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JEE Mains · Maths · STD 12 - 10. vector algebra

ધારો કે \(\vec{a}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\) અને \(\overrightarrow{\mathrm{d}}=\vec{a} \times \overrightarrow{\mathrm{b}}\). જો \(\overrightarrow{\mathrm{c}}\) એવો સદિશ છે કે જેથી \(\vec{a} \cdot \overrightarrow{\mathrm{c}}=|\overrightarrow{\mathrm{c}}|\), \(|\overrightarrow{\mathrm{c}}-2 \vec{a}|^2=8\) અને \(\overrightarrow{\mathrm{d}}\) તથા \(\overrightarrow{\mathrm{c}}\) વચ્ચેનો ખૂણો \(\frac{\pi}{4}\) છે, તો \(|10-3 \overrightarrow{\mathrm{~b}} \cdot \overrightarrow{\mathrm{c}}|+|\overrightarrow{\mathrm{d}} \times \overrightarrow{\mathrm{c}}|^2\) = ___

  1. A 6
  2. B 7
  3. C 8
  4. D 9
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Correct Answer

(A) 6

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\begin{aligned} & \vec{a}=\hat{i}+\hat{j}+\hat{k} \\ & \vec{b}=2 \hat{i}+2 \hat{j}+\hat{k} \\ & \vec{d}=\vec{a} \times \vec{b} \\ & =-\hat{i}+\hat{j} \\ & |\vec{c}-2 \vec{a}|^2=8 \\ & |\vec{c}|^2+4|\vec{a}|^2-4 \vec{a} \cdot \vec{c}=8 \\ & |\vec{c}|^2+12-4|\vec{c}|=8 \\ &…

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