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

ધારો કે \(\overrightarrow{\mathrm{a}}=4 \hat{i}-\hat{j}+\hat{k}, \overrightarrow{\mathrm{b}}=11 \hat{i}-\hat{j}+\hat{k}\) અને \(\overrightarrow{\mathrm{c}}\) એવો સદીશ છે કે જેથી \((\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times(-2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}})\). જો \((2 \vec{a}+3 \vec{b}) \cdot \vec{c}=1670\) હોય, તો \(|\vec{c}|^2=\) ...........

  1. A \(1627\)
  2. B \(1618\)
  3. C \(1600\)
  4. D \(1609\)
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Correct Answer

(B) \(1618\)

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Detailed explanation

\( (\vec{a}+\vec{b}) \times \vec{c}-\vec{c} \times(-2 \vec{a}+3 \vec{b})=0 \) \( (\vec{a}+\vec{b}) \times \vec{c}+(-2 \vec{a}+3 \vec{b}) \times \vec{c}=0 \) \( \Rightarrow(\vec{a}+\vec{b})-2 \vec{a}+3 \vec{b}) \times \vec{c}=0 \)…
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