Home / IBDP Maths AHL 3.16 vector product of two vectors: AA HL Paper 1- Exam Style Questions

IBDP Maths AHL 3.16 vector product of two vectors: AA HL Paper 1- Exam Style Questions

IBDP Maths AHL 3.16 vector 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 \(L_2\).

Diagram of intersecting lines L₁ and L₂

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

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 tangent to its base. The volume of the cone is \(90π\sqrt{3}\) cubic units.

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

▶️ Answer/Explanation
Solution (a)

Using the dot product formula for the angle between vectors:

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

Calculate the dot product:

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

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 the formula with \(\cos\left(\frac{π}{3}\right) = \frac{1}{2}\):

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

Multiply both sides by 2:

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

Key Insight: This relationship connects the parameter \(t\) with the geometric configuration of the lines through the angle condition.

[4 marks]

Solution (b)

Square both sides of the equation from part (a):

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

Simplify to quadratic form:

\(6t^2 – 12t – 18 = 0\)

Divide by 6:

\(t^2 – 2t – 3 = 0\)

Factorize:

\((t – 3)(t + 1) = 0\)

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

\(t = 3\)

Verification: Substituting \(t = 3\) back into the original equation confirms the solution is valid.

[3 marks]

Solution (c)

The shortest distance from point A to line \(L_1\) is given by:

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

Calculate the cross product with \(t = 3\):

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

Magnitude of cross product:

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

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

\(d = \frac{6\sqrt{3}}{\sqrt{2}} = 3\sqrt{6}\)

Geometric Interpretation: This distance represents the perpendicular distance from point A to line \(L_1\), which will be the radius of the cone’s base in part (e).

[4 marks]

Solution (d)

The normal vector to plane \(\Pi\) is the cross product of the direction vectors of \(L_1\) and \(L_2\):

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

\(\overrightarrow{L_2} = \overrightarrow{PA} = \begin{pmatrix}6 \\ 0 \\ 6\end{pmatrix}\) (using \(t=3\))

Cross product calculation:

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

Simplified normal vector:

\(\begin{pmatrix}1 \\ -1 \\ -1\end{pmatrix}\) or any scalar multiple

Note: The normal vector is perpendicular to both lines and defines the orientation of the plane containing them.

[3 marks]

Solution (e)

Using the cone volume formula \(V = \frac{1}{3}πr^2h\) with \(V = 90π\sqrt{3}\):

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

Substitute to find height \(h\):

\(\frac{1}{3}π(54)h = 90π\sqrt{3} ⇒ h = 5\sqrt{3}\)

The vertex lies along the normal direction from A. First, find the unit normal vector:

\(\mathbf{\hat{n}} = \frac{1}{\sqrt{3}}\begin{pmatrix}1 \\ -1 \\ -1\end{pmatrix}\)

Two possible vertex positions:

1. \(\mathbf{A} + h\mathbf{\hat{n}} = \begin{pmatrix}6 \\ 8 \\ 3\end{pmatrix} + 5\sqrt{3}\left(\frac{1}{\sqrt{3}}\begin{pmatrix}1 \\ -1 \\ -1\end{pmatrix}\right) = \begin{pmatrix}11 \\ 3 \\ -2\end{pmatrix}\)

2. \(\mathbf{A} – h\mathbf{\hat{n}} = \begin{pmatrix}6 \\ 8 \\ 3\end{pmatrix} – 5\sqrt{3}\left(\frac{1}{\sqrt{3}}\begin{pmatrix}1 \\ -1 \\ -1\end{pmatrix}\right) = \begin{pmatrix}1 \\ 13 \\ 8\end{pmatrix}\)

Geometric Interpretation: The two positions correspond to the cone opening in opposite directions along the normal vector, both maintaining the same volume and base radius.

