Question
Download Solution PDFसदिश n = 5i + 3j + 8k को x - अक्ष के चारों ओर 90° के एक कोण से घुमाइए। तो घूर्णित सदिश (Hn) क्या होगी?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसंकल्पना:
X - अक्ष के चारों ओर घूर्णन आव्यूह निम्न द्वारा ज्ञात किया जाता है
R(x, θ) = \(\left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&{\cos\theta }&{ - \sin\theta }\\ 0&{\sin\theta }&{\cos\theta } \end{array}} \right]\)
गणना:
R(x, 90°) = \(\left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&{\cos90}&{ - \sin90}\\ 0&{\sin90}&{\cos90} \end{array}} \right]\)
R(x, 90°) = \(\left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&0&{ - 1}\\ 0&1&0 \end{array}} \right]\)
अब, सदिश n = 5i + 3j + 8k के चारों ओर घुमाने पर
= \(\left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&0&{ - 1}\\ 0&1&0 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 5\\ 3\\ 8 \end{array}} \right]\)
= \(\left[ {\begin{array}{*{20}{c}} 5\\ { - 8}\\ 3 \end{array}} \right]\)
= \(\left[ {\begin{array}{*{20}{c}} 5\\ -8\\ { 3}\\ 1 \end{array}} \right]\)
Last updated on Jul 2, 2025
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