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IBDP Maths AHL 3.13 Scalar product of two vectors AA HL Paper 1- Exam Style Questions

IBDP Maths AHL 3.13 Scalar product of two vectors AA HL Paper 1- Exam Style Questions- New Syllabus

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Question

Two lines, \(L_1\) and \(L_2\), intersect at point P. Point A\((2t, 8, 3)\), where \(t > 0\), lies on \(L_2\). This is shown in the following diagram.

Diagram of two intersecting lines L₁ and L₂

The acute angle between the two lines is \(\frac{\pi}{3}\).

The direction vector of \(L_1\) is \(\begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix}\) and \(\overrightarrow{PA} = \begin{pmatrix}2t \\ 0 \\ 3+t\end{pmatrix}\)

(a) Show that \(4t = \sqrt{10t^2+12t+18}\)

(b) Find the value of \(t\).

(c) Hence or otherwise, find the shortest distance from A to \(L_1\).

A plane \(\Pi\) contains \(L_1\) and \(L_2\)

(d) Find a normal vector to \(\Pi\).

The base of a right cone lies in \(\Pi\) centered at A such that \(L_1\) is a tangent to its base. The volume of the cone is \(90\pi\sqrt{3}\) cubic units.

(e) Find the two possible positions of the vertex of the cone.

▶️ Answer/Explanation
Solution (a)

Using the dot product formula:

\(\overrightarrow{PA} \cdot \overrightarrow{L_1} = |\overrightarrow{PA}||\overrightarrow{L_1}|\cos\left(\frac{\pi}{3}\right)\)

Calculate dot product:

\(\begin{pmatrix}2t \\ 0 \\ 3+t\end{pmatrix} \cdot \begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix} = 2t\)

Calculate magnitudes:

\(|\overrightarrow{PA}| = \sqrt{(2t)^2 + 0^2 + (3+t)^2} = \sqrt{5t^2 + 6t + 9}\)

\(|\overrightarrow{L_1}| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2}\)

Substitute into formula:

\(2t = \sqrt{5t^2 + 6t + 9} \times \sqrt{2} \times \frac{1}{2}\)

Simplify to show:

\(4t = \sqrt{10t^2 + 12t + 18}\) [4 marks]

Solution (b)

Square both sides:

\(16t^2 = 10t^2 + 12t + 18\)

Simplify:

\(6t^2 – 12t – 18 = 0\)

Divide by 6:

\(t^2 – 2t – 3 = 0\)

Factorize:

\((t – 3)(t + 1) = 0\)

Since \(t > 0\), the solution is:

\(t = 3\) [3 marks]

Solution (c)

Shortest distance formula:

\(d = \frac{|\overrightarrow{PA} \times \overrightarrow{L_1}|}{|\overrightarrow{L_1}|}\)

Calculate cross product:

\(\overrightarrow{PA} \times \overrightarrow{L_1} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 6 & 0 \\ 1 & 1 & 0 \end{vmatrix} = -6\mathbf{i} + 6\mathbf{j} + 6\mathbf{k}\)

Magnitude:

\(\sqrt{(-6)^2 + 6^2 + 6^2} = \sqrt{108} = 6\sqrt{3}\)

Divide by \(|\overrightarrow{L_1}| = \sqrt{2}\):

\(d = \frac{6\sqrt{3}}{\sqrt{2}} = 3\sqrt{6}\) [4 marks]

Solution (d)

Normal vector is cross product of direction vectors:

\(\overrightarrow{L_1} = \begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix}\)

\(\overrightarrow{L_2} = \overrightarrow{PA} = \begin{pmatrix}6 \\ 0 \\ 6\end{pmatrix}\)

Cross product:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 0 \\ 6 & 0 & 6 \end{vmatrix} = 6\mathbf{i} – 6\mathbf{j} – 6\mathbf{k}\)

Simplified normal vector:

\(\begin{pmatrix}1 \\ -1 \\ -1\end{pmatrix}\)

[3 marks]

Solution (e)

Cone volume formula:

\(V = \frac{1}{3}\pi r^2 h = 90\pi\sqrt{3}\)

From part (c), radius \(r = 3\sqrt{6}\)

Solve for height \(h\):

\(\frac{1}{3}\pi (54) h = 90\pi\sqrt{3} \Rightarrow h = 5\sqrt{3}\)

Vertex positions:

1. \(\begin{pmatrix}6 \\ 8 \\ 3\end{pmatrix} + 5\sqrt{3} \times \frac{1}{\sqrt{3}}\begin{pmatrix}1 \\ -1 \\ -1\end{pmatrix} = \begin{pmatrix}11 \\ 3 \\ -2\end{pmatrix}\)

2. \(\begin{pmatrix}6 \\ 8 \\ 3\end{pmatrix} – 5\sqrt{3} \times \frac{1}{\sqrt{3}}\begin{pmatrix}1 \\ -1 \\ -1\end{pmatrix} = \begin{pmatrix}1 \\ 13 \\ 8\end{pmatrix}\)

[6 marks]

✅ Key Concepts:
  1. Vector operations: dot product and cross product
  2. Distance from point to line in 3D space
  3. Plane geometry and normal vectors
  4. Cone volume and geometric properties
Question

Consider the vectors \(\mathbf{a} = 6\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}\), \(\mathbf{b} = -3\mathbf{j} + 4\mathbf{k}\).

a. (i) Find the cosine of the angle between vectors \(\mathbf{a}\) and \(\mathbf{b}\).

