If the coordinate of vector p at base {abc} is (2010,2011,2012), then the coordinate of vector p at base {a, b+c, b-c} is? If the coordinates of vector p at base {abc} are (2010,2011,2012), then the coordinates of vector p at base {a, b+c, b-c} are?

P =2010a+2011b+2012c.(both vectors)
P=xa+y (b+c)+z (b-c)=xa+(y+z) b+(y-z) c.(x, y, z are numbers)
Therefore, x=2010, y+z=2011, y-z=2012
Solution: x=2010, y=2011.5, z=-0.5

P =2010a+2011b+2012c.(both vectors)
P=xa+y (b+c)+z (b-c)=xa+(y+z) b+(y-z) c.(x, y, z are numbers)
Therefore, x =2010, y+z =2011, y-z =2012
Solution x =2010, y =2011.5, z =-0.5

Vector a=(1,1) Vector b=(2, y) If |a+b|=a*b Then y =?

Vector a=(1,1), vector b=(2, y), if |a+b|=a•b, then y=? A+b=(3,1+y),—a+b—=√[9+(1+y)2]=√(y2+2y+10) a•b=2+y,√(y2+2y+10)=2+y is known, y2+2y+10=4+4y+y2 is obtained by square root removal sign, so 2y=6, y=3....

Linear Correlation of Vector Groups Let b1=(m-1) a1+3a2+a3, b2=a1+(m+1) a2+a3, b3=-a (m+1) a2+a3, b3=-a1-(m+1) a2+(m-1) a3. Linear correlation? Linear Correlation of Vector Groups Let b1=(m-1) a1+3a2+a3, b2=a1+(m+1) a2+a3, b3=-a1-(m+1) a2+(m-1) a3. Let b1, b2, b3 be linearly independent of m. Linear correlation?

Let x1b1+x2b2+x3b3=0, then there is [(m-1) x1+x2-x3] a1+[3x1+(m+1) x2-(m+1) x3] a2+[ x1+x2+(m-1) x3] a3=0. Since the vector group a1, a2, a3 is linearly independent, there is a system of equations (1)(m-1) x1+x2-x3=03x1+(m+1) x2-(m+1) x3=0x1+x2+(m-1) x3=0 if the vector group...

How to find the normal vector with the three coordinates of a plane? Urgent! How to find the normal vector with the three coordinates of a plane? The fire!

How to Find the Normal Vector of Plane in Space Vector Such as title

Two normal vectors a=(x1, y1, z1), b=(x2, y2, z2) of a plane are known, where x1,x2,y1,y2,z1,z2 are all known
Let the plane normal vector be n=(x, y, z)
N is the normal vector rule of the plane
N*a=0x*x1+y*y1+z*z1=0
N*b=0x*x2+y*y2+z*z2=0
Two equations, three unknowns x, y, z
So let & amp; apos; s give one of them, for example, let & amp; apos; s give x=1(can & amp; apos; t be 0), and get the value of y, z, then we can get a normal vector of the plane, because there are countless normal vectors of the plane, and the module can be arbitrary, so we can assume that.

Specific Points for Finding the Normal Vector of Plane Specific Points for Finding Normal Vector of Plane The Specific Points of Finding the Plane Normal Vector

I tell you that the lower plane vector high school method is any two non-parallel vectors a (x, y, z) b (m, n, p)
Try to calculate two equations for vector c (r, s, t) a*c=0b*c=0, let one of the letters r be 0, and find out the other two unknowns s, t. The method in university is to multiply by a, b cross directly.

I tell you that the lower plane vector high school method is any two non-parallel vectors a (x, y, z) b (m, n, p)
Try to calculate two equations for vector c (r, s, t) a*c=0b*c=0. Let r be 0 letters r be 0, and find out the other two unknowns s, t. The method in university is to multiply by a, b cross directly.

If the coordinate of vector p at base {abc} is (2010,2011,2012), then the coordinate of vector p at base {a, b+c, b-c} is? If the coordinates of vector p at base {abc} are (2010,2011,2012), then the coordinates of vector p at base {a, b+c, b-c} are?