Question

The values of x for which the angle between $\vec{a}=2x^{2}\hat{i}+4x\hat{j}+\hat{k}$ and $\vec{b}=7\hat{i}-2\hat{j}+x\hat{k}$ is obtuse are}

• A. $x<0$
• B. $x>0$
• C. $0<x<1/2$
• D. $x>1/2$

Hint:

Acute: $\vec{a} \cdot \vec{b} > 0$; Right: $\vec{a} \cdot \vec{b} = 0$; Obtuse: $\vec{a} \cdot \vec{b} < 0$.

Answer 1

The Correct Option is C

Step 1: Concept
Angle $\theta$ is obtuse if $\cos \theta < 0$, which implies $\vec{a} \cdot \vec{b} < 0$.

Step 2: Analysis
$\vec{a} \cdot \vec{b} = (2x^2)(7) + (4x)(-2) + (1)(x) = 14x^2 - 8x + x = 14x^2 - 7x$.

Step 3: Calculation
$14x^2 - 7x < 0 \implies 7x(2x - 1) < 0$.
The roots are $x=0$ and $x=1/2$.
Using the interval method, the expression is negative for $x \in (0, 1/2)$.

Step 4: Conclusion
Hence, $0 < x < 1/2$. Final Answer: (C)