Help in Vectors ( H2 MATHS )

lavastar SG

1a ) The triangle ABC has its verticles at the points A (0,0,1) , B(1,0,1) , C(2,1,3). Find the area of the triangle

b) THe parallelogram ABCD has three of its verticles at the points (1,2,-1), (1,3,2) and (-1,3,-1) . Find the area of the parallelogram. ( Ans : 7 )

For this question, how can i know which verticles correspond to which side of the figure? no diagrams are given.. Confused

2. Given the points A ( 1,2,-1), B(3,4,0) , C(1,5,-2) and that O is the origin, find the shortest distance of B on the line AC. ( ans : 0.5 X sq root 26)

For this question , im unsure of what the question wans me to find.

Thanks for any help Smile

popikachu SG

Erm...
I don't understand the question as well...
i dunno why got 3 numbers in the brackets de...
lol

but... for question 2 find the shortest distance from B to the line AC lor...
find the area for the triangle first then times 2 then divide by length of AC...
^^

eagle SG

1a)
Can use formula I think

1/2 * |
0 1 2 0 1
0 0 1 0 0
1 1 3 1 1
|

Area = 1/2 * {(0 + 1 + 0) - (0 + 0 + 0) } = 1/2 units square
Cannot really remember... If wrong, then have to use the following method

a) Find length of AB
b) Find midpoint of AB
c) Find dist from midpoint AB to C
d) multiply length of AB with dist from midpoint AB to C and then divide by 2

eagle SG

By inspection, we can see that (1,2,-1) is beside (1,3,2) and (1,2,-1) is also beside (-1,3,-1). So is area of triangle = 1/2 of parallelogram

Find area of this triangle (shoelace formula) =
1/2 * |
1 1 -1 1 1
3 2 3 3 2
2 -1 -1 2 -1 |

= 1/2 * | (-2 + 6 + 3) - (-4 + -3 + -3) | = 17/2

Area of parallelogram = 17

Note: Check... I not very sure of this answer too...

Try and visualise

Or

Draw out in 2D plane (x,y), then visualise the z vector yourself

eagle SG

Treat AC as a line
B as a point

Find shortest distance from B to some point on AC
Got formula for A levels, but I forgot, so I shall proceed to derive it

Let the pt of intersection be M
So BM must be perpendicular to AC.

So angle BAM = angle BAC = cos inverse ( BA.CA / ( BA * CA ) )

BA= (-2, -2, -1)
BA = 3

CA = (0, -3, 1)
CA = sqrt (10)

And

so BA.CA= 0 + 6 + -1 = 5

cos of angle BAM = 5 / 3(sqrt 10)

Length BM = length BA * cos angle BAM
= 5 / sqrt (10)

Somehow I quite sure of my answer, but it's different from your answer... Sad

SBS261P SG

de answer given by the qn is correct la =)

i use long method, i noe gt shorter one is use cross product but i use dot product all the way

First i find vector AC which is (0 3 -1) [write it vertically]
den find unit vector AC which is 1/sqrt10 x (0 3 -1) [write it vertically]

den i find vector AB which is (2 2 1) [write it vertically]

den i find de vector AM (M is where B meets AC perpendicularly) by using

(vector AB . unit vector AC) times (unit vector AC)

and find that it is (0 1.5 -0.5) [write it vertically]

den find vector BM (use vector AM minus vector AB) to get (-2 -0.5 -1.5) [write it vertically]

den find de length which is sqrt (2^2 + 0.5^2 + 1.5^2) = sqrt 6.5 = 0.5 x sq rt 26 =D

forgot how to use de cross product alr. sorry ah in NS now de brain a bit slowded Laughing

SBS261P SG

it shld be sin angle BAM, so i think tt's where ur answer went off =)

eagle SG

yup, you are right.

I did everything mentally, thus very prone to careless Crying or Very sad

SBS3625D SG

Haiya, draw the diagram and you can see better.

Use the formula 0.5 x sin........ to find the area.

What fuck. Forgot the formula already. Siao liao!

eagle SG

can derive formula one

Last time all my vectors tutorial, I derive all these stupid formulas... Just get the basic formulae right can liao.

But then, I got quite a number of things to do, so do mentally then save time...

SBS3625D SG

I got it.

0.5 x ab sin c.

Where abc are the 3 points of the triangle. Very Happy