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: Write the dot product.
$\vec a\cdot\vec b=(4)(1)+(-1)(\alpha)+(\alpha)(-4)=4-\alpha-4\alpha=4-5\alpha.$
Step 2: Find the lengths.
$|\vec a|=\sqrt{16+1+\alpha^2}=\sqrt{\alpha^2+17}$ and $|\vec b|=\sqrt{1+\alpha^2+16}=\sqrt{\alpha^2+17}.$ They are equal.
Step 3: Use the cosine formula.
The angle satisfies $\cos\theta=\dfrac{\vec a\cdot\vec b}{|\vec a||\vec b|}=\dfrac{4-5\alpha}{\alpha^2+17}=-\dfrac27.$
Step 4: Cross multiply.
$7(4-5\alpha)=-2(\alpha^2+17)$, so $28-35\alpha=-2\alpha^2-34.$ Bring all to one side: $2\alpha^2-35\alpha+62=0.$
Step 5: Solve the quadratic.
Factor: $(2\alpha-31)(\alpha-2)=0$, giving $\alpha=2$ or $\alpha=\dfrac{31}{2}.$ Since $\alpha_1<\alpha_2$, we take $\alpha_1=2$ and $\alpha_2=\dfrac{31}{2}.$
Step 6: Compute the asked value.
$\alpha_1+2\alpha_2=2+2\cdot\dfrac{31}{2}=2+31=33.$ \[ \boxed{33} \]