IBDP Maths analysis and approaches Topic: AHL 3.17 Vector equation of a plane HL Paper 1

Ans:

1. Since the coordinates of the points \(P, Q\) and \(R\) are \((4, 1, –1), (3, 3, 5)\) and \((1, 0, 2c),\) respectively, the vectors \(\overrightarrow{QR}\) and \(\overrightarrow{PR}\) are given by
   \(\overrightarrow{QR}\) = \(−2\overrightarrow{i} − 3\overrightarrow{j} + (2c − 5)\overrightarrow{k}\)                                  
   \(\overrightarrow{PR}\) = \(−3\overrightarrow{i} − \overrightarrow{j} + (2c + 1)\overrightarrow{k}\)                                   
   \(\overrightarrow{QR}\) is perpendicular to \(\overrightarrow{PR}\) if and only if \(\overrightarrow{QR} ⋅\overrightarrow{PR} = 0\)
   \(i.e. 6 + 3 + (2c – 5)(2c +1) = 0  \)                         
   ⇒ \(4c^ 2 – 8c + 4 = 0\)⇒ \((c – 1)^2 = 0\)⇒ \(c = 1\)                                           
2. \(\overrightarrow{PR}=-3\overrightarrow{i}-\overrightarrow{j}+3\overrightarrow{k}, \overrightarrow{PS}=-3\overrightarrow{i}+3\overrightarrow{k}\)                                                  
   \(\overrightarrow{PS}×\overrightarrow{PR}= \begin{vmatrix} \overrightarrow{i}&\overrightarrow{j}&\overrightarrow{k}\\ -3&0&3\\ -3&-1&3 \end{vmatrix}=3\overrightarrow{i}+3\overrightarrow{k}\)                                                       
3. The parametric equation of a line \(l\) which passes through the point \((3, 3, 5)\) and is parallel to the vector \(\overrightarrow{PR}\) is given by
   \(\overrightarrow{r}=(3\overrightarrow{i}+3\overrightarrow{j}+5\overrightarrow{k})+t(-3\overrightarrow{i}-\overrightarrow{j}+3\overrightarrow{k})\)                                    
   \(=3(1-t)\overrightarrow{i}+(3-t)\overrightarrow{j}+(5+3t)\overrightarrow{k}(-∞<t<∞) \)                                    
4. Let \(P_1\) and \(P_2\) be points on the line \(l\) corresponding to \(t = 0\) and \(t = 1\), respectively.
   Hence, for \(\overrightarrow{r}= x\overrightarrow{i} + y \overrightarrow{j}+ z\overrightarrow{k} , x = 3(1 – t), y = (3 – t)\) and \(z = 5 + 3t\).
   Putting \(t = 0\) and \(t = 1,\) we get the coordinates of points \(P_1\) and \(P_2\) as (3, 3, 5) and (0, 2, 8), respectively.
   Vectors \(\overrightarrow{SP_1}\) , and \(\overrightarrow{SP_2}\) are given by \(\overrightarrow{SP_1}=2\overrightarrow{i}+2\overrightarrow{j}+3\overrightarrow{k}\) and \(\overrightarrow{SP_2}=-\overrightarrow{i}+\overrightarrow{j}+6\overrightarrow{k}\)                                          
   A vector perpendicular to both \(\overrightarrow{SP_1}\) and \(\overrightarrow{SP_2}\) is \(\overrightarrow{SP_1}×\overrightarrow{SP_2}=\begin{vmatrix} \overrightarrow{i}&\overrightarrow{j}&\overrightarrow{k}\\ 2&2&3\\ -1&1&6 \end{vmatrix}\)                                      
   \(= -9\overrightarrow{i}-15\overrightarrow{j}+4\overrightarrow{k}\)                                   
   Let \(T(x, y, z)\) be any point of the plane \(\pi\). Since \(S = (1, 1, 2), \overrightarrow{ST}=(x-1)\overrightarrow{i}+(y-1)\overrightarrow{j}+(z-2)\overrightarrow{k}\) is a vector in \(\pi .\) Hence \(\overrightarrow{TS}.\overrightarrow{n}=0\) \(i.e 9(x – 1) – 15(y – 1) + 4(z – 2) = 0\) ⇒ \(9x – 15y + 4z – 2 = 0\)                           
   OR
   \(\overrightarrow{QS}=(1-3)\overrightarrow{i}+(1-3)\overrightarrow{j}+(2-5)\overrightarrow{k}\)
   The equation of the plane containing the line \(l\) and passing through the point \(S\) is determined by \(l\) and the vector \(\overrightarrow{SQ}\) .
   Hence, the equation is: \(\overrightarrow{r}= \begin{pmatrix} 3\\ 3\\ 5\end{pmatrix}+\lambda \begin{pmatrix} -3\\ -1\\ 3\end{pmatrix}+\mu  \begin{pmatrix} -2\\ -2\\ 3\end{pmatrix}\)
5. Shortest distance \(\frac{|\overrightarrow{PQ}.\overrightarrow{n}|}{|\overrightarrow{n}|}=\frac{\left | \begin{pmatrix} -1\\ 2\\ 6 \end{pmatrix}. \begin{pmatrix} 9\\ -15\\ 4 \end{pmatrix} \right |}{\sqrt{322}}\)
   \(=\frac{15}{\sqrt{322}}\) Accept 0.836 (3 s.f.)