IB DP Maths Topic: SL 3.8 :Solving trigonometric equations in a finite interval, both graphically and analytically HL Paper 1

(a)  To find the value of \( 4t \), we begin by understanding that the dot product of two vectors gives us the product of their magnitudes and the cosine of the angle between them.
For vectors \( \overrightarrow{PA} \) and the direction vector of \( L_1 \), this can be expressed as:
\(
\overrightarrow{PA} \cdot \overrightarrow{L_1} = |\overrightarrow{PA}||\overrightarrow{L_1}|\cos(\theta)
\)
Given that the acute angle \( \theta \) between the two lines is:
\(
\frac{\pi}{3}
\)
and the direction vector of \( L_1 \) is \( (1, 1, 0) \), we can calculate the dot product of \( \overrightarrow{PA} \) and \( \overrightarrow{L_1} \) using the provided vectors:
\(
\overrightarrow{PA} = (2t, 0, 3 + t) \quad\text{and}\quad \overrightarrow{L_1} = (1, 1, 0)
\)
The dot product \( \overrightarrow{PA} \cdot \overrightarrow{L_1} \) is:
\(
2t \times 1 + 0 \times 1 + (3 + t) \times 0 = 2t
\)
Now, we calculate the magnitude of vector \( \overrightarrow{PA} \):
\(
|\overrightarrow{PA}| = \sqrt{(2t)^2 + 0^2 + (3 + t)^2}
\)
\(
= \sqrt{4t^2 + 0 + 9 + 6t + t^2}
\)
Simplifying:
\(
|\overrightarrow{PA}| = \sqrt{5t^2 + 6t + 9}
\)
The magnitude of vector \( \overrightarrow{L_1} \) is:
\(
|\overrightarrow{L_1}| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2}
\)
Now, substitute the values into the dot product equation:
\(
2t = \sqrt{5t^2 + 6t + 9} \times \sqrt{2} \times \cos\left(\frac{\pi}{3}\right)
\)
Since:
\(
\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}
\)
the equation simplifies to:
\(
2t = \frac{1}{2} \times \sqrt{5t^2 + 6t + 9} \times \sqrt{2}
\)
Multiplying both sides by 2 to eliminate the fraction yields:
\(
4t = \sqrt{(5t^2 + 6t + 9) \times 2}
\)
Squaring both sides to eliminate the square root gives:
\(
(4t)^2 = (5t^2 + 6t + 9) \times 2
\)
Expanding both sides:
\(
16t^2 = 10t^2 + 12t + 18
\)
Subtracting \( 10t^2 \) from both sides:
\(
6t^2 = 12t + 18
\)
Dividing through by 6:
\(
t^2 = 2t + 3
\)
Thus:
\(
4t = 4\sqrt{(t^2)} = 4\sqrt{(2t + 3)}
\)
Substituting \( t^2 = 2t + 3 \) into our earlier expression for \( 4t \) gives:
\(
4t = \sqrt{10(2t + 3) + 12t + 18}
\)
Simplifying inside the square root:
\(
4t = \sqrt{(20t + 30 + 12t + 18)}
\)
\(
= \sqrt{(32t + 48)}
\)
\(
= \sqrt{(10t^2 + 12t + 18)}
\)

(b) The problem requires us to find the value of \( t \) where two lines \( L_1 \) and \( L_2 \) intersect at point \( P \).
To find this, we use the dot product of the direction vector of \( L_1 \), \( \mathbf{1} \), and the vector \( \overrightarrow{PA} \), with the given acute angle between the two lines.
The dot product is given by:
\(
1(2t) + 1(0) + 0(3 + t) = 2t
\)
Since the dot product also equals the product of the magnitudes of the two vectors and the cosine of the angle between them, we set up the equation:
\(
2t = ||\mathbf{1}|| \times ||\overrightarrow{PA}|| \times \cos\left(\frac{\pi}{3}\right)
\)
After calculating the magnitudes, the equation simplifies to:
\(
2t = \sqrt{2} \times \sqrt{4t^2 + (3 + t)^2} \times \frac{1}{2}
\)
Squaring both sides to eliminate the square roots, we solve the resulting quadratic equation.
The solutions obtained are \( t = -1 \) and \( t = 3 \).
Since the question specifies \( t > 0 \), the valid solution is:
\(
t = 3
\)

(c) Distance required  = \(\frac{|v\times \underset{PA}{\rightarrow}|}{|v|}\)

Shortest distance is \(\sqrt{54}\)

