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

Let \(\vec{a}=4 \hat{i}-\hat{j}+\hat{k}, \vec{b}=11 \hat{i}-\hat{j}+\hat{k}\) and \(\vec{c}\) be a vector such that \((\vec{a}+\vec{b}) \times \vec{c}=\vec{c} \times(-2 \vec{a}+3 \vec{b}) \text {. }\) If \((2 \vec{a}+3 \vec{b}) \cdot \vec{c}=1670\), then \(|\vec{c}|^2\) is equal to :

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

Answer & Solution

Correct Answer

(B) \(1618\)

Step-by-step Solution

Detailed explanation

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