\((3\vec{i}+4\vec{j}), (\vec{i}-\vec{j}+\vec{k})\) నాభిశ్రుతిల మధ్య కోణాన్ని వాటి నాభిశ్రుతి ఉత్పత్తిని ఉపయోగించి కనుగొనండి:

  1. \(\sin \theta = \frac{\sqrt{74}}{5\sqrt{3}}\)
  2. \(\sin \theta= \frac{74}{\sqrt{3}}\)
  3. \(\sin \theta = \frac{\sqrt{74}}{5}\)
  4. \(\frac{74}{5}\)

Answer (Detailed Solution Below)

Option 1 : \(\sin \theta = \frac{\sqrt{74}}{5\sqrt{3}}\)
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Detailed Solution

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భావన:

రెండు నాభిశ్రుతిలు క్రాస్/నాభిశ్రుతి లబ్ధం ఇలా నిర్వచించబడింది:

\({\rm{ \vec{A} \times \vec{B} = }}\left| {\rm{A}} \right|{\rm{ \times }}\left| {\rm{B}} \right|{\rm{ \times sin}}\;{\rm{\theta }} \times \rm ̂{n}\)

ఇక్కడ θ అనేది \({\rm{⃗ A}}\;{\rm{and}}\;{\rm{⃗ B}}\) మధ్య కోణం.

ఇక్కడ \(\rm ̂ n\) అనేది యూనిట్ నాభిశ్రుతి 

\(\rm \vec A = a_1̂ i +a_2̂ j+ a_3̂ k\) మరియు \(\rm \vec B = b_1̂ i +b_2̂ j+b_3 ̂ k\) అయితే, వాటి క్రాస్ ఉత్పత్తి:

\(\rm \vec A\times\vec B=\begin{vmatrix} \rm ̂ i & \rm ̂ j & \rm ̂ k \\ \rm a_1 & \rm a_2 & \rm a_3 \\ \rm b_1 & \rm b_2 & \rm b_3\end{vmatrix}\) .

లెక్కింపు:

వీలు,

\(\vec{a}\ =\ (3\vec{i}+4\vec{j})\)

\(\vec{b}\ =\ (\vec{i}-\vec{j}+\vec{k})\)

\(\rm \vec a\times\vec b=\begin{vmatrix} \rm ̂ i & \rm ̂ j & \rm ̂ k \\ \rm 3 & \rm 4 & \rm 0 \\ \rm 1 & \rm -1 & \rm 1\end{vmatrix}\)

= î(4 + 0) - ĵ (3 - 0) + k̂(- 3 - 4)

\((3\vec{i}+4\vec{j})\times (\vec{i}-\vec{j}+\vec{k}) = 4\vec{i}-3\vec{j}-7\vec{k}\)

ఇప్పుడు,

\(|4\vec{i}-3\vec{j}-7\vec{k}|=\ \sqrt{4^2\ +\ 3^2\ +\ 7^2}\)

\(|4\vec{i}-3\vec{j}-7\vec{k}|=\ \sqrt{74}\)

\(\Rightarrow \ \sqrt{74}\ =|\vec{a}||\vec{b}| \sin \theta\)

\(\Rightarrow \ \sqrt{74}\ = 5\sqrt{3}\sin \theta\)

అందువలన,

\(\sin \theta = \frac{\sqrt{74}}{5\sqrt{3}}\)

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