Need lots of help with vectors

• Han chiang

  4 Jan 11, 15:43

  Hi forumers, I am doing revision on the topic vectors and I encountered some difficulties. I cannot visualise or draw out a diagram to even get started..

  c is a constant for lamda which I don't know how to type.

  1) The position vectors of A, B with respect to origin O are a, b respectively.

  i) Prove that p = ca + (1 - k)b is collinear with A and B.   I can do this

  iii) The point R has position vector r given by r = 3a + 5b. Find the position vector of the point where OR meets AB (produced if necessary)

  iv) Prove that the value of c for which OP and AB are perpendicular is given by 

       c = [b . (b - a) ] / [ (b - a) . (b - a) ]

  Ans iii: 3a/8 + 5b/8

  2) line 1: r = (i + 3j + 2k) + c(i - j + 7k)

      line 2: x = 1, (y - 3)/ -4 = (z - 2)/ 3

      The point P lies on line1 with position vector (ai + j + 16k). The point Q lies on line2     such that PQ is perpendicular to line2.

  i) Prove that a = 3    I can do this.

  ii) Find the position vector of the point Q.

  Ans ii: (i - 5j + 8k)

  And I am abit confused with position vectors and the other vectors. I know position vector is like the P in question 2 above. What about line1 and line 2 ? Are they also position vectors?

  I do not expect full solutions to be provided, but I need help on how to draw a diagram for the vectors. 3D is so hard to visualise, not like the cartesian graphs that I am always used to..

  Thanks.

   

• wee_ws

  4 Jan 11, 16:10

  Hi,

    Suppose OR meets AB at X. Then OX = mu OR, where OX and OR are vectors.

    X, A, and B are collinear, so OX = lambda a + (1 - lambda) b for some value of lambda.

    Now lambda a + (1 - lambda) b = mu (3 a + 5 b), from which lambda and mu can be found.

    Equation of line 1 is expressed in vector form, i.e., r = a + lambda d, where

    r represents the position vector of a general point on line 1,

    a represents the position vector of a specific point on line 1,

    and d represents the direction vector.

    We could compare this form with our familiar equation of a line in two-dimensional space, i.e., y = mx + c. This line passes through the y-intercept (a specific point) and has gradient m (quite similar to the direction vector). In vector form, we will obtain

    r = (0, c) + lambda (a, b), where b/a = m.

    Equation of line 2 is expression in cartesian form, which can be converted to the vector form and vice-versa. Line 2 passes through the point (1, 3, 2) and has direction vector (0, -4, 3).

    Thanks.

  Cheers,
  Wen Shih

• Han chiang

  4 Jan 11, 16:27

  Huh blurred @_@ why OX is lambda a + ( 1 - lambda )b ? I got a + lambda(b - a)

  And after to the next equation where I equate the above one to the mu OR. There are 2 unknowns in the equation, how to find the mu and lambda??

• wee_ws

  4 Jan 11, 18:25

  Hi,

    We are using the result of (i).

    The values of lambda and mu can be found by comparing the coefficients of a and b.

    Lines and planes are most crucial for this topic, so put effort in those areas.

    Thanks.

  Cheers,
  Wen Shih

• Han chiang

  4 Jan 11, 19:02

  Ahhh xie xie ni.. For most questions I need to draw my own diagram. So the vectors A or B can be anywhere as long as it follow the question? e.g. If question say OP meet AB somewhere I have to just make sure that they intersect?

  If the question says vector AB is perpendicular to a line or to a vector. Then i must do the dot product? But how to dot lambda a + (1 - lambda)b and a vector or line together? I have quite alot of perpendicular questions that I don't know how to do the dot product. My notes don't have vectors perpendicular to line =( 

• wee_ws

  5 Jan 11, 05:49

  Hi,

    Yes, to your first three questions :)

    The dot product of two vectors (a1, a2, a3) and (b1, b2, b3) is given by

    (a1)(b1) + (a2)(b2) + (a3)(b3). For example, (1, 2, 3) . (0, -5, 1) = -7.

    For perpendicular vectors, their dot product is zero. For example, (1, 2, 3) is perpendicular to (1, -1/2, 0) since their dot product is zero.

    Where lambda is involved, the dot product leads us to an equation in which lambda can be found.

    Thanks.

  Cheers,
  Wen Shih