The point A with coordinates (-3, 7, 2) lies on a line \( l_1 \) The point B also lies on the line \( l_1 \) Given that \( \vec{AB} = \begin{pmatrix} 4 \ \ -6 \ \ 2 \end{pmatrix} \) (a) find the coordinates of point B - Edexcel - A-Level Maths Pure - Question 8 - 2018 - Paper 9

To find the cosine of the angle PAB, we first need the vectors ( \vec{AP} ) and ( \vec{AB} ).

The coordinates of point P are (9, 1, 8). Therefore, we can find ( \vec{AP} ) as follows:

[ \vec{AP} = P - A = \begin{pmatrix} 9 - (-3) \ \ 1 - 7 \ \ 8 - 2 \end{pmatrix} = \begin{pmatrix} 12 \ \ -6 \ \ 6 \end{pmatrix} ]

Now, we find the dot product of ( \vec{AP} ) and ( \vec{AB} ):

[ \vec{AP} \cdot \vec{AB} = 12 \times 4 + (-6) \times (-6) + 6 \times 2 = 48 + 36 + 12 = 96 ]

Next, we need the magnitudes of ( \vec{AP} ) and ( \vec{AB} ):

[ |\vec{AP}| = \sqrt{12^2 + (-6)^2 + 6^2} = \sqrt{144 + 36 + 36} = \sqrt{216} = 6\sqrt{6} ]

[ |\vec{AB}| = \sqrt{4^2 + (-6)^2 + 2^2} = \sqrt{16 + 36 + 4} = \sqrt{56} = 2\sqrt{14} ]

Now we can calculate ( \cos \theta ):

[ \cos \theta = \frac{\vec{AP} \cdot \vec{AB}}{|\vec{AP}| |\vec{AB}|} = \frac{96}{(6\sqrt{6})(2\sqrt{14})} = \frac{96}{12\sqrt{84}} = \frac{8}{\sqrt{84}} = \frac{8}{2\sqrt{21}} = \frac{4}{\sqrt{21}} ]

Thus, ( \cos \theta = \frac{4}{\sqrt{21}} ).

The area of triangle PAB can be calculated using the formula:

[ \text{Area} = \frac{1}{2} |\vec{AP} \times \vec{AB}| ]

First, we compute the cross product ( \vec{AP} \times \vec{AB} ):

[ \vec{AP} = \begin{pmatrix} 12 \ \ -6 \ \ 6 \end{pmatrix}, \quad \vec{AB} = \begin{pmatrix} 4 \ \ -6 \ \ 2 \end{pmatrix} ]

Using the determinant to find the cross product:

[ \vec{AP} \times \vec{AB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 12 & -6 & 6 \ 4 & -6 & 2 \end{vmatrix} = \hat{i}( -6 \cdot 2 - 6 \cdot (-6)) - \hat{j}(12 \cdot 2 - 6 \cdot 4) + \hat{k}(12 \cdot (-6) - (-6) \cdot 4) ]

[ = \hat{i}( -12 + 36) - \hat{j}(24 - 24) + \hat{k}( -72 + 24) = \hat{i}(24) + \hat{j}(0) + \hat{k}(-48) = \begin{pmatrix} 24 \ \ 0 \ \ -48 \end{pmatrix} ]

Now we calculate its magnitude:

[ |\vec{AP} \times \vec{AB}| = \sqrt{24^2 + 0^2 + (-48)^2} = \sqrt{576 + 0 + 2304} = \sqrt{2880} = 12\sqrt{20} = 12 \times 2\sqrt{5} = 24\sqrt{5} ]

Finally, we find the area:

[ \text{Area} = \frac{1}{2} \times 24\sqrt{5} = 12\sqrt{5} ]

Thus, the area of triangle PAB is ( 12\sqrt{5} ).

The line ( l_2 ) passes through point P and is parallel to line ( l_1 ). We know that:

1. The direction vector of line ( l_1 ) is ( \vec{AB} = \begin{pmatrix} 4 \ \ -6 \ \ 2 \end{pmatrix} ).
2. The coordinates of point P are (9, 1, 8).

Thus, the vector equation for line ( l_2 ) can be written as:

[ \vec{r} = \begin{pmatrix} 9 \ \ 1 \ \ 8 \end{pmatrix} + t \begin{pmatrix} 4 \ \ -6 \ \ 2 \end{pmatrix} ]

Where ( t ) is a scalar parameter. Therefore, the vector equation of line ( l_2 ) is:

[ \vec{r} = \begin{pmatrix} 9 + 4t \ \ 1 - 6t \ \ 8 + 2t \end{pmatrix} ]

For point Q on line ( l_2 ), we denote its position as follows:

[ Q = \begin{pmatrix} 9 + 4t \ \ 1 - 6t \ \ 8 + 2t \end{pmatrix} ]

Since the line segment AP is perpendicular to the line segment BQ, we can use the condition that the dot product must be zero:

[ \vec{AP} \cdot \vec{BQ} = 0 ]

Where:

[ \vec{BQ} = Q - B = \begin{pmatrix} (9 + 4t) - 1 \ \ (1 - 6t) - 1 \ \ (8 + 2t) - 4 \end{pmatrix} = \begin{pmatrix} 8 + 4t \ \ -6t \ \ 4 + 2t \end{pmatrix} ]

Now, substituting into the dot product equation gives:

[ \vec{AP} = \begin{pmatrix} 12 \ \ -6 \ \ 6 \end{pmatrix} ]

So, we have:

[ \begin{pmatrix} 12 \ \ -6 \ \ 6 \end{pmatrix} \cdot \begin{pmatrix} 8 + 4t \ \ -6t \ \ 4 + 2t \end{pmatrix} = 0 ]

Calculating the dot product yields:

[ 12(8 + 4t) - 6(-6t) + 6(4 + 2t) = 0 ]

[ 96 + 48t + 36t + 24 + 12t = 0 ]

[ 96 + 96t + 24 = 0 \Rightarrow 120 + 96t = 0 \Rightarrow 96t = -120 \Rightarrow t = -\frac{120}{96} = -\frac{5}{4} ]

Now we can find the coordinates of Q by substituting ( t = -\frac{5}{4} ) into the equation for Q:

[ Q = \begin{pmatrix} 9 + 4(-\frac{5}{4}) \ \ 1 - 6(-\frac{5}{4}) \ \ 8 + 2(-\frac{5}{4}) \end{pmatrix} = \begin{pmatrix} 9 - 5 \ \ 1 + \frac{30}{4} \ \ 8 - \frac{10}{4} \end{pmatrix} = \begin{pmatrix} 4 \ \ 1 + 7.5 \ \ 8 - 2.5 \end{pmatrix} = \begin{pmatrix} 4 \ \ 8.5 \ \ 5.5 \end{pmatrix} ]

Thus, the coordinates of point Q are (4, 8.5, 5.5).