Question

The set of value of \(x\) for which the angle between the \(\overrightarrow a = 2x²\hat i + 4x \hat j + \hat k\) and \(\overrightarrow b =7\hat i-2\hat j + x\hat k\) is obtuse is :

• A. \((0,\frac{1}{2})\)
• B. \((0,\frac{1}{3})\)
• C. \((\frac{1}{2},\frac{1}{3})\)
• D. (0,1)

Answer 1

The Correct Option is A

To determine the set of values of \( x \) for which the angle between vectors \(\overrightarrow{a} = 2x^2 \hat{i} + 4x \hat{j} + \hat{k}\) and \(\overrightarrow{b} = 7 \hat{i} - 2 \hat{j} + x \hat{k}\) is obtuse, we use the dot product. Vectors form an obtuse angle when their dot product is less than zero. The dot product is given by:

\[ \overrightarrow{a} \cdot \overrightarrow{b} = (2x^2)(7) + (4x)(-2) + (1)(x) \]

Calculating the dot product:

\[ \overrightarrow{a} \cdot \overrightarrow{b} = 14x^2 - 8x + x = 14x^2 - 7x \]

For the angle to be obtuse:

\[ 14x^2 - 7x < 0 \]

Factor the inequality:

\[ 7x(2x - 1) < 0 \]

Critical points:

• \( 7x = 0 \Rightarrow x = 0 \)
• \( 2x - 1 = 0 \Rightarrow x = \frac{1}{2} \)

Test intervals around the critical points:

• For \( x \in (-\infty, 0) \): both \( 7x < 0 \) and \( 2x - 1 < 0 \), so product is positive.
• For \( x \in (0, \frac{1}{2}) \): \( 7x > 0 \) and \( 2x - 1 < 0 \), so product is negative.
• For \( x \in (\frac{1}{2}, \infty) \): both \( 7x > 0 \) and \( 2x - 1 > 0 \), so product is positive.

Thus, the set of values for which the angle is obtuse is:

\( x \in (0, \frac{1}{2}) \)