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COMEDK · Maths · 33. Vector Algebra

If \(\vec{a}, \vec{b}, \vec{c}\) are three vectors such that \(a \neq 0\), \(|\vec{a}| = |\vec{c}| = 1, |\vec{b}| = 4\) and \(|\vec{b} \times \vec{c}| = \sqrt{15}\). If \(\vec{b} - 2\vec{c} = \lambda\vec{a}\) then \(\lambda^2\) equals:

  1. A \(1\)
  2. B \(4\)
  3. C \(-4\)
  4. D \(24\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(24\)

Step-by-step Solution

Detailed explanation

Using Lagrange's identity: \(|\vec{b} \times \vec{c}|^2 + (\vec{b} \cdot \vec{c})^2 = |\vec{b}|^2|\vec{c}|^2\) \(15 + (\vec{b} \cdot \vec{c})^2 = 16 \Rightarrow \vec{b} \cdot \vec{c} = \pm 1\) From \(\vec{b} - 2\vec{c} = \lambda\vec{a}\), squaring both sides:…