3 and 4 .Determinants and Matrices
normal

જો ત્રણ વાસ્તવિક સંખ્યાઓ  $p$ , $q$ , $r$ એ $\left[ {p\,\,q\,\,r} \right]\left[ {\begin{array}{*{20}{c}}
  2&p&q \\ 
  { - 3}&q&{ - p + r} \\ 
  {12}&r&{ - q + 3r} 
\end{array}} \right] = \left[ {5\,\,\,b\,\,c} \right]$ નું પાલન કરે છે તો $(b + c)$ ની ન્યૂનતમ કિમત મેળવો.

A

$\frac{{25}}{{157}}$

B

$\frac{{25}}{{49}}$

C

$\frac{{25 \times 271}}{{{{\left( {49} \right)}^2}}}$

D

$\frac{{25 \times 589}}{{{{\left( {157} \right)}^2}}}$

Solution

$2 p-3 q+12 r=5$

$\mathrm{b}=\mathrm{p}^{2}+\mathrm{q}^{2}+\mathrm{r}^{2}$

$c=p q-q p+q r-q r+3 r^{2}=3 r^{2}$

$b+c=p^{2}+q^{2}+4 r^{2}$

use : $(2\hat i – 3\hat j + 6\hat k) \cdot (p\hat i – q\hat j + 2r\hat k) \le \sqrt {{2^2} + {3^2} + {6^2}} \sqrt {{p^2} + {q^2} + 4{r^2}} $

$\frac{{25}}{{49}} \le {p^2} + {q^2} + 4{r^2}$

Standard 12
Mathematics

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