[6 marks]

✅ Key Concepts:
  1. Vector Geometry: Dot products relate angles between vectors, while cross products find normal vectors.
  2. 3D Distance: The shortest distance from a point to a line uses the cross product formula.
  3. Plane Geometry: A plane’s orientation is determined by its normal vector, found via the cross product of two direction vectors.
  4. Cone Properties: Volume depends on both base area and height, with the height direction determined by the plane’s normal.
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 \(\overrightarrow {{\text{AB}}} \) and \(\overrightarrow {{\text{AC}}} \). [2 marks]

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

▶️ Answer/Explanation
Solution (a)

To find the vectors between points:

\(\overrightarrow {{\text{AB}}} = \text{B} – \text{A} = \left( {\begin{array}{*{20}{c}} -3 – 1 \\ 1 – 2 \\ 4 – 1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { – 4} \\ { – 1} \\ 3 \end{array}} \right)\)

\(\overrightarrow {{\text{AC}}} = \text{C} – \text{A} = \left( {\begin{array}{*{20}{c}} 5 – 1 \\ -1 – 2 \\ 2 – 1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 4 \\ { – 3} \\ 1 \end{array}} \right)\)[2 marks]

Solution (b)

Step 1: Find the normal vector to the plane using the cross product:

\(\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} = \left| {\begin{array}{*{20}{c}} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -4 & -1 & 3 \\ 4 & -3 & 1 \end{array}} \right| = \mathbf{i}((-1)(1)-(3)(-3)) – \mathbf{j}((-4)(1)-(3)(4)) + \mathbf{k}((-4)(-3)-(-1)(4))\)

Calculating each component:
i-component: \((-1)(1) – (3)(-3) = -1 + 9 = 8\)
j-component: \(-[(-4)(1) – (3)(4)] = -[-4-12] = 16\)
k-component: \((-4)(-3) – (-1)(4) = 12 + 4 = 16\)

Thus, the normal vector is \(\left( {\begin{array}{*{20}{c}} 8 \\ 16 \\ 16 \end{array}} \right)\) which simplifies to \(\left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ 2 \end{array}} \right)\) when divided by 8

Step 2: Use the point-normal form to find the plane equation:

Using point A(1,2,1) and normal vector \(\mathbf{n} = \left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ 2 \end{array}} \right)\):

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

Expanding: \(x – 1 + 2y – 4 + 2z – 2 = 0\)

Combine like terms: \(x + 2y + 2z = 7\)

Question

For non-zero vectors \(\boldsymbol{a}\) and \(\boldsymbol{b}\), show that:

(a)(i) If \(|\boldsymbol{a} – \boldsymbol{b}| = |\boldsymbol{a} + \boldsymbol{b}|\), then \(\boldsymbol{a}\) and \(\boldsymbol{b}\) are perpendicular. [4 marks]

(a)(ii) \(|\boldsymbol{a} \times \boldsymbol{b}|^2 = |\boldsymbol{a}|^2|\boldsymbol{b}|^2 – (\boldsymbol{a} \cdot \boldsymbol{b})^2\). [4 marks]

The points A, B and C have position vectors \(\boldsymbol{a}\), \(\boldsymbol{b}\) and \(\boldsymbol{c}\).

(b)(i) Show that the area of triangle ABC is \(\frac{1}{2}|\boldsymbol{a} \times \boldsymbol{b} + \boldsymbol{b} \times \boldsymbol{c} + \boldsymbol{c} \times \boldsymbol{a}|\). [4 marks]

(b)(ii) Hence, show that the shortest distance from B to AC is \(\frac{|\boldsymbol{a} \times \boldsymbol{b} + \boldsymbol{b} \times \boldsymbol{c} + \boldsymbol{c} \times \boldsymbol{a}|}{|\boldsymbol{c} – \boldsymbol{a}|}\). [3 marks]

▶️ Answer/Explanation
Solution (a)(i)

Given \(|\boldsymbol{a} – \boldsymbol{b}| = |\boldsymbol{a} + \boldsymbol{b}|\), we square both sides:

\((\boldsymbol{a} – \boldsymbol{b}) \cdot (\boldsymbol{a} – \boldsymbol{b}) = (\boldsymbol{a} + \boldsymbol{b}) \cdot (\boldsymbol{a} + \boldsymbol{b})\)

Expanding both sides:

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

Simplifying:

\(-4\boldsymbol{a} \cdot \boldsymbol{b} = 0 \Rightarrow \boldsymbol{a} \cdot \boldsymbol{b} = 0\)

Since the dot product is zero, the vectors are perpendicular.

