Question

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

• A. \( 0 \text{ or } \frac{1}{2} \)
• B. \( x>\frac{1}{2} \)
• C. \( \left( 0, \frac{1}{2} \right) \)
• D. \( \left[ 0, \frac{1}{2} \right] \)

Hint:

The angle between two non-zero vectors \( \vec{a} \) and \( \vec{b} \) is obtuse if and only if their dot product \( \vec{a} \cdot \vec{b} \) is negative. Remember the formula for the dot product in terms of components: if \( \vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k} \) and \( \vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k} \), then \( \vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \). Solving the resulting inequality gives the range of the required variable.

Answer 1

The Correct Option is C

The angle \( \theta \) between two vectors \( \vec{a} \) and \( \vec{b} \) is obtuse if \( \cos \theta<0 \). This occurs when the dot product \( \vec{a} \cdot \vec{b}<0 \). The dot product of \( \vec{a} = 2x^2 \hat{i} + 4x \hat{j} + \hat{k} \) and \( \vec{b} = 7 \hat{i} - 2 \hat{j} + x \hat{k} \) is: $$\vec{a} \cdot \vec{b} = (2x^2)(7) + (4x)(-2) + (1)(x)$$ $$\vec{a} \cdot \vec{b} = 14x^2 - 8x + x$$ $$\vec{a} \cdot \vec{b} = 14x^2 - 7x$$ For the angle to be obtuse, we need \( \vec{a} \cdot \vec{b}<0 \): $$14x^2 - 7x<0$$ Factor out \( 7x \): $$7x(2x - 1)<0$$ The critical points are found by setting each factor to zero: $$7x = 0 \implies x = 0$$ $$2x - 1 = 0 \implies x = \frac{1}{2}$$ We analyze the sign of \( 7x(2x - 1) \) in the intervals \( (-\infty, 0) \), \( (0, \frac{1}{2}) \), and \( (\frac{1}{2}, \infty) \). \begin{itemize} \item For \( x<0 \) (e.g., \( x = -1 \)): \( 7(-1)(2(-1) - 1) = -7(-3) = 21>0 \) \item For \( 0<x<\frac{1}{2} \) (e.g., \( x = \frac{1}{4} \)): \( 7(\frac{1}{4})(2(\frac{1}{4}) - 1) = \frac{7}{4}(-\frac{1}{2}) = -\frac{7}{8}<0 \) \item For \( x>\frac{1}{2} \) (e.g., \( x = 1 \)): \( 7(1)(2(1) - 1) = 7(1) = 7>0 \) \end{itemize} The inequality \( 14x^2 - 7x<0 \) is satisfied when \( 0<x<\frac{1}{2} \). Therefore, the values of \( x \) for which the angle between the vectors is obtuse are in the interval \( \left( 0, \frac{1}{2} \right) \).