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NDA 2010
Exam
NDA
Year
2010
Subject
Mathematics
Chapter
Vector Algebra

NDA 2010 · Mathematics · Vector Algebra

aligned & If & a = i - k & b =x i + j +(1-x) k & c =y i +x j...

 If a⃗=i^−k^b⃗=xi^+j^+(1−x)k^c⃗=yi^+xj^+(1+x−y)k^\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 a⃗⋅(b⃗×c⃗)\vec{a} \cdot(\vec{b} \times \vec{c}) depends on

Options

  1. Axx only
  2. Byy only
  3. Cboth xx and yy
  4. Dneither xx nor yy

Correct answer

D. neither xx nor yy

Step-by-step solution

a⃗⋅b⃗×c⃗\vec{a} \cdot \vec{b} \times \vec{c} Now, \vec{b} \times \vec{c}=\left|i^j^k^x11−xyx(1+x−y)\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ x & 1 & 1-x \\ y & x & (1+x-y)\end{array}\right|=i^(1+x−y−x+x2)−j^(x+x2−xy−y+xy)+k^(x2−y)\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)−j^(x+x2−y)-\hat{j} \left(x+x²-y\right)+\hatk(x2−y)hat{k}\left(x^2-y\right) Now a⃗⋅(b⃗×$vecc)\vec{a} \cdot(\vec{b} \times \$ vec{c})=(i^−k^){ hati(1−y+x2)−j^(x+x2−y)+k^(x2−y)}=1−y+x2−(x2−y)=1−y+x2−x2+y=1\begin{aligned}& =(\hat{i}-\hat{k})\left\{\ hat{i}\left(1-y+x²\right)-\hat{j}\left(x+x²-y\right)+\hat{k}\left(x^2-y\right)\right\} \\ =1-y+x²-\left(x^2-y\right)\\ & =1-y+x²-x²+y \\ & =1 \end{aligned}$$
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