Hence the required value is $33$.

Final Answer: (C)