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IBDP Maths AHL 3.12 Concept of a vector AA HL Paper 1- Exam Style Questions

IBDP Maths AHL 3.12 Concept of a vector AA HL Paper 1- Exam Style Questions- New Syllabus

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Question

The diagram below shows a circle with centre O. The points A, B, C lie on the circumference of the circle and [AC] is a diameter.

Circle diagram with vectors

Let \( \overrightarrow{OA} = \boldsymbol{a} \) and \( \overrightarrow{OB} = \boldsymbol{b} \).

a. Write down expressions for \( \overrightarrow{AB} \) and \( \overrightarrow{CB} \) in terms of the vectors \( \boldsymbol{a} \) and \( \boldsymbol{b} \). [2]

b. Hence prove that angle \( ABC \) is a right angle. [3]

▶️ Answer/Explanation
Solution (a)

Using vector geometry:

\( \overrightarrow{AB} = \overrightarrow{OB} – \overrightarrow{OA} = \boldsymbol{b} – \boldsymbol{a} \) (A1)

\( \overrightarrow{CB} = \overrightarrow{OB} – \overrightarrow{OC} = \boldsymbol{b} – (-\boldsymbol{a}) = \boldsymbol{b} + \boldsymbol{a} \) (A1)

[2 marks]

Solution (b)

To prove angle \( ABC \) is right-angled:

Calculate the dot product:

\( \overrightarrow{AB} \cdot \overrightarrow{CB} = (\boldsymbol{b} – \boldsymbol{a}) \cdot (\boldsymbol{b} + \boldsymbol{a}) \) (M1)

\( = |\boldsymbol{b}|^2 – |\boldsymbol{a}|^2 \) (A1)

Since \( |\boldsymbol{a}| = |\boldsymbol{b}| \) (both are radii of the circle):

\( \overrightarrow{AB} \cdot \overrightarrow{CB} = 0 \) (R1)

Therefore, \( \overrightarrow{AB} \) is perpendicular to \( \overrightarrow{CB} \), proving angle \( ABC \) is a right angle. (AG)

[3 marks]

Question

The points A(1, 2, 1), B(−3, 1, 4), C(5, −1, 2) and D(5, 3, 7) are the vertices of a tetrahedron.

a. Find the vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \). [2]

b. Find the Cartesian equation of the plane \( \prod \) that contains the face ABC. [4]

▶️ Answer/Explanation
Solution (a)

Find the vectors:

\( \overrightarrow{AB} = \begin{pmatrix} -3 – 1 \\ 1 – 2 \\ 4 – 1 \end{pmatrix} = \begin{pmatrix} -4 \\ -1 \\ 3 \end{pmatrix} \) (A1)

\( \overrightarrow{AC} = \begin{pmatrix} 5 – 1 \\ -1 – 2 \\ 2 – 1 \end{pmatrix} = \begin{pmatrix} 4 \\ -3 \\ 1 \end{pmatrix} \) (A1)

[2 marks]

Solution (b)

Find the normal vector to the plane:

\( \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \\ -4 & -1 & 3 \\ 4 & -3 & 1 \end{vmatrix} = \begin{pmatrix} 8 \\ 16 \\ 16 \end{pmatrix} \) (M1A1)

Simplify the normal vector:

\( \boldsymbol{n} = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} \) (divided by 8)

Find the plane equation using point A(1, 2, 1):

\( 1(x-1) + 2(y-2) + 2(z-1) = 0 \)

Simplify to get the Cartesian equation:

\( x + 2y + 2z = 7 \) (A1)

[4 marks]

✅ Final Answer:
The Cartesian equation of the plane is \( x + 2y + 2z = 7 \)
Question

In the diagram below, [AB] is a diameter of the circle with center O. Point C is on the circumference of the circle. Let \( \overrightarrow{OB} = \boldsymbol{b} \) and \( \overrightarrow{OC} = \boldsymbol{c} \).

Circle with diameter AB and point C

a. Find an expression for \( \overrightarrow{CB} \) and for \( \overrightarrow{AC} \) in terms of \( \boldsymbol{b} \) and \( \boldsymbol{c} \). [2]

b. Hence prove that angle \( ACB \) is a right angle. [3]

▶️ Answer/Explanation
Solution (a)

Vector expressions:

\( \overrightarrow{CB} = \overrightarrow{OB} – \overrightarrow{OC} = \boldsymbol{b} – \boldsymbol{c} \) (A1)

\( \overrightarrow{AC} = \overrightarrow{OC} – \overrightarrow{OA} = \boldsymbol{c} – (-\boldsymbol{b}) = \boldsymbol{b} + \boldsymbol{c} \) (A1)

[2 marks]

Solution (b)

To prove angle \( ACB \) is right-angled:

Calculate the dot product:

\( \overrightarrow{AC} \cdot \overrightarrow{CB} = (\boldsymbol{b} + \boldsymbol{c}) \cdot (\boldsymbol{b} – \boldsymbol{c}) \) (M1)

Expand using dot product properties:

\( = |\boldsymbol{b}|^2 – |\boldsymbol{c}|^2 \) (A1)

Since \( |\boldsymbol{b}| = |\boldsymbol{c}| \) (both are radii of the circle):

\( \overrightarrow{AC} \cdot \overrightarrow{CB} = 0 \) (R1)

Therefore, \( \overrightarrow{AC} \) is perpendicular to \( \overrightarrow{CB} \), proving angle \( ACB \) is a right angle. (AG)

[3 marks]

✅ Geometric Proof:
This confirms Thales’ theorem that any angle inscribed in a semicircle is a right angle.
Question

The vertices of a triangle ABC have coordinates given by A\((-1, 2, 3)\), B\((4, 1, 1)\) and C\((3, -2, 2)\).

a. (i) Find the lengths of the sides of the triangle.

