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Find the angle between Tangents \( T_{1}=2 i+2 j-3 k \) and \( T_{2}=-2 i+2 j-3 k \) |
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Answer» \(\vec T_1 = 2\hat i + 2\hat j - 3\hat k\) \(\vec T_2 = -2\hat i + 2\hat j - 3\hat k\) \(|\vec T_1|=\sqrt{2^2+2^2+(-3)^2}\) = \(\sqrt{4+4+9}=\sqrt{17}\) \(|\vec T_2|=\sqrt{-2^2+2^2+(-3)^2}\) = \(\sqrt{4+4+9}=\sqrt{17}\) Let angle between T1 and T2 is θ. \(\therefore\) θ = \(\frac{\vec T_1.\vec T_2}{|\vec T_1||\vec T_2|}\) = \(\frac{(2\hat i+2\hat j-3\hat k).(-2\hat i+2\hat j-3\hat k)}{\sqrt{17}{\sqrt{17}}}\) \(=\theta = cos^{-1}\frac1{17}\) Hence, angle between both tangents is cos-1\(\frac9{17}\). |
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