maths question help !! (1 Viewer)

[poptarts12345]

How do I do this question :

In triangle abc, d is the midpoint of bc and angle bac is 90 degrees. Use the vector method to show that length da= length bd.

 

[Everwinter]

Hope this helps, if you have any part which you don't understand, you can always ask. It's an interesting question, because I didn't believe this statement to be true when I first saw your post. I rushed this so some of the vectors do not have the dash.

 

[Everwinter]

The thought process for this one, is first to draw a diagram of course. The question requires length, so to convert a vector to length, one effective way is the dot product, which the dot product of itself is equal to the distance squared, e.g. x • x = |x|^2. So in the R.T.P. I do that to make the final answer easier to find. Since the information contains 90°, and this probably only works for 90°, knowing that when a is perpendicular to c, a • c = 0, I quickly find vector AD in respect to a and c, this is also the reason why c is pointing outwards, because dot product only work this way and we don't want c to be negative. I believe the process for this is quite clear in my response.

Then as I initially planned, I tried to find the equation that is equivalent to |AD|^2, which is the dot product of the alternate equation for vector AD which I found earlier. I expanded this, since there is 1/4, I know I am probably in the right track, so I turned a • c to zero, because what I was initially going for. This left me |a|^2 + |c|^2, it may be a bit difficult to realise the use of Pythagoras here, but when you look back to the R.T.P. this should be fairly obvious. And after that, boom finished.

 

[poptarts12345]

Hope this helps, if you have any part which you don't understand, you can always ask. It's an interesting question, because I didn't believe this statement to be true when I first saw your post. I rushed this so some of the vectors do not have the dash.

thanks ur a bloody legend

 

[CM_Tutor]

This question is related to an important theorem in circle geometry, that the angle in a semicircle is a right angle. That is:

(a) if AB is the diameter of a circle and C is any point on the circle (other than A or B) then triangle ABC is right-angled at C. A, B, and C must be equidistant from O, the midpoint of AB, as OA, OB, and OC are equal radii.

(b) if ABC is a triangle right-angled at C, then C lies on the circle with AB as diameter, and the equidistant result again follows.

It came up as a vector proof in several MX1 Trials in 2020 and you should make sure you can do it in the form of (a). That is, as:

AB is the diameter of a circle with centre O and C lies on the circle. Let OA be the vector a and OC be the vector b. By finding vector forms for the sides AC and BC, or otherwise, show that angle ACB is a right angle.