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Given that a normal vector n=(-2.2.1) point A (-1.3.0) of plane a is in a, then p (-2.1.4) is more than
There is a formula: d=|n*PA|/|n|, the numerator is the absolute value of the number product of two vectors, and the denominator is the module of the normal vector.
Because vector PA=(1,2,4), so n*PA=-10, and |n|=3, so d=10/3.
Let vector a=(10,-4), vector b=(3,1), vector c=(-2,3) 1. Verifying that the vectors b, c can serve as a set of bases for all vectors in the same plane; 2. Represent vector a with vector b, c
0
M is known to be a point in the triangle ABC, and the vector A B * vector AC =2 root numbers 3, the angle BAC =30°, if For triangle MBC, triangle MCA and triangle MAB, the area is 1/2, x, y, and the minimum value is 1/x+4/y. M is known to be a point in the triangle ABC and the vector A B * vector AC =2 root numbers 3, the angle BAC =30°, if The areas of triangle MBC, triangle MCA and triangle MAB are 1/2, x, y, and find the minimum value of 1/x+4/y.
AB*VECTOR AC =2 ROOT NUMBER 3
AB*AC=|AB||AC|cos BAC=2 root number 3
|AB||AC|* root 3/2=2 root 3
|AB||AC|=4
Triangle area=1/2*4*sinBAC=1
1/2+X+Y=1
X+Y=1/2
Minimum value of 1/x +4/y =18(x =1/6, y =1/3)
AB*VECTOR AC =2 NOS 3
AB*AC=|AB||AC|cos BAC=2 root number 3
|AB||AC|*Root 3/2=2Root 3
|AB||AC|=4
Triangle area=1/2*4*sinBAC=1
1/2+X+Y=1
X+Y=1/2
Minimum value of 1/x +4/y =18(x =1/6, y =1/3)
AB*VECTOR AC =2 ROOT NUMBER 3
AB*AC=|AB||AC|cos BAC=2 root number 3
|AB||AC|*Root 3/2=2Root 3
|AB||AC|=4
Triangle area=1/2*4*sinBAC=1
1/2+X+Y=1
X+Y=1/2
Minimum value of 1/x +4/y =18(x =1/6, y =1/3)
[Vector a+(m times vector b)] is parallel to [2 times vector a-(4 times vector b)], then the value of m is?
A+mb=2a-4b=2(a-2b) m=-2
Given the space vector a=(2,-1,3) b=(1, m,-1), if a is vertical b, then the value of m is
2*1+(-1)*M+3*(-1)=0m=-1
Given that vector n=(1,0-1) is perpendicular to plane α and α passes through point A (2,3,1), then the distance from point P (4,3,2) to α is?
D=|AP·n|/|n|=1/√2=√2/2
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