Question

If the angle between the vectors $\vec{a}=x\hat{i}+3\hat{j}+\hat{k}$ and $\vec{b}=x\hat{i}-x\hat{j}+2\hat{k}$ is acute, then $x$ lies is

• A. $(-\infty,-1)\cup(1,\infty)$
• B. $(-\infty,-1)\cup(2,\infty)$
• C. $(-\infty,1)\cup(2,\infty)$
• D. $(-\infty,0)\cup(1,\infty)$
• E. $(-\infty,0)\cup(2,\infty)$

Hint:

Logic Tip: The "Wavy Curve Method" is perfect here. Mark roots 1 and 2 on a number line. Start with a '+' sign on the rightmost interval $(2, \infty)$, alternate to '-' for $(1, 2)$, and back to '+' for $(-\infty, 1)$. We want the regions $>0$ (the '+' regions).

Answer 1

The Correct Option is C

Concept:
The dot product of two vectors $\vec{a}$ and $\vec{b}$ is defined as $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$, where $\theta$ is the angle between them. If the angle $\theta$ is acute ($0 \le \theta<90^{\circ}$), then $\cos\theta>0$. Since magnitudes are always positive, this implies that their dot product must be strictly greater than zero ($\vec{a} \cdot \vec{b}>0$).
Step 1: Compute the dot product of the two vectors.
Given: $\vec{a} = x\hat{i} + 3\hat{j} + \hat{k}$ $\vec{b} = x\hat{i} - x\hat{j} + 2\hat{k}$ $$\vec{a} \cdot \vec{b} = (x)(x) + (3)(-x) + (1)(2)$$ $$\vec{a} \cdot \vec{b} = x^2 - 3x + 2$$
Step 2: Apply the condition for an acute angle.
Set the dot product greater than zero: $$x^2 - 3x + 2>0$$
Step 3: Factor the quadratic expression.
Find two numbers that multiply to $2$ and add to $-3$. These are $-1$ and $-2$: $$(x - 1)(x - 2)>0$$
Step 4: Determine the intervals.
The roots of the equation are $x = 1$ and $x = 2$. These divide the real number line into three intervals: $(-\infty, 1)$, $(1, 2)$, and $(2, \infty)$. Since the parabola $y = x^2 - 3x + 2$ opens upwards (positive leading coefficient), it is positive outside the roots and negative between them. Thus, the inequality $(x - 1)(x - 2)>0$ holds true when: $$x<1 \text{ or } x>2$$ In interval notation, this is: $$(-\infty, 1) \cup (2, \infty)$$