Question

Find the equation of a line (in vector and cartesian form) that passes through the point of intersection of lines \( \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} \) and \( \frac{x - 4}{5} = \frac{y - 1}{2} = z \) and is parallel to the vector \( 3\hat{i} + 2\hat{j} - 8\hat{k} \).

Hint:

• Always check the intersection point against all three coordinates to ensure the lines actually intersect.
• "Parallel to a vector" means the line takes that vector as its direction ratios.

Answer 1

Concept:
• To find the intersection of two lines, express general points in terms of parameters \( r \) and \( s \).
• A line passing through \( \vec{a} \) parallel to \( \vec{b} \) is \( \vec{r} = \vec{a} + \lambda \vec{b} \).

Step 1: Find the point of intersection
Line 1: \( x = 2r+1, y = 3r+2, z = 4r+3 \)
Line 2: \( x = 5s+4, y = 2s+1, z = s \)
Equating \( z \): \( s = 4r + 3 \)
Equating \( y \): \( 3r + 2 = 2s + 1 \implies 3r + 2 = 2(4r + 3) + 1 \implies 3r + 2 = 8r + 7 \)
\( -5 = 5r \implies r = -1 \)
Then \( s = 4(-1) + 3 = -1 \).
Check \( x \): \( 2(-1)+1 = -1 \) and \( 5(-1)+4 = -1 \). (Matches)
Intersection point \( P(-1, -1, -1) \).

Step 2: Write the equation in Vector form
Point \( \vec{a} = -\hat{i} - \hat{j} - \hat{k} \), Direction \( \vec{b} = 3\hat{i} + 2\hat{j} - 8\hat{k} \).
\( \vec{r} = (-\hat{i} - \hat{j} - \hat{k}) + \lambda(3\hat{i} + 2\hat{j} - 8\hat{k}) \)

Step 3: Write the equation in Cartesian form
Passing through \( (-1, -1, -1) \) with DRs \( (3, 2, -8) \):
\( \frac{x + 1}{3} = \frac{y + 1}{2} = \frac{z + 1}{-8} \)