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3 and 4 .Determinants and Matrices
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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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