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यदि \(\overrightarrow{p}=i-2j+3k\) तथा \(\overrightarrow{q}=3i+3j+k\), तब \(\left(\overrightarrow{p}-\overrightarrow{q}\right)^2\) बराबर है:

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  1. \(\overrightarrow{p}-\overrightarrow{q}\)
  2. \(\overrightarrow{p}+\overrightarrow{q}\)
  3. \(\left|\overrightarrow{p}\right|^2-\left|\overrightarrow{q}\right|^2\)
  4. \(\left(\overrightarrow{p}+\overrightarrow{q}\right)^2\)

Answer (Detailed Solution Below)

Option 4 : \(\left(\overrightarrow{p}+\overrightarrow{q}\right)^2\)
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प्रयुक्त संकल्पना:

किसी भी सदिश के लिए, \(\vec{r}\) 
\( \vec{r}^2=\vec{r} \cdot \vec{r}\) और 
\( \begin{aligned} &\hat{j} \cdot \hat{i}=\hat{j} \cdot \hat{j}=\hat{k} \cdot \hat{k}=1 \\ &\hat{i} \cdot \hat{j}=\hat{j} \cdot \hat{k}=\hat{k} \cdot \hat{i}=0 \end{aligned}\)
 

गणना:

\( \Rightarrow \vec{p}-\vec{q}=\hat{i}-3 \hat{i}-2 \hat{j}-3 \hat{j}+3 \hat{k}-\hat{k} \)
\(\Rightarrow \vec{p}-\vec{q}=-2 \hat{i}-5 \hat{j}+2 \hat{k} \)
\( \Rightarrow (\vec{p}-\vec{q})^2=(-2 \hat{i}-5 \hat{j}+2 \hat{k}) \cdot(-2 \hat{i}-5\hat{ j}+2 \hat{k}) \)
\(\Rightarrow (\vec{p}-\vec{q})^2=4+25+4=33\)
​\(\Rightarrow \vec{p}-\vec{q}=\hat{i}-3 \hat{i}-2 \hat{j}-3 \hat{j}+3 \hat{k}-\hat{k} \)​
 
\(\Rightarrow \vec{p}+\vec{q}=\hat{i}+3 \hat{i}-2 \hat{j}+3 \hat{j}+3 \hat{k}+\hat{k} \)
\(\Rightarrow \vec{p}+\vec{q}=4 \hat{i}+ \hat{j}+4 \hat{k} \)
\(\Rightarrow (\vec{p}+\vec{q})^2=(4 \hat{i}+ \hat{j}+4 \hat{k} ) \cdot(4 \hat{i}+ \hat{j}+4 \hat{k} ) \)
\(\Rightarrow (\vec{p}+\vec{q})^2=16+1+16=33\)
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