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MHT CET · Maths · Vector Algebra

\(\bar{a}, \overline{\mathrm{~b}}, \overline{\mathrm{c}}\) are nonzero vectors such that \(\bar{a}\) is perpendicular to \(\overline{\mathrm{b}}\) and \(\overline{\mathrm{c}}\), \(|\bar{a}|=1,|\bar{b}|=2,|\bar{c}|=1\) and \(\bar{b} \cdot \bar{c}=1\). There is nonzero vector \(\bar{d}\) coplanar with \(\bar{a}+\overline{\mathrm{b}}\) and \(2 \overline{\mathrm{~b}}-\overline{\mathrm{c}}\). If \(\overline{\mathrm{d}} \cdot \bar{a}=1\), then \(|\overline{\mathrm{d}}|^2=\)
(Note that \(x\) and \(y\) are parameters involved when we write \(d=x(a+b)+y(2 b-c))\)

  1. A \(13 y^2+14 y+5\)
  2. B \(y^2+14 y+5\)
  3. C \(y^2-14 y-5\)
  4. D \(y^2-14 y+5\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(13 y^2+14 y+5\)

Step-by-step Solution

Detailed explanation

\( \bar{d} = x(\bar{a} + \overline{b}) + y(2\overline{b} - \overline{c}) = x\bar{a} + (x+2y)\overline{b} - y\overline{c} \) \( \overline{d} \cdot \bar{a} = 1 \)