MCQ
જો $\hat{a}, \hat{b}, \hat{c}$ એ એકમ સદિશો હોય તો $\left | \hat{a}+\hat{b} \right |^2+\left | \hat{b}+\hat{c} \right |^2+\left | \hat{c}+\hat{a} \right |^2$ ની ન્યુનતમ કિમત મેળવો.
- A$1$
- B$3$
- C$9$
- D$12$
$=(\hat{a}+\hat{b}) \cdot (\hat{a}+\hat{b})$
$=|\hat{a}|^{2}+|\hat{b}|^{2}+2 \hat{a} . \hat{b}$
$=2+2 \hat{a} . \hat{b}$
Similarly , $|\hat{b}+\hat{c}|^{2}=2+2 \hat{b} \cdot \hat{c}$
and $|\hat{a}+\hat{c}|^{2}=2+2 \hat{c} . \hat{a}$
Given that
$\Rightarrow|\hat{a}+\hat{b}|^{2}+|\hat{b}+\hat{c}|^{2}+|\hat{a}+\hat{c}|^{2}=6+2(\hat{a} \cdot \hat{b}+\hat{b} \cdot \hat{c}+\hat{c}, \hat{a}) \ldots . .(1)$
Now, $|\widehat{a}+\hat{b}+\hat{c}|^{2}=|\widehat{a}|^{2}+|\hat{b}|^{2}+|\hat{c}|^{2}+2(\hat{a} \cdot \hat{b}+\hat{b} \cdot \hat{c}+\hat{c}, \hat{a})$
$\Rightarrow|\widehat{a}+\hat{b}+\hat{c}|^{2}=3+2(\hat{a} \cdot \hat{b}+\hat{b} \cdot \hat{c}+\hat{c} \cdot \hat{a})$
Hence, $|\widehat{a}+\widehat{b}+\hat{c}|^{2} \geq 0$
$\Rightarrow 3+2(\hat{a} . \hat{b}+\hat{b} \cdot \hat{c}+\hat{c} \hat{a}) \geq 0$
$\Rightarrow 2(\hat{a} . \hat{b}+\hat{b} \cdot \hat{c}+\hat{c} \cdot \hat{a}) \geq-3$
Using the above result in equation (1) , we get the minumum value of the expression in question as 3
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(જ્યાં $C$ એ સંક્લ્યકારક અચળાંક છે)