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Given that point D is the midpoint of the edge BC of the triangle, there is a point P in the plane where the triangle ABC is located, which satisfies vector PA + vector BP + vector CP =0, The module of vector PA is divided by the module of vector PD is equal to λ, and the value of λ is obtained.
Because "vector PA + vector BP + vector CP=0", P is the center of gravity of the triangle ABC, and AD is a center line of the triangle, so AP: PD=2:1, so λ=2
Given that D is the midpoint of the edge BC of △ABC, there is a point P in the plane of △ABC, which satisfies PA+ BP+ CP=0, set | AP | | PD|=λ, then the value of λ is ______.
By
PA+
BP+
CP=0, deformed to
PA =
PB+
PC is known from the parallelogram rule of vector addition, PA must be the diagonal of the parallelogram with PB and PC as the adjacent side,
D is the midpoint of BC, so P, D and A are collinear, and D is the midpoint of PA.
Also|
AP|
|
PD|=λ, so λ=2
Therefore, the answer is 2
By
PA+
BP+
CP=0, deformed to
PA =
PB+
PC is known from the parallelogram rule of vector addition, PA must be the diagonal of the parallelogram with PB and PC as the adjacent side,
D is the midpoint of BC, so P, D and A are collinear, and D is the midpoint of PA.
Also|
AP |
|
PD|=λ, so λ=2
Therefore, the answer is 2
In known triangle ABC, AB=8, AC=3, BC=7, A is the center of circle, diameter PQ=4, find the maximum and minimum values of vector BP×CQ The angle between the two vectors is given when the maximum value is taken.
In triangle ABC, cosine theorem shows: cosA=(64+9-49)/(2×3×8)=1/2.(vector AB)*(vector AC)=|AB AC cosA=12. Obviously, the module length of vectors PA and QA is 2, and the included angle is 180o, i.e. AQ=-AP. BP*CQ=(BA+AP)*(CA+AQ)=BA*CA+BA*AQ+AP*CA+AP*AQ*AQ*AQ*.
Circle C:(x-0.5)^2+(y-3)^3=37/4-m. Intersect with line: x+2y-3=0 at P, Q, vector OP times vector OQ=0 to find constant m Circle C:(x-0.5)^2+(y-3)^3=37/4-m. Intersect with line: x+2y-3=0 at P, Q, vector OP times vector OQ=0 to obtain constant m
Let M (3-2y, y), C (0.5,3) be the midpoint of PQ, then MC⊥PQ,(y-3)/(2.5-2y)=2, y=1.6, M (-0.2,1.6)
OP⊥OQ, OM=MP=root (CP^2-CM^2)
13/5=37/4-M-41/20m=23/5
Given that the circle C: x2+y2+x-6y+m=0 intersects with the straight line l: x+2y-3=0 at P and Q, O is the origin, if the vector OP·vector OQ=0. (1) Fact the value of m; (2) If R (x, y) is a point on circle C, find the maximum and minimum values of x+y-5/6m. Given that the circle C: x2+y2+x-6y+m=0 intersects the straight line l: x+2y-3=0 at two points P and Q, O is the origin, if the vector OP·vector OQ=0. (1) Fact the value of m; (2) If R (x, y) is a point on circle C, find the maximum and minimum values of x+y-5/6m.
(1) Substitute x=3-2y into the equation of circle to obtain (3-2y)^2+y^2+(3-2y)-6y+m=0,
5Y^2-20y+12+m=0,
Let P (x1, y1), Q (x2, y2),
Then y1+y2=4, y1*y2=(12+m)/5,
Therefore x1*x2=(3-2y1)(3-2y2)=9-6(y1+y2)+4y1y2=4/5*(12+m)-15,
Since OP*OQ=0, x1x2+y1y2=0,
I.e.12+m-15=0,
The result is m=3.
(2) The equation of the circle obtained from (1) is (x+1/2)^2+(y-3)^2=25/4,
Therefore the center of the circle is (-1/2,3), radius r =5/2,
Let t=x+y-5/6*m, then the linear equation is x+y-t-5/2=0,
It is known that that line has a common point with the circle so that the distance from the cent of the circle to the line does not exceed the radius,
I.e.|-1/2+3-t-5/2|/√2<=5/2,
|T |<=5√2/2,
Solution:-5√2/2<=t <=5√2/2,
Therefore, the maximum value is 5√2/2 and the minimum value is -5√2/2.
Given that the circle x^2+y^2+x-6y+m=0 and the straight line x+2y-3=0 intersect the two points P and Q and the vector OQ multiplies the vector OP=0(O is the coordinate origin) How hard it is to get the coordinates and radius of the circle -- Given circle x^2+y^2+x-6y+m=0 and straight line x+2y-3=0 intersected at P and Q and vector OQ multiplied by vector OP=0(O is coordinate origin) How hard it is to find the coordinates and radius of the circle -- Given that the circle x^2+y^2+x-6y+m=0 and the straight line x+2y-3=0 intersect the two points P and Q and the vector OQ multiplies the vector OP=0(O is the coordinate origin) Ask for the circle center coordinates and radius of this question is very hard ah --
Oh,my god, answer looks like it's gon na be hard,
(1) Coordinate of circle center, and convert the circle equation into:
(X+1/2)^2+(y-3)^2=9+1/4+m
So center coordinates (-1/2,3)
The first question is soy sauce, right? .
The middle point M of PQ of the line segment is connected with MC and PC. It is easily known that MC⊥PQ (C is the center of circle)
Since the straight line of PQ is x+2y-3=0,
According to the geometric relation, PC=PQ, M is the midpoint of PQ, and M is the midpoint of PQ in the isosceles triangle, so CM is the vertical line (three lines in one) on the edge of PQ, two lines are vertical, and the product of slope is -1.
Therefore, the linear MC equation is 2x-y+D=0
Substitute the point C coordinate into the straight line MC equation to obtain:
2*(-1/2)-3+D=0, solution D=4
Therefore, the linear MC equation is 2x-y+4=0
Simultaneous equation 2x-y+4=0, x+2y-3=0 Find the intersection point of straight line PQ and straight line MC, i.e. PQ midpoint M
Solvable x=-1, y=2
So midpoint M is (-1,2)
Then |OM|=√5,|CM|=√(1/4+1)=√5/2
Because of OP⊥OQ,
In Rt△OPQ, from the point M is the midpoint on the bevel PQ:
|PQ|=2|OM|=2√5
I.e.|MP|=√5
Then in Rt△CMP,|CM|=√5/2, beveled edge|CP|=r=√(37/4-m)
From Pythagorean theorem |CP|2=|MP|2+|CM|2:
37/4-M =5+5/4=25/4
Solution m=3
So radius:√(9+1/4+3)=7/2
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