Write down vector equations of the lines AD and BC and find the position vector of the point of intersection.Points A,B,C,D in a plane have position vectors a = 6i + 8j, b = 3/2a, c = 6i + 3j, d =...
Write down vector equations of the lines AD and BC and find the position vector of the point of intersection.
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)`
`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)`
`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.
`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.