Question
Download Solution PDFयदि \(\vec{a}=2 \hat{i}+5 \hat{j}+3 \hat{k}, \vec{b}\) = \(3 \hat{i}+3 \hat{j}+6 \hat{k}\) और \(\vec{c}=2 \hat{i}+7 \hat{j}+4 \hat{k}\), है, तो \(|(\vec{a}-\vec{b}) \times(\vec{c}-\vec{a})|\) का नाम ________ है।
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFस्पष्टीकरण:
\(\vec{a}=2 \hat{i}+5 \hat{j}+3 \hat{k}\), \(\vec{b}=3 \hat{i}+3 \hat{j}+6 \hat{k}\), \(\vec{c}=2 \hat{i}+7 \hat{j}+4 \hat{k}\)
इसलिए, \(\vec a-\vec{b}=2 \hat{i}+5 \hat{j}+3 \hat{k}-(3 \hat{i}+3 \hat{j}+6 \hat{k})\) = \(-\hat{i}+2 \hat{j}-3 \hat{k}\)
और \(\vec c-\vec{a}=2 \hat{i}+7 \hat{j}+4 \hat{k}-(2 \hat{i}+5 \hat{j}+3 \hat{k})\) = \(2 \hat{j}+ \hat{k}\)
इसलिए, \((\vec{a}-\vec{b}) \times(\vec{c}-\vec{a})\) = \(\begin{vmatrix}\hat i&\hat j&\hat k\\-1&2&-3\\0&2&1\end{vmatrix}\) = \(8\hat i+\hat j-2\hat k\)
अतः \(|(\vec{a}-\vec{b}) \times(\vec{c}-\vec{a})|\) = \(\sqrt{64+1+4}\) = \(\sqrt{69}\)
अतः (5) सही है।
Last updated on May 25, 2025
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