The question in the image reads as follows:

June 2019

Given that lines l₁ and l₂ have vector equations:

l₁: r = 2i + 3j + λ(3i - 5j)
l₂: r = 3i + 4j + μ(i + j)

where λ and μ are scalar parameters.

(i) Find the position vector of the point of intersection of l₁ and l₂.
(ii) Find the cosine and angle between l₁ and l₂.

Let me solve this systematically:

(i) Point of intersection

For the lines to intersect, their position vectors must be equal. That is, equating $r$ for $l_1$ and $l_2$:
$2i + 3j + \lambda(3i - 5j) = 3i + 4j + \mu(i + j)$

Expanding and collecting components of $i$ and $j$, we get:

$(2 + 3\lambda)i + (3 - 5\lambda)j = (3 + \mu)i + (4 + \mu)j$

From this, we equate components of $i$ and $j$:

1. $2 + 3\lambda = 3 + \mu$
2. $3 - 5\lambda = 4 + \mu$

Now solve these two equations simultaneously:

Step 1: Solve for $\mu$ in terms of $\lambda$ using equation (1):

$\mu = 2 + 3\lambda - 3 = -1 + 3\lambda$

Step 2: Substitute $\mu = -1 + 3\lambda$ into equation (2):

$3 - 5\lambda = 4 + (-1 + 3\lambda)$ Simplify: $3 - 5\lambda = 4 - 1 + 3\lambda$ $3 - 5\lambda = 3 + 3\lambda$ $-5\lambda - 3\lambda = 3 - 3$ $-8\lambda = 0$ $\lambda = 0$

Step 3: Substitute $\lambda = 0$ into $\mu = -1 + 3\lambda$:

$\mu = -1 + 3(0) = -1$

Step 4: Find the point of intersection:

Substitute $\lambda = 0$ into $l_1: r = 2i + 3j + \lambda(3i - 5j)$: $r = 2i + 3j$

Verify using $l_2$ with $\mu = -1$:
Substitute $\mu = -1$ into $l_2: r = 3i + 4j + \mu(i + j)$: $r = 3i + 4j - (i + j) = (3 - 1)i + (4 - 1)j = 2i + 3j$

Hence, the point of intersection is:
$\boxed{2i + 3j}$

(ii) Cosine and angle between the lines

The direction vectors of the lines are:

• For $l_1$: $\mathbf{d_1} = 3i - 5j$
• For $l_2$: $\mathbf{d_2} = i + j$

The cosine of the angle $\theta$ between the lines is given by: $\cos\theta = \frac{\mathbf{d_1} \cdot \mathbf{d_2}}{\|\mathbf{d_1}\| \|\mathbf{d_2}\|}$

Step 1: Find $\mathbf{d_1} \cdot \mathbf{d_2}$:

$\mathbf{d_1} \cdot \mathbf{d_2} = (3i - 5j) \cdot (i + j)$ $= (3 \cdot 1) + (-5 \cdot 1) = 3 - 5 = -2$

Step 2: Find $\|\mathbf{d_1}\|$ and $\|\mathbf{d_2}\|$:

$\|\mathbf{d_1}\| = \sqrt{3^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34}$ $\|\mathbf{d_2}\| = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}$

Step 3: Find $\cos\theta$:

$\cos\theta = \frac{-2}{\sqrt{34} \cdot \sqrt{2}} = \frac{-2}{\sqrt{68}} = \frac{-2}{2\sqrt{17}} = \frac{-1}{\sqrt{17}}$

Step 4: Find the angle $\theta$:

$\theta = \cos^{-1}\left(\frac{-1}{\sqrt{17}}\right)$

Final Answers:

1. Point of intersection: $\mathbf{2i + 3j}$
2. Cosine and angle:
   $\cos\theta = \frac{-1}{\sqrt{17}}, \quad \theta = \cos^{-1}\left(\frac{-1}{\sqrt{17}}\right)$

Would you like further explanation or additional steps?

Here are 5 related questions to expand your understanding:

1. How do we check if two lines in 3D space are parallel?
2. What is the formula for finding the shortest distance between two skew lines?
3. How do we calculate the sine of the angle between two vectors?
4. How can we verify if two lines intersect in 3D space instead of 2D?
5. What is the significance of the dot product in vector geometry?

Tip: Always verify the direction vectors when solving vector-related problems to avoid calculation errors.