Question

If the angle between a = \(2y^2 \hat{i} + 4y \hat{j} + \hat{k}\)  and  b = \(7\hat{i} - 2\hat{j} + y\hat{k}\)  is obtuse, then:

• A. \(0 < y < \frac{1}{2}\)
• B. \(-1 < y < -\frac{1}{2}\)
• C. \(\frac{1}{2} < y < 1\)
• D. \(-\frac{1}{2} < y < 0\)

Answer 1

The Correct Option is A

The objective is to find the range of \(y\) for which the angle between vectors \( \mathbf{a} = 2y^2 \hat{i} + 4y \hat{j} + \hat{k} \) and \( \mathbf{b} = 7\hat{i} - 2\hat{j} + y\hat{k} \) is obtuse. An obtuse angle is characterized by a negative dot product of the vectors.

The dot product is computed as follows: \[ \mathbf{a} \cdot \mathbf{b} = (2y^2)(7) + (4y)(-2) + (1)(y) \] \[ = 14y^2 - 8y + y \] \[ = 14y^2 - 7y \]

For the angle to be obtuse, the dot product must be negative: \[ 14y^2 - 7y < 0 \]

Factoring out \(y\) yields: \[ 7y(2y - 1) < 0 \]

The critical points where the expression's sign changes are \(y = 0\) and \(y = \frac{1}{2}\). Analyzing the intervals defined by these points:

• For \( (-\infty, 0) \), the expression \(7y(2y-1) > 0\).
• For \( (0, \frac{1}{2}) \), the expression \(7y(2y-1) < 0\).
• For \( (\frac{1}{2}, \infty) \), the expression \(7y(2y-1) > 0\).

Therefore, the condition for an obtuse angle is met when \(0 < y < \frac{1}{2}\).