(ii) Find \(\cos \angle BAC\). [6]

b. (i) Show that \(\overrightarrow{BC} \times \overrightarrow{CA} = -7\mathbf{i} – 3\mathbf{j} – 16\mathbf{k}\).

(ii) Hence, show that the area of the triangle ABC is \(\frac{1}{2}\sqrt{314}\). [5]

c. Find the Cartesian equation of the plane containing the triangle ABC. [3]

d. Find a vector equation of (AB). [2]

The point D on (AB) is such that \(\overrightarrow{OD}\) is perpendicular to \(\overrightarrow{BC}\) where O is the origin.

e. (i) Find the coordinates of D.

(ii) Show that D does not lie between A and B. [5]

▶️ Answer/Explanation
Solution (a)

(i) Finding side lengths:

\(\overrightarrow{AB} = \begin{pmatrix} 4-(-1) \\ 1-2 \\ 1-3 \end{pmatrix} = \begin{pmatrix} 5 \\ -1 \\ -2 \end{pmatrix}\)

\(|\overrightarrow{AB}| = \sqrt{5^2 + (-1)^2 + (-2)^2} = \sqrt{25 + 1 + 4} = \sqrt{30}\)

\(\overrightarrow{BC} = \begin{pmatrix} 3-4 \\ -2-1 \\ 2-1 \end{pmatrix} = \begin{pmatrix} -1 \\ -3 \\ 1 \end{pmatrix}\)

\(|\overrightarrow{BC}| = \sqrt{(-1)^2 + (-3)^2 + 1^2} = \sqrt{1 + 9 + 1} = \sqrt{11}\)

\(\overrightarrow{CA} = \begin{pmatrix} -1-3 \\ 2-(-2) \\ 3-2 \end{pmatrix} = \begin{pmatrix} -4 \\ 4 \\ 1 \end{pmatrix}\)

\(|\overrightarrow{CA}| = \sqrt{(-4)^2 + 4^2 + 1^2} = \sqrt{16 + 16 + 1} = \sqrt{33}\)

(ii) Finding \(\cos \angle BAC\):

\(\overrightarrow{AB} \cdot \overrightarrow{AC} = 5 \times 4 + (-1) \times (-4) + (-2) \times (-1) = 20 + 4 + 2 = 26\)

\(\cos \angle BAC = \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{|\overrightarrow{AB}||\overrightarrow{AC}|} = \frac{26}{\sqrt{30} \times \sqrt{33}} = \frac{26}{\sqrt{990}} = \frac{26}{3\sqrt{110}}\)

[6 marks]

Solution (b)

(i) Cross product calculation:

\(\overrightarrow{BC} \times \overrightarrow{CA} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & -3 & 1 \\ -4 & 4 & 1 \end{vmatrix}\)

\(= \mathbf{i}[(-3)(1)-(1)(4)] – \mathbf{j}[(-1)(1)-(1)(-4)] + \mathbf{k}[(-1)(4)-(-3)(-4)]\)

\(= -7\mathbf{i} – 3\mathbf{j} – 16\mathbf{k}\)

(ii) Area calculation:

Area \(= \frac{1}{2}|\overrightarrow{BC} \times \overrightarrow{CA}| = \frac{1}{2}\sqrt{(-7)^2 + (-3)^2 + (-16)^2} = \frac{1}{2}\sqrt{49 + 9 + 256} = \frac{1}{2}\sqrt{314}\)

[5 marks]

Solution (c)

Plane equation:

Using normal vector \(\mathbf{n} = -7\mathbf{i} – 3\mathbf{j} – 16\mathbf{k}\) and point A\((-1, 2, 3)\):

\(-7(x+1) – 3(y-2) – 16(z-3) = 0\)

Simplified: \(7x + 3y + 16z = 47\)

[3 marks]

Solution (d)

Vector equation of AB:

\(\mathbf{r} = \begin{pmatrix} -1 \\ 2 \\ 3 \end{pmatrix} + t\begin{pmatrix} 5 \\ -1 \\ -2 \end{pmatrix}\) or

\(\mathbf{r} = (-\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}) + t(5\mathbf{i} – \mathbf{j} – 2\mathbf{k})\)

[2 marks]

Solution (e)

(i) Coordinates of D:

Parametric form: \(D(-1 + 5t, 2 – t, 3 – 2t)\)

Condition: \(\overrightarrow{OD} \cdot \overrightarrow{BC} = 0\)

\((-1+5t)(-1) + (2-t)(-3) + (3-2t)(1) = 0\)

Solution: \(t = -\frac{1}{2}\)

Coordinates: \(D\left(-\frac{7}{2}, \frac{5}{2}, 4\right)\)

(ii) Position of D:

Since \(t = -0.5 \notin [0,1]\), D is not between A and B

[5 marks]

✅ Key Takeaways:
  1. Vector magnitudes: \(|\mathbf{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}\)
  2. Dot product: \(\mathbf{a} \cdot \mathbf{b} = a_xb_x + a_yb_y + a_zb_z\)
  3. Cross product: \(\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{vmatrix}\)
  4. Plane equation: \(n_x(x-x_0) + n_y(y-y_0) + n_z(z-z_0) = 0\)
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