Let S be the set of all $\mathrm{a} \in \mathrm{R}$ for which the angle between the vectors $\overrightarrow{\mathrm{u}}=\mathrm{a}\left(\log _{\mathrm{e}} \mathrm{b}\right) \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{v}}=\left(\log _{\mathrm{e}} \mathrm{b}\right) \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \mathrm{a}\left(\log _{\mathrm{e}} \mathrm{b}\right) \hat{\mathrm{k}},(\mathrm{b}>1)$ is acute. Then S is equal to

(A) $\left(-\infty,-\frac{4}{3}\right)$

(C) $\left(-\frac{4}{3}, 0\right)$

(D) $\left(\frac{12}{7}, \infty\right)$

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Vector Algebra

Let S be the set of all

a \in R

for which the angle between the vectors

u = a (lo g_{e} b) \hat{i} - 6 \hat{j} + 3 \hat{k}

and

v = (lo g_{e} b) \hat{i} + 2 \hat{j} + 2 a (lo g_{e} b) \hat{k}, (b > 1)

is acute. Then S is equal to

Answer: (B)

For the angle between the vectors to be acute their dot product must stay positive, and here

u \cdot v = a (lo g_{e} b)^{2} + 6 a (lo g_{e} b) - 12

. Writing

t = lo g_{e} b

, which is strictly positive because

b > 1

, this becomes

a t^{2} + 6 a t - 12

, and the requirement is that it remain positive for every admissible

b

, that is for every

t > 0

. As

b

approaches

1

from above,

t \to 0^{+}

and the expression tends to

- 12

, which is negative whatever value

a

takes. Since the dot product is forced negative for values of

b

just larger than

1

, the angle can never stay acute throughout, so no real

a

qualifies and the set

S

is empty.

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Single correct option