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: Understanding the Question:
The angle between two vectors is obtuse if their dot product is negative ($\cos \theta < 0 \implies \vec{a} \cdot \vec{b} < 0$).

Step 2: Key Formula or Approach:
1. Dot product $\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3$.
2. Condition for obtuse angle: $\vec{a} \cdot \vec{b} < 0$.

Step 3: Detailed Explanation:
Given $\vec{a} = 2x^2 \hat{i} + 4x \hat{j} + \hat{k}$ and $\vec{b} = 7 \hat{i} - 2 \hat{j} + x \hat{k}$.
Compute the dot product:
\[ \vec{a} \cdot \vec{b} = (2x^2)(7) + (4x)(-2) + (1)(x) \] \[ \vec{a} \cdot \vec{b} = 14x^2 - 8x + x = 14x^2 - 7x \] For the angle to be obtuse:
\[ 14x^2 - 7x < 0 \] \[ 7x(2x - 1) < 0 \] This inequality holds when $x$ lies between the roots of the corresponding equation $7x(2x-1) = 0$.
The roots are $x = 0$ and $x = 1/2$.
Thus, the condition is $0 < x < \frac{1}{2}$.

Step 4: Final Answer:
The values of $x$ are $0 < x < \frac{1}{2}$.