Q174 / 270 · NDA 2010 (Phase 2) · Mathematics

$\begin{aligned} & \text { If } \\ & \vec{a}=\hat{i}-\hat{k} \\ & \vec{b}=x \hat{i}+\hat{j}+(1-x) \hat{k} \\ & \vec{c}=y \hat{i}+x \hat{j}+(1+x-y) \hat{k} \end{aligned}$ then $\vec{a} \cdot(\vec{b} \times \vec{c})$ depends on

  • A
    $x$ only
  • B
    $y$ only
  • C
    both $x$ and $y$
  • Correct answer
    D
    neither $x$ nor $y$

Solution

$\vec{a} \cdot \vec{b} \times \vec{c}$ Now, $\vec{b} \times \vec{c}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ x & 1 & 1-x \\ y & x & (1+x-y)\end{array}\right|$ $\begin{array}{r} =\hat{i}\left(1+x-y-x+x^2\right)-\hat{j}\left(x+x^2-x y-y+x y\right) \\ +\hat{k}\left(x^2-y\right) \end{array}$ $=\hat{i}\left(1-y+x^2\right)-\hat{j}\left(x+x^2-y\right)+\hat{k}\left(x^2-y\right)$ Now $\vec{a} \cdot(\vec{b} \times \vec{c})$ $\begin{aligned} & =(\hat{i}-\hat{k})\left\{\hat{i}\left(1-y+x^2\right)-\hat{j}\left(x+x^2-y\right)+\hat{k}\left(x^2-y\right)\right\} \\ & =1-y+x^2-\left(x^2-y\right) \\ & =1-y+x^2-x^2+y \\ & =1 \end{aligned}$

Hence, the correct answer is option: 'D'.

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