If $\quad \overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}},…

If $\quad \overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \quad \overrightarrow{\mathbf{c}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{d}}=\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}$, then observe the following
The correct match of List-I to List-II $\begin{array}{llll}\text { i } & \text { ii } & \text { iii } & \text { iv }\end{array}$
  1. $\begin{array}{llll}\text { C } & \text { A } & \text { B } & \text { F }\end{array}$
  2. $\begin{array}{llll}\text { C } & \text { A } & \text { F } & \text { E }\end{array}$
  3. $\begin{array}{llll}\text { A } & \text { C } & \text { B } & \text { F }\end{array}$
  4. $\begin{array}{llll}\text { A } & \text { C } & \text { F } & \text { D }\end{array}$

Solution

Given vectors are $ \begin{aligned} \overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \overrightarrow{\mathbf{b}} & =\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \overrightarrow{\mathbf{c}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}} \\ \text { and } \quad \overrightarrow{\mathbf{d}} & =\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}} \end{aligned} $ and $ \overrightarrow{\mathbf{d}}=\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}} $ $ \text { (i) Now, } \begin{aligned} \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} & =(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}) \\ & =1-1+1=1 \\ \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{d}} & =(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}) \\ & =1+1-1=1 \\ \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} & =\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{d}} \end{aligned} $ (ii) $ \text { (ii) Now, } \begin{aligned} \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}} & =(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}) \\ & =1-1-1=-1 \\ \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{d}} & =(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}) \\ & =1-1-1=-1 \\ \therefore \quad \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}} & =\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{d}} \end{aligned} $ $ \text { (iii) } \begin{aligned} {\left[\begin{array}{ll} \overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}} \end{array}\right] } & =\left|\begin{array}{rrr} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right| \\ & =1(1-1)-1(-1-1)+1(1+1) \\ & =0+2+2=4 \end{aligned} $ (iv) $ \begin{aligned} \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}} & =\left|\begin{array}{rrr} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right| \\ & =\hat{\mathbf{i}}(1-1)-\hat{\mathbf{j}}(-1-1)+\hat{\mathbf{k}}(1+1) \\ & =2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \end{aligned} $ Hence, option (2) is correct

Asked in: AP EAMCET 2008

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