Find the values of $c$ for which the vectors $\vec a=(c\log_2x) \hat i-6\hat j+3\hat k$ and $\vec b=(\log_2x)\hat i-2\hat j+(2c\log_2x)\hat k$ make an obtuse angle for any $x {\epsilon} (0,\infty)$

I found out the dot product, and found $cos\theta<0$, whereby I got a quadratic inequality. Since, the inequality was less than zero, I used the discriminant as not having any solution to find the value of $c \epsilon (0,\frac{4}{3})$. However, the answer did not match, and I am a bit confused. Please help.

  • $\begingroup$ What is the answer? $\endgroup$ – Teddy38 Oct 10 '17 at 12:51
  • $\begingroup$ The answer is $(-\frac{4}{3},0)$ $\endgroup$ – Aakhi Chatterjee Oct 10 '17 at 12:52
  • $\begingroup$ I think the answer is wrong. If we take $c=-1$, which is in the region you gave, then we have $$\cos\theta=\frac{-[\log_2x]^2-6[\log_2x]+12}{\text{something positive}}.$$ But the quadratic $-Y^2-6Y+12=0$ has distinct roots, which means the quadratic can take positive values. Something's not right... $\endgroup$ – Teddy38 Oct 10 '17 at 12:58
  • $\begingroup$ I feel the same, that's the reason I am a bit confused. $\endgroup$ – Aakhi Chatterjee Oct 10 '17 at 12:58
  • $\begingroup$ Yeah, unless I've made a silly mistake I think the answer is wrong. $\endgroup$ – Teddy38 Oct 10 '17 at 12:59

we have $$\cos(\phi)=\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}$$ so we get $$\vec{a}\cdot \vec{b}=c(\log_2 x)^2+12+6c\log_2 x$$ can you finish? the discriminat is $$9-\frac{12}{c}$$ it must be $$9-\frac{12}{c}<0$$ and we get $$0<c<\frac{4}{3}$$ as you stated

  • $\begingroup$ I got this exactly, and since $cos\phi<0$ , I solved the discriminant and my solution for $c$ came out to be $(0,\frac{4}{3})$. P.S. I took $D<0$ $\endgroup$ – Aakhi Chatterjee Oct 10 '17 at 12:57
  • $\begingroup$ you solved it for $c$? $\endgroup$ – Dr. Sonnhard Graubner Oct 10 '17 at 12:58
  • $\begingroup$ Yes sir. I did, and found the above answer. I would request you to verify if you are getting the same as me. $\endgroup$ – Aakhi Chatterjee Oct 10 '17 at 12:59
  • $\begingroup$ So, I guess the given answer $-\frac{4}{3}<c<0$ is incorrect? $\endgroup$ – Aakhi Chatterjee Oct 10 '17 at 13:06
  • $\begingroup$ if we assume $$c<0$$ so we get $$c>\frac{4}{3}$$ this is impossible, yes $\endgroup$ – Dr. Sonnhard Graubner Oct 10 '17 at 13:07

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