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COMEDK · Maths · 33. Vector Algebra

Let a, b, c be three vector such that \(a \neq 0\) and \(\vec{a} \times \vec{b}=2 \vec{a} \times \vec{c},|a|=|c|=1,|b|=4\) and \(|\vec{b} \times \vec{c}|=\sqrt{15}\). If \(\vec{b}-2 \vec{c}=\lambda \vec{a}\) then \(\lambda\) equals to

  1. A 1
  2. B \(-1\)
  3. C 2
  4. D \(-4\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(-4\)

Step-by-step Solution

Detailed explanation

Given \(\vec{a} \times \vec{b} = 2 \vec{a} \times \vec{c}\), we can rewrite this as \(\vec{a} \times (\vec{b} - 2 \vec{c}) = \vec{0}\). It is given that \(\vec{b} - 2 \vec{c} = \lambda \vec{a}\). Substituting this into the cross product equation, we get…