Question

The lines l and m have vector equations
r=i+j+k+s(i−j+2k) and r=4i+6j+k+t(2i+2j+k)

(ii) Calculate the acute angle between the lines.

Answer 1

the direction vector for line l\[\vec{l}=\vec{i}-\vec{j}+2\vec{k}\]and for line m\[\vec{m}=2\vec{i} + 2\vec{j} +\vec{k}\]to find the angle use the dot product between the two vectors\[\cos \theta=\frac{\vec{l}.\vec{m}}{\left| \vec{l} \right|\left| \vec{m} \right|}\]

Answer 2

Can you show this in full? I need to go now though :/

Answer 3

\[\cos \theta = \frac{\left( \vec{i}-\vec{j}+2\vec{k} \right).\left( 2\vec{i}+2\vec{j}+\vec{k} \right)}{\sqrt{1^2 +(-1)^2 + 2^2}\sqrt{2^2 + 2^2 + 1}}\]\[\cos \theta = \frac{2-2+2}{3\sqrt{6}}=\frac{2}{3\sqrt{6}}\]\[\theta =\cos^{-1} \left( \frac{2}{3\sqrt{6}} \right)=74.21^{\circ}\]