Question
Download Solution PDFa = 2î − 3ĵ − k̂, b = −î + k̂, c = 2ĵ − k̂ అయితే, (a+b) మరియు (b+c) లను వికర్ణాలుగా కలిగిన సమాంతర చతుర్భుజం యొక్క వైశాల్యం ఎంత?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFఉపయోగించిన భావన:
సమాంతర చతుర్భుజం యొక్క వికర్ణాలు \(\vec{d}_1\) మరియు \(\vec{d}_2\) అయితే, సమాంతర చతుర్భుజం యొక్క వైశాల్యం \(\frac{1}{2}\left|\vec{d}_1 \times \vec{d}_2\right|\)
గణన:
\(\vec{a}+\vec{b}=2 \hat{i}-3 \hat{j}-\hat{k}-\hat{i}+\hat{k} \)
\( \Rightarrow \vec{a}+\vec{b}=\hat{i}-3 \hat{j} \)
మరియు \(\vec{b}+\vec{c}=-\hat{i}+\hat{k}+2 \hat{j}-\hat{k} \)
\( \Rightarrow \vec{b}+\vec{c}=-\hat{i}+2 \hat{j} \)
సమాంతర చతుర్భుజం యొక్క వైశాల్యం \(=\frac{1}{2}|(\vec{a}+\vec{b}) \times(\vec{b}+\vec{c})| \)
\( \Rightarrow \text { area }=\frac{1}{2}\left|\begin{array}{ccc} \hat{2}& \hat{j}& \hat{k} \\ 1 & -3 & 0 \\ -1 & 2 & 0 \end{array}\right|\)
\( \Rightarrow \text { area }=\frac{1}{2}|\hat{i}(0-0)-j(0-0)+\hat{k}(2-3)| \)\( \Rightarrow \text { area }=\frac{1}{2}|-\hat{k}|=\frac{1}{2}\)
Last updated on Feb 14, 2024
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