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Given that a and b are non-collinear non-zero vectors, if there is a vector c such that c is parallel to a and c is parallel to b, then c equals?
C is zero vector ~
Is the direction vector of the straight line ax+by+c=0(abc is not equal to 0)? What is the normal vector? Given the direction vector (3,2), find the slope? Given the normal vector (3,2), find the slope? What is the direction vector of the straight line ax+by+c=0(abc is not equal to 0)? What is the normal vector? Given the direction vector (3,2), find the slope? Given the normal vector (3,2), find the slope?
0
In the triangle ABC, the opposite sides of the angle ABC are respectively abc, which satisfies 2 times the vector AB multiplied by the square-(b+c) of the vector AC=a Find the size of angle A. In the triangle ABC, the opposite sides of the angle ABC are respectively abc, satisfying 2 times the vector AB multiplied by the square of the vector AC=a-(b+c) Find the size of angle A. In the triangle ABC, the opposite sides of the angle ABC are respectively abc, satisfying 2 times the vector AB multiplied by the square-(b+c) of the vector AC=a Find the size of angle A.
0
Given vector abc, satisfying |a|=1,|b|=2, c=a+b, c is vertical a, then the angle between a and b is equal to
Multiply both sides of the equation by a
Then ac=a^2+ab=|a |^2+ab
Ac =0 because a and c are perpendicular
|A |^2=1
So ab=-1
Given by |a|=1,|b|=2,|a||b|=2
Cos=ab/(|a||b|)=-1/2
So an angle between a and b is 120 degrees
Multiply both sides of the equation by a
Then ac=a^2+ab=|a |^2+ab
Because a and c are perpendicular, ac =0
|A |^2=1
So ab=-1
Given by |a|=1,|b|=2,|a||b|=2
Cos=ab/(|a||b|)=-1/2
So an angle between a and b is 120 degrees
Multiply both sides of the equation by a
Then ac=a^2+ab=|a |^2+ab
Because a and c are perpendicular, ac =0
|^2=1
So ab=-1
Given by |a|=1,|b|=2,|a||b|=2
Cos=ab/(|a||b|)=-1/2
So an angle between a and b is 120 degrees
In △ABC, the edges of angles A, B and C are a, b and c, respectively, and satisfy cos (A/2)=2√5/5. Multiply vector AB by vector AC equal to 3.1, find the area of △ABC.2 If c=1, find the value of a. In △ABC, the edges of angles A, B, C are a, b, c, respectively, and satisfy cos (A/2)=2√5/5. Multiply vector AB by vector AC equal to 3.1, find the area of △ABC.2 If c=1, find the value of a. In △ABC, the edges of angles A, B, C are a, b, c, respectively, and satisfy cos (A/2)=2√5/5. Multiply vector AB by vector AC equal to 3.1. Find the area of △ABC 2. If c=1, find the value of a.
From the question
1CosA=2[ cos (A/2)]^2-1=3/5, sinA=4/5.
Vector AB*AC=|AB AC|*3/5=3,
S△ABC=(1/2)|AB AC|sinA=2.
2 From 1, b=5.
A^2=25+1-10*3/5=20,
A=2√5
In △ABC, A (-1,1), B (3,1), C (2,5), the bisector of the internal angle of angle A intersects the opposite edge of D, then the vector The AD coordinates are equal to ___.
Let D (x, y) be AC=(2+1)2+(5-1)2=5, AB=4. According to the bisector theorem of triangle inner angle, BD=45DC, i.e. BD=45DC.(x-3, y-1)=45(2-x,5-y), x-3=45(2-x) y-1=45(5-y), x=229y=259, AD=(329,169). The answer is:(329,169).
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