IBDP Maths AHL 3.15- Different types of lines AA HL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a)
Let the common parameter for \(L_2\) be \(t\).
From the equations of the two lines:
\(5+s=-1+2t\)
\(4-s=7+t\)
\(2+s=-5+3t\)
Solving simultaneously gives
\(t=1,\quad s=-4\).
Substitute into either line:
\( \mathbf r= \begin{pmatrix} 5\\4\\2 \end{pmatrix} -4 \begin{pmatrix} 1\\-1\\1 \end{pmatrix} = \begin{pmatrix} 1\\8\\-2 \end{pmatrix} \)
The same point is obtained from \(L_2\), confirming the intersection.
✅ Answer: \( \boxed{\begin{pmatrix}1\\8\\-2\end{pmatrix}} \)
(b)(i)
The plane contains both direction vectors:
\( \mathbf b= \begin{pmatrix} 1\\-1\\1 \end{pmatrix}, \qquad \mathbf c= \begin{pmatrix} 2\\1\\3 \end{pmatrix} \)
Using the point of intersection:
\( \boxed{ \mathbf r= \begin{pmatrix} 1\\8\\-2 \end{pmatrix} +\lambda \begin{pmatrix} 1\\-1\\1 \end{pmatrix} +\mu \begin{pmatrix} 2\\1\\3 \end{pmatrix} } \)
This uses one point together with two non-parallel direction vectors.
(b)(ii)
The normal vector is
\( \mathbf n= \begin{pmatrix} -4\\-1\\3 \end{pmatrix} \)
Using point \( (1,8,-2) \):
\( -4(x-1)-(y-8)+3(z+2)=0 \)
Simplifying:
\( -4x-y+3z=-18 \)
or equivalently
\( \boxed{4x+y-3z=18} \)
The normal vector determines the orientation of the plane.
(c)(i)
On the \(y\)-axis, \(x=0,\;z=0\).
\(y=18\)
✅ \(q=18\)
(c)(ii)
On the \(z\)-axis, \(x=0,\;y=0\).
\(-3z=18\)
\(z=-6\)
✅ \(r=-6\)
(d)
\( \overrightarrow{PQ}= \begin{pmatrix} -4.5\\18\\0 \end{pmatrix}, \qquad \overrightarrow{PR}= \begin{pmatrix} -4.5\\0\\-6 \end{pmatrix} \)
\( \overrightarrow{PQ}\times\overrightarrow{PR} = \begin{pmatrix} -108\\-27\\81 \end{pmatrix} \)
Magnitude:
\( \left| \overrightarrow{PQ}\times\overrightarrow{PR} \right| = \sqrt{(-108)^2+(-27)^2+81^2} = 27\sqrt{26} \)
Hence
\( \text{Area} = \frac12(27\sqrt{26}) = \boxed{\frac{27\sqrt{26}}{2}} \) square units.
The area of a triangle is half the magnitude of the cross product of two sides.
(e)
The normal line passes through \( (1,8,-2) \) and has direction \( (-4,-1,3) \).
\( \boxed{ \mathbf r= \begin{pmatrix} 1\\8\\-2 \end{pmatrix} + \gamma \begin{pmatrix} -4\\-1\\3 \end{pmatrix} } \)
(f)
Substitute \(S(-11,5,7)\):
\( -11=1-4\gamma \)
\( 5=8-\gamma \)
\( 7=-2+3\gamma \)
Each equation gives
\( \boxed{\gamma=3} \)
The point satisfies all three coordinate equations consistently.
(g)
The height of the pyramid is
\( |PS| = \sqrt{(-12)^2+(-3)^2+9^2} = 3\sqrt{26} \)
Volume of pyramid:
\( V = \frac13 \times \frac{27\sqrt{26}}2 \times 3\sqrt{26} \)
\( = \boxed{351} \) cubic units.
The volume equals one-third of the base area multiplied by the perpendicular height.