ABCD A1B1C1D1, AB 5, AD 3, AA1 7, B BAD 60, BAA1 DAA1 45, find the length of AC1. vector method .

25+9+49+2*|Vector ab|*|Vector ad|*cos "Vector ab, vector ad>+2*|Vector ab|*|Vector aa1|*cos《Vector AB, vector AA1>+2*|Vector aa1|*|Vector ad|*COS "Vector AA1, Vector AD>

The length of AC1 = (98+56 2).

The addition of vectors translates into length with squares.

AC1 is the body diagonal, which is added according to the vector.

AC1=AB+BC+CC1=AB+AD+AA1 vectors add to transform the growth degree: |ac1|=|ab+ad+aa1|, the things in it are squared first and then squared (equivalent to finding the absolute value). So |ac1|=sqrt[(ab+ad+aa1)*(ab+ad+aa1)]=sqrt[ab 2+ad 2+aa1 2+2ab*ad+2ad*aa1+2aa1*ab] Note that this is a vector multiplication.

Here you have to make the inside perfectly squared, 98 + 56 sqrt2 = 98 + 2 * 28 sqrt2

It should be A+B*sqrt2

where a 2 + 2 * b 2 = 98

a*b=28

Obtain: 2b 4-98*b 2+28*28=0, which means b 4-49*b 2+28*14=0

.But b does not have a rational number solution. Regretful...

So the answer is sqrt[98+56*sqrt2] or the question data is wrong...

Summary. Dear, please send the original question**Yo? The teacher will answer in detail! Thank you ha! <>3.The known vector ab, aa=-40, |a|=10, 1b1=8, then vector.

Dear, please send the original question**Yo? The teacher will answer in detail! Thank you ha! <>Okay, handwritten.

Where is minus 40 not -40 ha Look clearly.

There should be a mistake in the front of the question, and that is a b

a A cannot be negative.

Do the math. Equal to 40 and count the rest unchanged Thank you.

It can't be negative.

The multiplication of these two is equivalent to the square of the modulus.

I mean change that negative sign to a positive sign and count it.

It's not right to change it to a positive sign.** It can only be a b, I'll write it down, you can see it again.

So, the answer is c

Known vectors +a=(-3,4)+,60+,ab=10+,c=(2,m)+,ac.-|1) Seek |a|;-2 Answer, hello, according to the question, there are the following answers: 1

a| =52.Since the unknown quantity m cannot be determined, the length of c cannot be found. According to the definition of vector, the length of vector a is its modulus length, which can be obtained by the Pythagorean theorem.

Specifically, there are: |a|=sqrt((-3) 2 + 4 2) =5 At the same time, the known vector a and its inner product can be obtained by the inner product formula of the vector: =a c = 3)(2) +4(m) =6 + 4m substitute the known =60 into the above equation, and the solution is :

m = 21 but this result cannot be used to calculate the length of the vector c, because c does not satisfy ac=10. Therefore, there is no definite answer to question 2. Hope this helps<>

11.Known vectors +a=(-3,4)+,60+,ab=10+,c=(2,m)+,ac.-|1) Seek |a|;-2

1) Seek |a|;-2 Answer, hello, according to the question, there are the following answers: 1a| =52.

Since the unknown quantity m cannot be determined, the length of c cannot be found. According to the definition of vector, the length of vector a is its modulus length, which can be obtained by the Pythagorean theorem. Specifically, there are: |

a|=sqrt((-3) 2 + 4 2) =5 At the same time, the known vector a and its inner product can be obtained by the inner product formula of the vector: =a c = 3)(2) +4(m) =6 + 4m substitute the known =60 into the above equation, and the solution is: m = 21, but the result of the cherry head beam cannot be used to calculate the length of the vector c, because the c parallel ridge transport at this time does not satisfy ac=10.

Therefore, there is no definite answer to question 2. Hope this helps<>

In the two-dimensional dimension space, the length of the vector can also be called the modulus length of the vector or the magnitude of the vector. It represents the length of the line segment formed from the start to the end point of the vector. The length of the vector can be found by the Pythagorean theorem.

In the vector space of three dimensions and above, the length of the vector can also be used to guess the matching as its modulus length by analogy, but more complex calculation methods are required. <>

11 questions. Take a look at the picture I posted.

Hello, the symbol can't be sent, it can only be sent like this.

It is possible to establish coordinates.

Let the grasp of the celestial a(0,0) b(2,-1) c(x,y) then the vector ac=(x,y) vector bc=(x-2,y+1) is known by n*ac=7

So 2x+y=7

It is easy to calculate n*bc=2x+y-3

Lianli above two nano-roller type.

There is n*bc=4

Therefore, Duan Qi chose the C answer.

Solution: (1) Let the coordinates of point d be (x,y).

then ab=(1,2),ad=(x+1,y).

ab*ad=x+1+2y=5 1)

x+1)^2+y^2=100 2)

When the solution is x=- 19 under the root number, y=2 + 19 under the root number;

x = 19 under the root number, y=2 - 19 under the root number

Therefore, the coordinates of point d are (19,2 under the root number + 19 under the root number) or (19 under the root number, 2 - 19 under the root number).

2) If point d is in the second quadrant, that is, when the coordinates of point d are (19 under the root number, 2 + 19 under the root number).

ad=(1 - 19 under the root number, 2 + 19 under the root number).

3) Vector ae = (m, 2), if 3 * vector ab + vector ac is perpendicular to vector ae, 3 * vector ab + vector ac = 3 * (1, 2) + (2, 1) = (1, 7), perpendicular to vector ae, there is m*1 + 2 * 7 = 0

m=-14, so the vector ae coordinates are (-14,2).