KCET · Maths · Vector Algebra
If \(\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}}\) are unit vectors along the positive direction of \(\mathrm{x}, \mathrm{y}\) and \(\mathrm{z}\)-axes, then a false statement in the following is
- A \(\sum \hat{\mathbf{i}} \times(\hat{\mathbf{j}}+\hat{\mathbf{k}})=\overrightarrow{\mathbf{0}}\)
- B \(\sum \hat{\mathbf{i}} \times(\hat{\mathbf{j}} \times \hat{\mathbf{k}})=\overrightarrow{\mathbf{0}}\)
- C \(\Sigma \hat{\mathbf{i}} \cdot(\hat{\mathbf{j}} \times \hat{\mathbf{k}})=\overrightarrow{\mathbf{0}}\)
- D \(\Sigma \hat{\mathbf{i}} \cdot(\hat{\mathbf{j}}+\hat{\mathbf{k}})=\overrightarrow{\mathbf{0}}\)
Answer & Solution
Correct Answer
(C) \(\Sigma \hat{\mathbf{i}} \cdot(\hat{\mathbf{j}} \times \hat{\mathbf{k}})=\overrightarrow{\mathbf{0}}\)
Step-by-step Solution
Detailed explanation
Given, \(\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}}\) are unit vectors along the positive direction of \(x, y\) and \(z\)-axes, then
(a) \(\Sigma \hat{\mathbf{i}} \times(\hat{\mathbf{j}}+\hat{\mathbf{k}})\)
\[
\begin{aligned}
&=\hat{\mathbf{i}} \times(\hat{\mathbf{j}}+\hat{\mathbf{k}})+\hat{\mathbf{j}} \times(\hat{\mathbf{k}}+\hat{\mathbf{i}})+\hat{\mathbf{k}} \times(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \\
&=\hat{\mathbf{k}}-\hat{\mathbf{j}}+\hat{\mathbf{i}}-\hat{\mathbf{k}}+\hat{\mathbf{j}}-\hat{\mathbf{i}} \\
&=0
\end{aligned}
\]
(b) \(\Sigma \hat{\mathbf{i}} \times(\hat{\mathbf{j}} \times \hat{\mathbf{k}})=\hat{\mathbf{i}} \times(\hat{\mathbf{j}} \times \hat{\mathbf{k}})+\hat{\mathbf{j}} \times(\hat{\mathbf{k}} \times \hat{\mathbf{i}})\)
\[
\begin{aligned}
&+\hat{\mathbf{k}} \times(\hat{\mathbf{i}} \times \hat{\mathbf{j}}) \\
&\times \hat{\mathbf{k}})
\end{aligned}
\]
\[
\begin{aligned}
&=(\hat{\mathbf{i}} \times \hat{\mathbf{i}})+(\hat{\mathbf{j}} \times \hat{\mathbf{j}})+(\hat{\mathbf{k}} \times \hat{\mathbf{k}}) \\
&=0+0+0 \\
&=0
\end{aligned}
\]
(c) \(\Sigma \hat{\mathbf{i}} \cdot(\hat{\mathbf{j}} \times \hat{\mathbf{k}})=\Sigma(\hat{\mathbf{i}} \cdot \hat{\mathbf{i}})=\Sigma(1)\)
\[
=1+1+1=3
\]
(d) \(\Sigma \hat{\mathbf{i}} \cdot(\hat{\mathbf{j}}+\hat{\mathbf{k}})=\Sigma(\hat{\mathbf{i}} \cdot \hat{\mathbf{j}}+\hat{\mathbf{i}} \cdot \hat{\mathbf{k}})\)
\[
=\Sigma(0+0)=0
\]
(a) \(\Sigma \hat{\mathbf{i}} \times(\hat{\mathbf{j}}+\hat{\mathbf{k}})\)
\[
\begin{aligned}
&=\hat{\mathbf{i}} \times(\hat{\mathbf{j}}+\hat{\mathbf{k}})+\hat{\mathbf{j}} \times(\hat{\mathbf{k}}+\hat{\mathbf{i}})+\hat{\mathbf{k}} \times(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \\
&=\hat{\mathbf{k}}-\hat{\mathbf{j}}+\hat{\mathbf{i}}-\hat{\mathbf{k}}+\hat{\mathbf{j}}-\hat{\mathbf{i}} \\
&=0
\end{aligned}
\]
(b) \(\Sigma \hat{\mathbf{i}} \times(\hat{\mathbf{j}} \times \hat{\mathbf{k}})=\hat{\mathbf{i}} \times(\hat{\mathbf{j}} \times \hat{\mathbf{k}})+\hat{\mathbf{j}} \times(\hat{\mathbf{k}} \times \hat{\mathbf{i}})\)
\[
\begin{aligned}
&+\hat{\mathbf{k}} \times(\hat{\mathbf{i}} \times \hat{\mathbf{j}}) \\
&\times \hat{\mathbf{k}})
\end{aligned}
\]
\[
\begin{aligned}
&=(\hat{\mathbf{i}} \times \hat{\mathbf{i}})+(\hat{\mathbf{j}} \times \hat{\mathbf{j}})+(\hat{\mathbf{k}} \times \hat{\mathbf{k}}) \\
&=0+0+0 \\
&=0
\end{aligned}
\]
(c) \(\Sigma \hat{\mathbf{i}} \cdot(\hat{\mathbf{j}} \times \hat{\mathbf{k}})=\Sigma(\hat{\mathbf{i}} \cdot \hat{\mathbf{i}})=\Sigma(1)\)
\[
=1+1+1=3
\]
(d) \(\Sigma \hat{\mathbf{i}} \cdot(\hat{\mathbf{j}}+\hat{\mathbf{k}})=\Sigma(\hat{\mathbf{i}} \cdot \hat{\mathbf{j}}+\hat{\mathbf{i}} \cdot \hat{\mathbf{k}})\)
\[
=\Sigma(0+0)=0
\]
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