L_1 is defined by (x+1)/2=(y−3)/3=1−z So we can write this as ((2t−1),(3t+3),(1−t))=r L_2 passes through (5,4,2) and intersects with L_1 at right angles.

Paulkenyo

Paulkenyo

Open question

2022-08-20

L 1 is defined by x + 1 2 = y 3 3 = 1 z
So we can write this as ( ( 2 t 1 ) , ( 3 t + 3 ) , ( 1 t ) ) = r
L 2 passes through (5,4,2) and intersects with L 1 at right angles.
I am asked to determine the point of intersections between those two lines.
I have tried to find the dot product of the directional vector of L1 and another directional vector (x,y,z) So I got this 2x+3y−z=0 and said that x=3,y=2 and z=12
So now L 2 has the following equation r=((5+3t),(4+2t),(2+12t)) However when i equate L 1 and L 2 to find t. The equations aren't satisfied, meaning that t isn't common.

Answer & Explanation

hyttnp

hyttnp

Beginner2022-08-21Added 4 answers

The point of intersection lies on L 1 , so we can express it as P(2t−1,3t+3,1−t) for some t R . We require that the vector from P ( 5 , 4 , 2 ) be orthogonal to the vector (2,3,−1), which is colinear with the line. That is,
( 2 t 1 , 3 t + 3 , 1 t ) ( 5 , 4 , 2 ) ( 2 , 3 , 1 )
This implies,
( 2 t 6 , 3 t 1 , 1 t ) ( 2 , 3 , 1 ) = 0
giving the value of t=1 and the required point as (1,6,0)

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