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

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### 1 Answer

The vector of position of the intercepting point of the 2 lines has to verify the vectorial equations of the lines.

Let the vector of position be 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)

Comparing, we'll get:

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