The point A with coordinates (-3, 7, 2) lies on a line I1 The point B also lies on the line I1 Given that \[ \vec{AB} = \begin{pmatrix} 4 \\ -6 \\ 2 \end{pmatrix} \] (a) find the coordinates of point B - Edexcel - A-Level Maths Pure - Question 1 - 2018 - Paper 9

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

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

Next, we calculate the dot product:
[ \vec{AP} \cdot \vec{AB} = (12)(4) + (-6)(-6) + (6)(2) = 48 + 36 + 12 = 96 ]

Now, we find the magnitudes of the vectors:

• ( ||\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} )

Using the dot product formula: [ \cos(PAB) = \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{4}{\sqrt{21}} ]

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

The area of triangle PAB can be found using the formula: [ \text{Area} = \frac{1}{2} ||\vec{AP} \times \vec{AB}|| ]

Calculating the cross product ( \vec{AP} \times \vec{AB} ): [ \vec{AP} = \begin{pmatrix} 12 \ -6 \ 6 \end{pmatrix}, \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} ]

Calculating: [ = \hat{i}((-6)(2) - (6)(-6)) - \hat{j}((12)(2) - (6)(4)) + \hat{k}((12)(-6) - (-6)(4)) ] [ = \hat{i}(-12 + 36) - \hat{j}(24 - 24) + \hat{k}(-72 + 24) ] [ = \hat{i}(24) - \hat{j}(0) + \hat{k}(-48) ]

So, ( \vec{AP} \times \vec{AB} = \begin{pmatrix} 24 \ 0 \ -48 \end{pmatrix} )

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

Thus, the area is: [ \text{Area} = \frac{1}{2} (24\sqrt{5}) = 12\sqrt{5} ]

Since line I2 passes through point P and is parallel to line I1, we first determine the direction vector of line I1 which is ( \vec{AB} ).

The position vector for point P is ( \vec{P} = \begin{pmatrix} 9 \ 1 \ 8 \end{pmatrix} ).

A vector equation for line I2 can be written as: [ \vec{r} = \vec{P} + t \vec{AB} ]

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

Thus, the vector equation for line I2 is: [ \vec{r} = \begin{pmatrix} 9 + 4t \ 1 - 6t \ 8 + 2t \end{pmatrix} ]

Since the line segment AP is perpendicular to the line segment BQ, we need to ensure:

• The direction vector of line AP is ( \vec{AP} = \begin{pmatrix} 12 \ -6 \ 6 \end{pmatrix})
• The direction vector of line BQ is ( \vec{BQ} = \vec{Q} - egin{pmatrix} 1 \ 1 \ 4 \end{pmatrix})

To satisfy the perpendicular condition, we apply the dot product: [ \vec{AP} \cdot \vec{BQ} = 0 ]

Substituting and simplifying gives us the coordinates of Q. Upon solving the system involving the components of both vectors, we ultimately find:

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