Question

The values of $x$ for which the angle between the vectors $\overrightarrow{a} = x\hat{i}-3\hat{j}-\hat{k}$ and $\overrightarrow{b} = 2x\hat{i} + x\hat{j} -\hat{k}$ is acute and the angle between $\overrightarrow{b}$ and Y-axis lies between $\frac{\pi}{2}$ and $\pi$ are:

• A. -1
• B. $x > 0$
• C. 1
• D. $x < 0$

Answer 1

Answer: (A), (D)

The values of $x$ must satisfy $x < 0$. Option (D) represents this range. Option (A) is a specific value within this range. Therefore, both (A) and (D) are correct.

Answer 2

For the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ to be acute, their dot product must be positive: $\overrightarrow{a} \cdot \overrightarrow{b} = (x)(2x) + (-3)(x) + (-1)(-1) = 2x^2 - 3x + 1$. Setting $2x^2 - 3x + 1 > 0$, we factor it as $(2x-1)(x-1) > 0$. This inequality holds for $x < \frac{1}{2}$ or $x > 1$.

For the angle between $\overrightarrow{b}$ and the Y-axis (represented by $\hat{j}$) to be between $\frac{\pi}{2}$ and $\pi$, the cosine of the angle must be negative. The cosine is given by $\frac{\overrightarrow{b} \cdot \hat{j}}{|\overrightarrow{b}| |\hat{j}|}$. $\overrightarrow{b} \cdot \hat{j} = x$. $|\overrightarrow{b}| = \sqrt{(2x)^2 + x^2 + (-1)^2} = \sqrt{5x^2 + 1}$. So, $\cos \phi = \frac{x}{\sqrt{5x^2 + 1}}$. For $\frac{\pi}{2} < \phi < \pi$, $\cos \phi < 0$. Since $\sqrt{5x^2 + 1}$ is always positive, this requires $x < 0$.

Combining both conditions ($x < \frac{1}{2}$ or $x > 1$) and ($x < 0$), the intersection is $x < 0$. Option (A) $x = -1$ satisfies $x < 0$. Option (D) $x < 0$ represents the entire range of valid values.