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 Concept:
Two vectors make an obtuse angle if their dot product is negative ($\vec{a} \cdot \vec{b} < 0$).

Step 2: Formula Application:
$\vec{a} \cdot \vec{b} = [c(\log_7 x)][\log_7 x] + (2)[3c(\log_7 x)] + (3)(-4)$. Let $t = \log_7 x$. The condition is: $ct^2 + 6ct - 12 < 0$.

Step 3: Explanation:
For a quadratic $at^2 + bt + d$ to be always negative, we must have $a < 0$ and Discriminant $D < 0$. 1. $c < 0$. 2. $D = (6c)^2 - 4(c)(-12) = 36c^2 + 48c < 0$. $12c(3c + 4) < 0$. Since $c < 0$, we must have $3c + 4 > 0 \implies c > -4/3$. Thus, $c \in (-4/3, 0)$.

Step 4: Final Answer:
The interval for $c$ is (-4/3, 0).