(d)  To solve for the normal vector to the plane \( \pi \) that contains lines \( L_1 \) and \( L_2 \), we must first understand that the normal vector is perpendicular to both direction vectors of \( L_1 \) and \( L_2 \).
The direction vector of \( L_1 \) is given as:
\(
\begin{pmatrix}
1 \\
1 \\
0
\end{pmatrix}
\)
and to find the direction vector for \( L_2 \), we can use the vector \( \overrightarrow{PA} \), where \( P \) is the point of intersection and \( A \) is a point on \( L_1 \), given by:
\(
\overrightarrow{PA} =
\begin{pmatrix}
2t \\
0 \\
3 + t
\end{pmatrix}
\)
The vector product (cross product) of these two vectors will give us the normal vector to the plane.
Let’s calculate the cross product of:
\(
\begin{pmatrix}
1 \\
1 \\
0
\end{pmatrix}\)
and
\(\begin{pmatrix}
2t \\
0 \\
3 + t
\end{pmatrix}
\)
The cross product is calculated as follows:
\(
\mathbf{n} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 1 & 0 \\
2t & 0 & 3 + t
\end{vmatrix}
\)
Expanding the determinant:
\(
= \mathbf{i}(1 \cdot (3 + t) – 0 \cdot 0) – \mathbf{j}(1 \cdot (3 + t) – 0 \cdot 2t) + \mathbf{k}(1 \cdot 0 – 1 \cdot 2t)
\)
Simplifying:
\(
= \mathbf{i}(3 + t) – \mathbf{j}(3 + t) – 2t\mathbf{k}
\)
Which gives the normal vector:
\(
\begin{pmatrix}
3 + t \\
-(3 + t) \\
-2t
\end{pmatrix}
\)
Since the vector product results in a vector that is normal to the plane containing both lines, the answer is:
\(
\begin{pmatrix}
3 + t \\
-(3 + t) \\
-2t
\end{pmatrix}
\)
However, the marking scheme indicates that we should award \( A0 \) for a final answer in coordinate form.
Hence, we need to express our vector in terms of \( \eta \), which is any scalar multiple of:
\(
\begin{pmatrix}
1 \\
-1 \\
-1
\end{pmatrix}
\)
Looking at our vector, we can see that it is indeed a scalar multiple of the given vector \( \eta \), where the scalar is \( 3 + t \).
Therefore, our final answer for the normal vector to the plane \( \pi \), taking into account the scalar multiple, is:
\(
\begin{pmatrix}
3 + t \\
-(3 + t) \\
-2t
\end{pmatrix}
\)
which is a scalar multiple of:
\(
\begin{pmatrix}
1 \\
-1 \\
-1
\end{pmatrix}
\)
and satisfies the condition:
\(
\mathbf{n} \neq
\begin{pmatrix}
1 \\
-1 \\
-1
\end{pmatrix}
\text{ or }
\begin{pmatrix}
-1 \\
-1 \\
-1
\end{pmatrix}
\)

(e)  We begin by using the volume formula for a cone, \( V = \frac{1}{3} \pi r^2 h \), and equating it to the given volume to find the height, \( h \):
\(
\frac{1}{3}\pi(3\sqrt{6})^2 h = 90\sqrt{3}\pi \implies h = 5\sqrt{3}
\)
The position vector of the vertex is given by \( \overrightarrow{OA} + \mu \mathbf{n} \), where \( \mu \) is a parameter and \( \mathbf{n} \) is a unit vector in the direction of the height of the cone.
Recognizing that the magnitude of the height vector \( \mu \mathbf{n} \) is equal to \( h \), we have:
\(
|\mu \mathbf{n}| = 5\sqrt{3} \implies \mu^2 = 75 \implies \mu = \pm 5 \quad\text{(accept \( \mu = 5 \))}
\)
Next, we attempt to find the cone’s height vector \( h\mathbf{n} \) by normalizing the direction vector of \( L_1 \), which yields:
\(
\mathbf{n} = \frac{1}{\sqrt{3}}\mathbf{i} – \frac{1}{\sqrt{3}}\mathbf{j} – \frac{1}{\sqrt{3}}\mathbf{k}
\)
Then, we calculate the two possible positions of the vertex by adding the height vector \( 5\mathbf{n} \) to the position vector of point \( A(2t, 8, 3) \):
\(
\text{Vertex} = \mathbf{A} \pm 5\mathbf{n} = (2t, 8, 3) \pm 5\left(\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right)
\)
\(
= \left(2t + \frac{5}{\sqrt{3}}, 8 – \frac{5}{\sqrt{3}}, 3 – \frac{5}{\sqrt{3}}\right)
\)
OR
\(
\left(2t – \frac{5}{\sqrt{3}}, 8 + \frac{5}{\sqrt{3}}, 3 + \frac{5}{\sqrt{3}}\right)
\)
\(
= (11, 3, -2) \text{ and } (1, 13, 8) \quad\text{(accept position vectors)}
\)