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

Answer: (A)

$0 < y < \frac{1}{2}$

Answer 2

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:

$0 < y < \frac{1}{2}$