Question

Find the angle between Tangents \( T_{1}=2 i+2 j-3 k \) and \( T_{2}=-2 i+2 j-3 k \)

Answer 1

\(\vec T_1 = 2\hat i + 2\hat j - 3\hat k\)

\(\vec T_2 = -2\hat i + 2\hat j - 3\hat k\) 

\(|\vec T_1|=\sqrt{2^2+2^2+(-3)^2}\) = \(\sqrt{4+4+9}=\sqrt{17}\)

\(|\vec T_2|=\sqrt{-2^2+2^2+(-3)^2}\) = \(\sqrt{4+4+9}=\sqrt{17}\)

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\(\frac9{17}\).