Question

Let a vector \[ \vec a=\sqrt{2}\,\hat i-\hat j+\lambda \hat k,\ \lambda>0, \] make an obtuse angle with the vector \[ \vec b=-\lambda^2\hat i+4\sqrt{2}\hat j+4\sqrt{2}\hat k \] and an angle \( \theta \), \(\frac{\pi}{6}<\theta<\frac{\pi}{2}\), with the positive \(z\)-axis. If the set of all possible values of \( \lambda \) is \((\alpha,\beta)-\{\gamma\}\), then\(\alpha+\beta+\gamma\) is equal to ____________. 

Hint:

Always apply dot-product conditions before angular constraints to simplify vector problems.

Answer 1

Correct Answer: 5

Step 1: Condition for obtuse angle.
For obtuse angle between \( \vec a \) and \( \vec b \): 
\[ \vec a\cdot\vec b<0 \] \[ (\sqrt2)(-\lambda^2)+(-1)(4\sqrt2)+\lambda(4\sqrt2)<0 \] \[ -\sqrt2\lambda^2-4\sqrt2+4\sqrt2\lambda<0 \] Dividing by \( \sqrt2 \): 
\[ -\lambda^2+4\lambda-4<0 \] \[ (\lambda-2)^2>0 \] Thus, 
\[ \lambda\neq 2 \]
Step 2: Angle with positive \(z\)-axis. 
\[ \cos\theta=\frac{\lambda}{\sqrt{2+1+\lambda^2}}=\frac{\lambda}{\sqrt{\lambda^2+3}} \] Given \( \frac{\pi}{6}<\theta<\frac{\pi}{2} \Rightarrow 0<\cos\theta<\frac{\sqrt3}{2} \). 
\[ 0<\frac{\lambda}{\sqrt{\lambda^2+3}}<\frac{\sqrt3}{2} \] Squaring: 
\[ \frac{\lambda^2}{\lambda^2+3}<\frac{3}{4} \] \[ 4\lambda^2<3\lambda^2+9 \Rightarrow \lambda^2<9 \Rightarrow 0<\lambda<3 \]
Step 3: Combine conditions. 
\[ \lambda\in(0,3)-\{2\} \] Thus, 
\[ \alpha=0,\ \beta=3,\ \gamma=2 \] \[ \alpha+\beta+\gamma=5 \]
Final Answer: 
\[ \boxed{5} \]