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A vector area problem ~ must be handed in tomorrow Given the vector AB = (6,1), the vector BC = (x, y), and the vector CD = (- 2, - 6). If the vector AC is perpendicular to the vector BD, find the values of X, y and the area of ABCD
The following two consecutive capital letters are vectors,
AC=BC+AB=(X+6,Y+1)
BD=BC+CD=(X-2,Y-6)
∵AC⊥BD,∴(X+6)(X-2)+(Y+1)(Y-6)=0
∴X1=-6,X2=2; Y1=6,Y2=-1
Ψ AC = (0,7), BD = (- 8,0) or AC = (8,0), BD = (0, - 7)
The area of ABCD = | AC | * | BD | / 2 = 28
Finding triangle area with vector~
It is known that the coordinates of the three vertices of a triangle are a (- 5, - 1) B (4,1) C (0,4)
How to find the area of triangle?
If the quadrilateral ABCD is a parallelogram, calculate the D coordinate of the point
The intersection of the line AB and the y-axis is at point P
S△ABC=S△PCB+S△PAC
The AB line is y = (2x + 1) / 9, P (0,1 / 9)
S△ABC=S△PCB+S△PAC=1/2 *(5+4)*1/9=1/2
Learn determinant, use determinant formula to calculate, very simple
DA=CB
(x,y)-(-5,-1)=(0,4)-(4,1)
(x,y)=(-4,3)+(-5,-1)=(-9,2)
Triangle area vector formula
Suddenly forgot to find the vector formula of triangle area ~ did not bring back the book ~ is doing homework said~
1/2absinaC
How can the area of a triangle be represented by the vectors corresponding to its two sides?
The corresponding vectors are a vector and B vector respectively
Let the angle between the two vectors be theta,
A vector · B vector
cosθ = ------------------
|A vector | ·| B vector|
Sin θ = √ (1-cos θ) & sup2;
√ [| a vector | & sup2; ·| B vector | & sup2; - (a vector · B vector) & sup2;]
= ---------------------------------------------
|A vector | ·| B vector|
Triangle area s = | a vector | ·| B vector | · sin θ / 2
√ [| a vector | & sup2; ·| B vector | & sup2; - (a vector · B vector) & sup2;]
= --------------------------------------------------
two
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