# What is the vector of position of the intersection point of the lines d1 and d2? r=2i+j+m(i+3j) r=6i-j+n(i-4j)

hala718 | High School Teacher | (Level 1) Educator Emeritus

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r= 2i + j + m(i+3j) .............(1)

r= 6i - j + n(i-4j)...............(2)

To find the point f intersection, then eq. (1) = eq.(2)

==> 2i + j + m(i+3j) = 6i - j + n(i-4j)

==>Let us expand brackets:

==> 2i + j + mi + 3mj = 6i - j + ni - 4nj

==> (2+m) i + (1+3m)j = (6+n)i + (-1-4n)j

==> (2+ m - 6 - n)i + (1+3m +1 + 4n) j = 0

==> (-4+m-n) i + (2+3m+4n) = 0

==> -4 + m - n = 0

==> m= n+4

==> 3m + 4n+2= 0

==> 3(n+4) + 4n = -2

==> 3n + 12 + 4n =- 2

===> 7n = -14

==> n = -2

==> m = n+ 4 = -2 + 4 = 2

==> m= 2

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

The vector of position of the intercepting point between the 2 lines has to verify the vectorial equations of the lines. We'll note the vector of position as p:

p =  2i+j+m(i+3j)

Since the point is located on the second line, the vector of position is also verifying the vectorial equation of the second line:

p = 6i-j+n(i-4j)

2i+j+m(i+3j) = 6i-j+n(i-4j)

We'll remove the brackets both sides:

2i + j + mi + 3mj = 6i - j + ni - 4nj

We'll combine like terms both sides:

2i + j + mi + 3mj = 6i - j + ni - 4nj

i(2 + m) + j(1 + 3m) = i(6+n) + j(-1 - 4n)

For expressions to be equivalent, the correspondent coefficients to the unit vectors, have to be equal.

﻿﻿2 + m﻿ = 6 + n

m - n = 4 (1)

1 + 3m = -1 - 4n

3m + 4n = -2 (2)

We'll multiply (1) by 4 and we'll add the resulting expression to (2):

4m - 4n + 3m + 4n = 16 - 2

We'll eliminate and combine like terms:

7m = 14

m = 2

We'll substitute m in (1):

﻿m - n = 4

2 - n = 4

n = -2

The vector of position of the intercepting point of the 2 lines is:

p = ﻿i(2 + 2) + j(1 + 3*2)

p = 4i + 7j