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 angle between two vectors \(\vec{a}\) and \(\vec{b}\) is obtuse if their dot product is negative, i.e., \(\vec{a} \cdot \vec{b} < 0\).

\[\vec{a} = 2y^{2}\hat{i} + 4y\hat{j} + \hat{k}, \quad \vec{b} = 7\hat{i} - 2\hat{j} + y\hat{k}\]

The dot product \(\vec{a} \cdot \vec{b}\) is:

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

Simplify each term:

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

For the angle to be obtuse, we require:

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

Factorize:

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

The critical points are \(y = 0\) and \(y = \frac{1}{2}\). Using a sign analysis for \(7y(2y - 1)\):

• For \(y \in (0, \frac{1}{2})\), \(7y > 0\) and \((2y - 1) < 0\), so the product is negative.
• For \(y < 0\) or \(y > \frac{1}{2}\), the product is non-negative.

Thus, the solution is:

\[0 < y < \frac{1}{2}\]

Hence, the correct answer is:

\[0 < y < \frac{1}{2}\]