Help in Vectors ( H2 MATHS )
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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..
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
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Erm...Originally posted by lavastar: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..
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
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...
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1a)Originally posted by lavastar: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
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 -
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 parallelogramOriginally posted by lavastar: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 )
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 visualiseOriginally posted by lavastar:For this question, how can i know which verticles correspond to which side of the figure? no diagrams are given..
Or
Draw out in 2D plane (x,y), then visualise the z vector yourself -
Treat AC as a lineOriginally posted by lavastar: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.
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...
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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
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it shld be sin angle BAM, so i think tt's where ur answer went off =)Originally posted by eagle:Length BM = length BA * cos angle BAM
= 5 / sqrt (10)( -
yup, you are right.Originally posted by SBS261P:it shld be sin angle BAM, so i think tt's where ur answer went off =)
I did everything mentally, thus very prone to careless
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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! -
can derive formula oneOriginally posted by SBS3625D: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!
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... -
I got it.Originally posted by eagle: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...
0.5 x ab sin c.
Where abc are the 3 points of the triangle.