4. (a) The parametric equation is r (-1, 2)+A(1,3) (1+, V2+3A). Equating r (x,y)

Question

4. (a) The parametric equation is r (-1, 2)+A(1,3) (1+, V2+3A). Equating r (x,y) (b) The parametric equation is r (3,0) + A(0,7) (3,7A). Equating r (r, y) gives z 3, so 5. (i) We have a = (3,2,-1) and Aj = (3,-7,3). Then the parametric equation of the line gives 1+A,y v2+ 3A and so X1(v- 2)/3, which gives the Cartesian equation y 3x+3+V2 this is the Cartesian equation of the line. through A and B is r- a AAB (3,2,-1)+A(3,-7,3) -(3+3,2-7A,-1+31). (x, y, z), then we can write Or, if r z = 3+3A y 2-7 2 1+3. For the Cartesian form, we can write 2 z +1 3 (ii) For the point P to lie on this line one-third of the way from A to B, we have A 1/3. So, substitute X-1/3 into (1) to get the position vector for the point P: p= (3,2,-1)+ 3,-7,3)= (4,-1,0).

Explanation / Answer

I think you are referring to the point (ii) of Question 5 solution. This is straight foward

Point A = (3,2,-1) and Point B = Vector AB + A = (6,-5,2)

Now if i join the line segment AB and find the coordinates of vector AB it will be B - A = (3,-7,3)

So for any point lying on the vector AB

r = a + lambda * AB = (3,2,-1) + lambda(3,-7,3)

Now if we put lambda=0, we get the point A and if we put the value of lambda=1, we get the point B as calculated above

We want the point which is 1/3rd in the way of A to B, i.e. 1/3 of A and 2/3 of B, so the value of lambda must be 1/3. which will give the points

r =  (3,2,-1) + 1/3 * (3,-7,3) = (4,-1/3,0)

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