Question

If the vectors $\vec{a} = c (\log_7 x) \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b} = (\log_7 x) \hat{i} + 3c (\log_7 x) \hat{j} - 4\hat{k}$ make obtuse angle for any x > 0, then c belongs to ______.

• A. (0, 3/4)
• B. (-3/4, 0)
• C. (-4/3, 0)
• D. (0, 4/3)

Hint:

When an inequality like $Ax^2 + Bx + C > 0$ must hold for ALL real $x$, the Discriminant ($\Delta = B^2 - 4AC$) is ALWAYS strictly negative. The sign of $A$ dictates the direction ($A > 0$ for always positive, $A < 0$ for always negative).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given two vectors containing a logarithmic variable $x$ and an unknown constant $c$. The angle between them is always obtuse regardless of the value of $x > 0$. We need to find the valid range for $c$.

Step 2: Key Formula or Approach:
If the angle $\theta$ between two vectors is obtuse ($90^\circ < \theta < 180^\circ$), their dot product must be strictly negative.
$$\vec{a} \cdot \vec{b} < 0$$
This condition must hold true for all possible values of the variable. We then apply quadratic inequality rules.

Step 3: Detailed Explanation:
First, calculate the dot product $\vec{a} \cdot \vec{b}$:
$$\vec{a} \cdot \vec{b} = [c (\log_7 x)][(\log_7 x)] + [2][3c (\log_7 x)] + [3][-4]$$
$$\vec{a} \cdot \vec{b} = c(\log_7 x)^2 + 6c(\log_7 x) - 12$$
We require this dot product to be strictly less than 0 for all $x > 0$.
Let $t = \log_7 x$. Because $x > 0$, the logarithmic term $t$ can take any real value ($t \in \mathbb{R}$).
The inequality becomes a standard quadratic in $t$:
$$ct^2 + 6ct - 12 < 0 \quad \text{for all } t \in \mathbb{R}$$
For a quadratic expression $At^2 + Bt + C < 0$ to be true for all real values, its parabola must open entirely downwards and never touch the axis. This requires two conditions:
1. Leading coefficient must be negative: $A < 0 \implies c < 0$.
2. Discriminant must be negative: $\Delta = B^2 - 4AC < 0$.
Calculate the discriminant:
$$(6c)^2 - 4(c)(-12) < 0$$
$$36c^2 + 48c < 0$$
Divide by 12:
$$3c^2 + 4c < 0$$
$$c(3c + 4) < 0$$
The roots of the corresponding equation are $c = 0$ and $c = -4/3$.
Using the wavy curve method for the inequality, $c$ must lie strictly between the roots:
$$-4/3 < c < 0$$
This range also perfectly satisfies the first condition ($c < 0$).

Step 4: Final Answer:
The constant $c$ belongs to the interval $(-4/3, 0)$, matching option (c).