(ii) Find \(\mathbf{a} \times \mathbf{b}\).

(iii) Hence find the Cartesian equation of the plane \(\Pi\) containing the vectors \(\mathbf{a}\) and \(\mathbf{b}\) and passing through the point \((1, 1, -1)\).

(iv) The plane \(\Pi\) intersects the \(x\)-\(y\) plane in the line \(l\). Find the area of the finite triangular region enclosed by \(l\), the \(x\)-axis and the \(y\)-axis. [11]

b. Given two vectors \(\mathbf{p}\) and \(\mathbf{q}\):

(i) Show that \(\mathbf{p} \cdot \mathbf{p} = |\mathbf{p}|^2\);

(ii) Hence, or otherwise, show that \(|\mathbf{p} + \mathbf{q}|^2 = |\mathbf{p}|^2 + 2\mathbf{p} \cdot \mathbf{q} + |\mathbf{q}|^2\);

(iii) Deduce that \(|\mathbf{p} + \mathbf{q}| \leq |\mathbf{p}| + |\mathbf{q}|\). [8]

▶️ Answer/Explanation
Solution (a)

(i) Cosine of angle:

Using dot product formula:

\(\mathbf{a} \cdot \mathbf{b} = (6)(0) + (3)(-3) + (2)(4) = -9 + 8 = -1\)

Magnitudes:

\(|\mathbf{a}| = \sqrt{6^2 + 3^2 + 2^2} = 7\)

\(|\mathbf{b}| = \sqrt{0^2 + (-3)^2 + 4^2} = 5\)

Therefore:

\(\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|} = \frac{-1}{35}\)

(ii) Cross product:

\(\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 6 & 3 & 2 \\ 0 & -3 & 4 \end{vmatrix}\)

\(= \mathbf{i}(3 \times 4 – 2 \times -3) – \mathbf{j}(6 \times 4 – 2 \times 0) + \mathbf{k}(6 \times -3 – 3 \times 0)\)

\(= 18\mathbf{i} – 24\mathbf{j} – 18\mathbf{k}\)

(iii) Plane equation:

Using normal vector \(\mathbf{n} = 18\mathbf{i} – 24\mathbf{j} – 18\mathbf{k}\) and point \((1,1,-1)\):

\(18(x-1) – 24(y-1) – 18(z+1) = 0\)

Simplify:

\(18x – 24y – 18z = 12\) or \(3x – 4y – 3z = 2\)

(iv) Triangular area:

When \(z = 0\) (x-y plane):

\(3x – 4y = 2\)

x-intercept: \(x = \frac{2}{3}\)

y-intercept: \(y = -\frac{1}{2}\)

Area of triangle:

\(\frac{1}{2} \times \frac{2}{3} \times \frac{1}{2} = \frac{1}{6}\)

[11 marks]

Solution (b)

(i) Dot product identity:

\(\mathbf{p} \cdot \mathbf{p} = |\mathbf{p}||\mathbf{p}|\cos 0 = |\mathbf{p}|^2\)

(ii) Magnitude expansion:

\(|\mathbf{p} + \mathbf{q}|^2 = (\mathbf{p} + \mathbf{q}) \cdot (\mathbf{p} + \mathbf{q})\)

\(= \mathbf{p} \cdot \mathbf{p} + 2\mathbf{p} \cdot \mathbf{q} + \mathbf{q} \cdot \mathbf{q}\)

\(= |\mathbf{p}|^2 + 2\mathbf{p} \cdot \mathbf{q} + |\mathbf{q}|^2\)

(iii) Triangle inequality:

From part (ii):

\(|\mathbf{p} + \mathbf{q}|^2 = |\mathbf{p}|^2 + 2\mathbf{p} \cdot \mathbf{q} + |\mathbf{q}|^2\)

Using \(\mathbf{p} \cdot \mathbf{q} \leq |\mathbf{p}||\mathbf{q}|\) (Cauchy-Schwarz):

\(\leq |\mathbf{p}|^2 + 2|\mathbf{p}||\mathbf{q}| + |\mathbf{q}|^2\)

Take square roots:

\(|\mathbf{p} + \mathbf{q}| \leq |\mathbf{p}| + |\mathbf{q}|\)

Vector triangle inequality diagram

[8 marks]

✅ Key Concepts:
  1. Vector operations: dot product and cross product
  2. Plane equations using normal vectors
  3. Geometric interpretation of vector relationships
  4. Triangle inequality for vectors
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