Question

The set of all real values of \( c \) so that the angle between the vectors
\( \vec{a} = c\hat{i} - 6\hat{j} + 3\hat{k} \) and \( \vec{b} = x\hat{i} + 2\hat{j} + 2c\hat{k} \) is an obtuse angle for all real \( x \), is:

• A. \( \left(0, \dfrac{4}{3} \right] \)
• B. \( \left(0, \dfrac{2}{3} \right] \)
• C. \( \left( -\dfrac{2}{3}, 0 \right) \)
• D. \( \left( -\dfrac{4}{3}, 0 \right) \)

Hint:

To ensure two vectors always form an obtuse angle, their dot product must be negative for all values. Use sign analysis on expressions involving parameters.

Answer 1

The Correct Option is D

Step 1: Use the dot product condition
Vectors \(\vec{a}\) and \(\vec{b}\) form an obtuse angle when \( \vec{a} . \vec{b}<0 \). \[ \vec{a} . \vec{b} = (c)(x) + (-6)(2) + (3)(2c) = cx - 12 + 6c \] Step 2: Require this expression to be negative for all real \( x \): \[ cx + 6c - 12<0
\text{for all } x \in \mathbb{R} \] Step 3: Analyze sign of linear expression in \( x \)
To ensure \( cx + 6c - 12<0 \) for all real \( x \), the linear function must always be negative. That can only happen if: - The coefficient of \( x \) (which is \( c \)) is zero \( \Rightarrow \) contradiction: expression becomes constant. - Or the coefficient \( c<0 \), and the expression is maximum at \( x \to -\infty \). Let’s find the maximum value of this function over all \( x \) when \( c<0 \). Since \( c<0 \), the function is decreasing. Maximum occurs at smallest \( x \Rightarrow x \to -\infty \), but to guarantee the expression always negative: Check discriminant of quadratic form: Ensure \( cx + 6c - 12<0 \Rightarrow \) function is always below x-axis. Step 4: Boundary of interval Set maximum value = 0 to find critical points: \[ \text{Let } cx + 6c - 12 = 0 \Rightarrow x = \frac{12 - 6c}{c} \] To make this never satisfied for real \( x \), the inequality \( cx + 6c - 12<0 \) must hold for all \( x \). It is known that such inequality \( cx + d<0 \) is always true for all \( x \) only when \( c<0 \) and the function's maximum value is negative. So: \[ \text{Maximum of } f(x) = cx + 6c - 12 \text{ occurs at } x = -\infty \Rightarrow f(x) \to -\infty \] But for edge case (transition point), let’s find values of \( c \) for which inequality holds for all \( x \): Let’s find when the expression crosses zero: \[ cx + 6c - 12 = 0 \Rightarrow x = \frac{12 - 6c}{c} \] This should not be real To ensure inequality is always<0, find range of \( c \) satisfying: \[ \text{Expression } cx + 6c - 12<0 \text{ for all } x \Rightarrow c<0 \text{ and } 6c - 12<0 \Rightarrow c<2 \] Now maximize \( f(x) \) at some extreme \( x \to \infty \) or \( x \to -\infty \). Pick specific values to test: Try \( c = -1 \Rightarrow f(x) = -x - 6 - 12 = -x - 18<0 \) — satisfied. Try \( c = -4/3 \Rightarrow f(x) = \left( -\dfrac{4}{3} \right)x + 6 . \left( -\dfrac{4}{3} \right) - 12 = -\dfrac{4}{3}x - 8 - 12 = -\dfrac{4}{3}x - 20 \) — always negative. Hence, solution: \( c \in \left( -\dfrac{4}{3}, 0 \right) \)