[4 marks]

Solution (a)(ii)

Starting with the cross product magnitude:

\(|\boldsymbol{a} \times \boldsymbol{b}|^2 = (|\boldsymbol{a}||\boldsymbol{b}|\sin\theta)^2 = |\boldsymbol{a}|^2|\boldsymbol{b}|^2\sin^2\theta\)

Now examine the right side:

\(|\boldsymbol{a}|^2|\boldsymbol{b}|^2 – (\boldsymbol{a} \cdot \boldsymbol{b})^2 = |\boldsymbol{a}|^2|\boldsymbol{b}|^2 – (|\boldsymbol{a}||\boldsymbol{b}|\cos\theta)^2\)

\(= |\boldsymbol{a}|^2|\boldsymbol{b}|^2(1 – \cos^2\theta) = |\boldsymbol{a}|^2|\boldsymbol{b}|^2\sin^2\theta\)

Thus both sides are equal.

[4 marks]

Solution (b)(i)

The area of triangle ABC is half the magnitude of the cross product of \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\):

\(\text{Area} = \frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}| = \frac{1}{2}|(\boldsymbol{b} – \boldsymbol{a}) \times (\boldsymbol{c} – \boldsymbol{a})|\)

Expanding the cross product:

\(= \frac{1}{2}|\boldsymbol{b} \times \boldsymbol{c} – \boldsymbol{b} \times \boldsymbol{a} – \boldsymbol{a} \times \boldsymbol{c} + \boldsymbol{a} \times \boldsymbol{a}|\)

Simplifying (noting \(\boldsymbol{a} \times \boldsymbol{a} = \boldsymbol{0}\) and \(-\boldsymbol{b} \times \boldsymbol{a} = \boldsymbol{a} \times \boldsymbol{b}\)):

\(= \frac{1}{2}|\boldsymbol{a} \times \boldsymbol{b} + \boldsymbol{b} \times \boldsymbol{c} + \boldsymbol{c} \times \boldsymbol{a}|\)

[4 marks]

Solution (b)(ii)

The shortest distance from B to AC is the perpendicular distance (height).

Area of triangle can also be expressed as: \(\frac{1}{2}|\overrightarrow{AC}| \times \text{height}\)

From part (i), we have: \(\frac{1}{2}|\overrightarrow{AC}| \times h = \frac{1}{2}|\boldsymbol{a} \times \boldsymbol{b} + \boldsymbol{b} \times \boldsymbol{c} + \boldsymbol{c} \times \boldsymbol{a}|\)

Solving for height \(h\):

\(h = \frac{|\boldsymbol{a} \times \boldsymbol{b} + \boldsymbol{b} \times \boldsymbol{c} + \boldsymbol{c} \times \boldsymbol{a}|}{|\overrightarrow{AC}|}\)

Since \(|\overrightarrow{AC}| = |\boldsymbol{c} – \boldsymbol{a}|\), we get the required result.

[3 marks]

✅ Key Concepts:
  1. Vector Perpendicularity: Two vectors are perpendicular if and only if their dot product \(\boldsymbol{a} \cdot \boldsymbol{b} = 0\).
  2. Cross Product Identity: The magnitude relationship \(|\boldsymbol{a} \times \boldsymbol{b}|^2 + (\boldsymbol{a} \cdot \boldsymbol{b})^2 = |\boldsymbol{a}|^2|\boldsymbol{b}|^2\).
  3. Triangle Area: Area \(= \frac{1}{2}|\boldsymbol{u} \times \boldsymbol{v}|\) for any two sides \(\boldsymbol{u}, \boldsymbol{v}\).
  4. Point-Line Distance: Distance \(= \frac{|\boldsymbol{a} \times \boldsymbol{b} + \boldsymbol{b} \times \boldsymbol{c} + \boldsymbol{c} \times \boldsymbol{a}|}{|\boldsymbol{c} – \boldsymbol{a}|}\).
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