The position vectors of points A,B,C,D are,

`hata = 6i+8j`

`hatb = 3/2a` ----> `hatb = 3/2(6i+8j) = 9i+12j`

`hatc = 6i+3j`

`hatd = 5/3c` -----> `hatd = 5/3(6i+3j) = 10i+5j`

**Line AD is given by,**

`r = a+t(d-a)`

`r = (6i+8j)+t((10i+5j)-(6i+8j))`

`r = (6i+8j)+t(4i-3j)`

Simplifyng,

`r = (6+4t)i+(8-3t)j`

**Line BC is given by,**

`p = b+s(c-b)`

`p = (9i+12j)+s((6i+3j)-(9i+12j))`

`p = (9i+12j)+s(-3i-9j)`

`P = (9i+12j)-s(3i+9j)`

Simplifying,

`p = (9-3s)i+(12-9s)j`

Now if these two lines, intersect each other,

`(6+4t)i+(8-3t)j = (9-3s)i+(12-9s)j`

The two i, j components on either side must be equal to each other separately.

Therefore,

`6+4t = 9-3s` and `8-3t = 12-9s`

This gives two equations to find t and s.

`4t+3s = 3` and `3t-9s = -4`

Solving these two equations you would get,

`t = 1/3` and `s = 5/9`

so the point of interesection can be found by substituting t value in AD line,

If the point of intersection is E,

`hate = (6+4(1/3))i+(8-3(1/3))j`

`hate = 22/3i+7j`

**The point of intesection of AD and BC is (22/3)i+7j.**

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