Given the module of vector a=10, vector b=(3,4) and vector a//b, find vector a

Given the module of vector a=10, vector b=(3,4) and vector a//b, find vector a

Let vector a=(x, y), a//b, so 4x-3y=0, module length of vector a=10, square of x + square of y=100, the simultaneous solution of vector a=(6,8)

Let vector a=(x, y), a//b, so that 4x-3y=0, the module length of vector a=10, the square of x+y=100, the simultaneous solution of vector a=(6,8)

In triangle ABC, vector AB*vector AC=absolute value (vector AB-vector AC)=2 Find the maximum area of triangle ABC is the size of angle A In the triangle ABC, the vector AB*vector AC=absolute value (vector AB-vector AC)=2 Find that the maximum area of the triangle ABC is the size of the angle A

Let A, B, C of triangle ABC be a, b, c.
Then vector AB·vector AC=cbcosA,
Vector AC-vector AB=vector BC,
Because vector AB·vector AC=|vector AC-vector AB|=3,
So cbcosA=2, a=2.
According to the cosine theorem, a^2=b^2+c^2-2cbcosA,
I.e.4=b^2+c^2-4, b^2+c^2=8.
So bc≤(b^2+c^2)/2=4,
And bc=2/cosA, so 2/cosA≤4, cosA≥1/2.
So 0

Let a=(2,1), b=(m,3), and the vector a parallel to the vector b, find the value of m Let a=(2,1), b=(m,3), and the vector a is parallel to the vector b to obtain the value of m

Here, the formula is set directly, and the two vectors are parallel, corresponding to the proportional value,2/m=1/3, so m=6

Given A (2,1), B (−3,−2), AM=2 3 AB, then the coordinate of point M is () A.(−1 2,−1 2) B.(−4) 3,−1) C.(1 3,0) D.(0,−1 5)

Set point M (x, y)
Ze

AM =(x−2, y−1),

AB =(−5,−3)
∵

AM=2
3

AB
∴
X−2=−10
3
Y−1=−2,
Solved
X=−4
3
Y =−1
The coordinates of point M are (−4
3,−1).
Therefore, B.

Set point M (x, y)
Ze

AM =(x−2, y−1),

AB =(−5,−3)
∵

AM=2
3

AB
∴
X−2=−10
3
Y−1=−2,
Jiede
X=−4
3
Y =−1
The coordinates of point M are (−4
3,−1).
Therefore, B.

Given that a normal vector of plane A is n (1,1,1) and the origin O (0,0,0) is in plane A, then the distance from point P (4,5,3) to A

4 Pcs.3

Let a vector =(10,-4), b vector =(3,1), c vector =(-2,3). )

Let vector a=xb+yc
A = x (3,1)+ y (-2,3)
=(3X-2y, x+3y)
I.e.3x-2y=10
X+3y=-4
Solution x=2, y=-2
So c=2b-2c