The position vectors of points \(A\), \(B\) and \(C\), referred to an origin \(O\), are \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\) respectively, where
\[\begin{align*} \mathbf{a} &= 3\mathbf{i} + 4 \mathbf{j} + 5\mathbf{k}, \\ \mathbf{b} &= 7\mathbf{i} - \mathbf{k}, \\ \mathbf{c} &= 5\mathbf{i} + 5 \mathbf{j}. \\ \end{align*}\]\(P\) is a point which is equidistant from the lines \(OA\), \(OB\) and \(OC\). Write down expressions for the cosine of the angle between \(OP\) and \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\) respectively and hence deduce an expression for a unit vector in the direction of \(\mathbf{OP}\).
Let \(A'\) be a point on \(OA\) such that \(A'P\) is perpendicular to \(OA\) and let \(B'\) and \(C'\) be the equivalent points on \(OB\) and \(OC\) respectively.
If we think about the three right angled triangles \(A'OP\), \(B'OP\) and \(C'OP\) we have hypotenuse \(OP\) common to each and \(P\) was defined to be equidistant from the lines \(OA\), \(OB\) and \(OC\). So \(OA'=OB'=OC'\) making the triangles congruent (RHS).
The question asks about the cosine of the angle \(\alpha\), which suggests using the scalar product. Let the vector \(OP\) be \(\mathbf{p} = \begin{pmatrix}p_i \\ p_j \\ p_k\end{pmatrix}\), so we have \(\mathbf{p}.\mathbf{a} = 3p_i+4p_j+5p_k\).
But \(\mathbf{p}.\mathbf{a} = |\mathbf{p}||\mathbf{a}|\cos \alpha\), so \(\cos \alpha = \dfrac{\mathbf{p}.\mathbf{a}}{|\mathbf{p}|\sqrt{50}}=\dfrac{3p_i+4p_j+5p_k}{|\mathbf{p}|\sqrt{50}}\). We have used the fact that \(|\mathbf{a}|=\sqrt{3^2+4^2+5^2} = \sqrt{50}\).
Since the three triangles are congruent, they each contain the same angle \(\alpha\). Also notice that \(|\mathbf{b}|=|\mathbf{c}|=\sqrt{50}\).
So for vector \(\mathbf{b}\) we have \(\cos \alpha= \dfrac{7p_i-p_k}{|\mathbf{p}|\sqrt{50}}\), and for \(\mathbf{c}\) we have \(\cos \alpha = \dfrac{5p_i+5p_j}{|\mathbf{p}|\sqrt{50}}\).
Combining these we can write \(3p_i+4p_j+5p_k = 7p_i-p_k= 5p_i+5p_j\).
Using the second and third parts gives \(p_k = 2p_i - 5p_j\), and substituting this into the first two parts gives \(3p_i+4p_j+5(2p_i - 5p_j) = 7p_i-(2p_i - 5p_j)\).
Thus \(p_j = \dfrac{4}{13} p_i\), and \(p_k = \dfrac{6}{13} p_i\), and so the vector \(\mathbf{p}\) is \(t \begin{pmatrix}13 \\ 4 \\ 6\end{pmatrix}\) for some scalar \(t\).
To find the unit vector in this direction, we divide by the magnitude \(\sqrt{13^2+4^2+6^2} = \sqrt{221}\), and so the unit vector in the direction of \(\mathbf{OP}\) is \[\hat{\mathbf{p}} = \frac{13}{\sqrt{221}} \mathbf{i} + \frac{4}{\sqrt{221}}\mathbf{j} + \frac{6}{\sqrt{221}}\mathbf{k}.\]
Given that the line \(OP\) meets the plane \(ABC\) at \(Q\), find \(\mathbf{OQ}\).
We can define the plane \(ABC\) from the three position vectors \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\) via \[\mathbf{r} = \mathbf{a} + \lambda(\mathbf{b}-\mathbf{a}) + \mu(\mathbf{c}-\mathbf{a}),\] where \(\lambda\) and \(\mu\) are scalar parameters.
If we start at the origin, \(\mathbf{a}\) takes us onto the plane. The vectors \((\mathbf{b}-\mathbf{a})\) and \((\mathbf{c}-\mathbf{a})\) are distinct vectors lying in the plane, so a linear combination of these will take us anywhere on the plane from \(\mathbf{a}\).
Thus the plane \(ABC\) is \[\mathbf{r} = (3+4\lambda+2\mu)\mathbf{i} + (4-4\lambda+\mu) \mathbf{j} + (5-6\lambda-5\mu)\mathbf{k}.\]
The line \(OP\) is given by \(\mathbf{r} = t (13\mathbf{i}+ 4\mathbf{j}+ 6\mathbf{k})\). To find the intersection we set these equal, then the three components give us three equations in three unknowns, \(\lambda\), \(\mu\) and \(t\).
Solving these simultaneously gives \(t = \dfrac{5}{13}\) and so \(\mathbf{OQ}\) = \(5\mathbf{i} + \frac{20}{13}\mathbf{j} +\frac{30}{13}\mathbf{k}\).Given that \(G\) is the centroid of the triangle \(ABC\), show that \(QG\) is parallel to the coordinate plane \(Oyz\).
The position vector of the centroid \(G\) of a triangle whose vertices have position vectors \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\) is given by \[\mathbf{g}=\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{3}.\] In our case \[\mathbf{g}=5 \mathbf{i} + 3 \mathbf{j} + \frac{4}{3} \mathbf{k}.\]
The direction of the line \(QG\) is given by the vector \(\mathbf{g}-\mathbf{OQ}\).
If this direction vector is parallel to the \(Oyz\) plane, its \(x\)-component will be zero. Hence we have to show that \(q_i = g_i\).
The \(x\)-components of \(\mathbf{g}\) and \(\mathbf{OQ}\) are both \(5\), so \(QG\) is indeed parallel to the plane.
