The angle between the vectors \(3\mathbf{i}-2\mathbf{j}\) and \(6\mathbf{i}+d\mathbf{j}-\sqrt{7}\mathbf{k}\) is \({{\cos }^{-1}}\left( \frac{6}{13} \right)\). Show that \(2{{d}^{2}}-117d+333=0\).[3]
With reference to the origin \(O\), the points \(A\), \(B\), \(C\) and \(D\) are such that \(\overrightarrow{OA}=\mathbf{a}\), \(\overrightarrow{OB}=\mathbf{b}\), \(\overrightarrow{AC}=5\mathbf{a}\) and \(\overrightarrow{BD}=3\mathbf{b}\). The lines \(AD\) and \(BC\) cross at \(E\) (see diagram).
Find \(\overrightarrow{OE}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).[6]
The point \(F\) divides the line \(CD\) in the ratio \(5:3\). Show that \(O\), \(E\) and \(F\) are collinear, and find \(OE:OF\).[4]
Video Solution:
Default solution
Answer:(a) \(2d^2-117d+333=0\). (b)(i) \(\overrightarrow{OE}=\frac{18}{23}\mathbf a+\frac{20}{23}\mathbf b\). (ii) \(O,E,F\) are collinear and \(OE:OF=8:23\).