IBDP Maths AHL 3.16 vector product of two vectors: AA HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a)
The line passes through \(P(-1,1,-13)\) and has direction vector \((7,1,2)\).
\( \boxed{\mathbf{r}= \begin{pmatrix} -1\\ 1\\ -13 \end{pmatrix} +\lambda \begin{pmatrix} 7\\ 1\\ 2 \end{pmatrix}} \)
Simply use the position vector of a point on the line together with its direction vector.
(b)
The direction vector of \(L_2\) is
\( \overrightarrow{AB} = \begin{pmatrix} 7-2\\ -6-(-4)\\ 1-2 \end{pmatrix} = \begin{pmatrix} 5\\ -2\\ -1 \end{pmatrix}. \)
Hence
\( \boxed{ \mathbf{s} = \begin{pmatrix} 2\\ -4\\ 2 \end{pmatrix} +\mu \begin{pmatrix} 5\\ -2\\ -1 \end{pmatrix}} \)
Using point \(B\) instead is also acceptable.
(c)
The direction vectors are
\( \begin{pmatrix} 7\\ 1\\ 2 \end{pmatrix}, \qquad \begin{pmatrix} 5\\ -2\\ -1 \end{pmatrix}. \)
These are not scalar multiples, so the lines are not parallel.
Solving the simultaneous equations obtained by equating the coordinates gives no common solution.
Therefore the lines do not intersect.
Hence
\( \boxed{L_1\text{ and }L_2\text{ are skew.}} \)
(d)
A general point on \(L_2\) is
\( N=(2+5\mu,\,-4-2\mu,\,2-\mu). \)
Therefore
\( \overrightarrow{PN} = \begin{pmatrix} 3+5\mu\\ -5-2\mu\\ 15-\mu \end{pmatrix}. \)
Since
\( \overrightarrow{AB} = \begin{pmatrix} 5\\ -2\\ -1 \end{pmatrix}, \)
\( \overrightarrow{PN}\cdot\overrightarrow{AB} =(3+5\mu)(5)+(-5-2\mu)(-2)+(15-\mu)(-1). \)
Simplifying,
\( \boxed{\overrightarrow{PN}\cdot\overrightarrow{AB}=10+30\mu.} \)
The dot product is zero when the two vectors are perpendicular.
(e)
The closest point occurs when
\( \overrightarrow{PN}\perp\overrightarrow{AB}. \)
Hence
\( 10+30\mu=0 \)
\( \mu=-\frac13. \)
Substituting into the equation of \(L_2\),
\( N= \left( 2-\frac53,\, -4+\frac23,\, 2+\frac13 \right). \)
Therefore
\( \boxed{ N\left( \frac13,\, -\frac{10}{3},\, \frac73 \right)} \)
The nearest point on a line is obtained by using the perpendicular condition.
(f)
The plane contains the vectors
\( \overrightarrow{OP} = \begin{pmatrix} -1\\ 1\\ -13 \end{pmatrix}, \qquad \overrightarrow{ON} = \begin{pmatrix} 1\\ -10\\ 7 \end{pmatrix}. \)
A normal vector is
\( \overrightarrow{OP}\times\overrightarrow{ON} = \begin{pmatrix} -123\\ -6\\ 9 \end{pmatrix}. \)
Since the plane passes through the origin, its equation is
\( -123x-6y+9z=0. \)
Therefore
\( \boxed{-123x-6y+9z=0.} \)
The cross product provides a vector perpendicular to the plane, which is then used to form the Cartesian equation.