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Given the space three points A (0,2,3), B (-2,1,6), C (1,-1,5)(1), find the area S of a parallelogram with vector A B, AC as a set of adjacent sides
1. AB=√14, AC=√14, BC=√14
ABC is a regular triangle
∠A=60°
Area of parallelogram=AB*AC*sinA=7√3
Vector AB=(3,4), vector AC=(1,-2), then the area of parallelogram with AB and AC as adjacent sides is
AB*AC=3-8=-5
The modulus of AB is 5
Module of AC is root number 5
Cosx=AB*AC/|AB||AC|=-1/Root 5
Sinx =2 nos.5/5
Area is 0.5*|AB||AC|*sinx=5
It is known that A (0,2) B (-2,1) C (1,-1) is the area of the parallelogram ABCD with the vector AB adjacent to the vector AC. Previous picture, please No answer? It is known that A (0,2) B (-2,1) C (1,-1) takes the area of the parallelogram ABCD with the vector AB adjacent to the vector AC. Please take the previous picture No answer? It is known that A (0,2) B (-2,1) C (1,-1) takes the area of the parallelogram ABCD with the vector AB adjacent to the vector AC. Previous picture, please No answer?
First, the area of a parallelogram can be divided into the sum of the areas of two triangles, so you just need to multiply the area of the triangle ABC by 2. Draw a graph on the coordinate axis, and the area of the triangle is 7/2, so the area of the parallelogram is 7
Given point A (6,1)(1,3)(3,1), then vector AB is projected on vector BC as 、、、、、、
Cross point A to make the vertical line of straight line BC intersect with D to find the length of BD.
Find the slope of the straight line BC, and then the negative reciprocal is the slope of the straight line AD. Let AD and BC intersect to find the coordinate of D (9/2,-1/2). Then B D uses the distance formula to find the second of seven
Given vector AB=(6,1), vector BC=(X, y), vector CD=(-2,-3) (1) If vector BC//vector DA, find the relation between x and y (2) if vector AC vertical vector BD, vector BC parallel vector DA, find the value of x, y and the area of quadrilateral ABCD
AD=AB+BC+CD=(4+x, y-2)
AC=AB+BC=(6+x, y+1)
BD=BC+CD=(x-2, y-3)
(1)
BC//AD
X (y-2)= y (x+4)
(2)
AC⊥BD
(X+6)(x-2)+(y+1)(y-3)=0
X=2; y=-1 or x=-6; y=3
AC=(8,0) or AC=(0,4)
BD=(0,-4) or BD=(-8,0)
Area of quadrilateral ABCD S=8*4÷2=16
Given three points A (0,2,3), B (-2,1,6), C (1,-1,5).
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