Find the vector and Cartesian Equations of the line through the point(1,2,-4) perpendicular to the two lines

Question

Find the vector equation of the line passing through (1, 2, -4) and perpendicular to the two lines

(x−8)/3 = (y+19)/−16 = (z−10)/7 and (x−15)/3 = (y−29)/8 = (z−5)/ −5

Answer 1

Step 1: 

Let the vector equation of the line passing through the point (1,2,−4)(1,2,−4) be

Where b₁ i + b₂ j + b₃ k is the vector parallel to the given line.

Step 2:

Let us consider the lines (x−8)/3 = (y+19)/−16 = (z−10)/7-----(2) 

The direction cosines of the lines are (3,−16,7)

The vector parallel to that line is 3i −16j + 7k 

But it is given the line (1) and (2) are perpendicular to each other. 

Therefore (b₁ i + b₂ j + b₃ k). (3i −16j + 7k) = 0

⇒3b₁ −16b₂ + 7b₃ =0-----(3)

Step 3: 

Consider the line: 

(x−11)/3 = (y−29)/8 = (x−5)/−5----(4) 

The vector which is parallel to the line is 5i + 3j − 5k 

The lines (1) and (4) are perpendicular to each other. 

Therefore (b₁ i + b₂ j + b₃ k).(3i + 8j −5k )= 0 

⇒3b₁ + 8b₂ − 5b₃ =0-----(5)

Step 4: 

On solving equ(3) and equ(5)

Step 5: 

Dividing throughout by 12 

⇒b₁/2 = b₂/3 = b₃/ 6

Therefore direction of the vector which is parallel to line (1) is 

2i + 3j + 6k 

Therefore Equation of the line (1) is