It is known that in the cuboid abcd-a1b1c1d1, ab = 1, ad = 2, Aa1 = 3, then the vector BD * vector AC1 is equal to

It is known that in the cuboid abcd-a1b1c1d1, ab = 1, ad = 2, Aa1 = 3, then the vector BD * vector AC1 is equal to

Let AB be the positive direction of x-axis, ad be the positive direction of y-axis, Aa1 be the positive direction of z-axis to establish the space rectangular coordinate system, then the vector BD = (- 1,2,0), the vector AC1 = (1,2,3) and then the question is, is your multiplication sign dot multiplication "·" or cross multiplication "×" If it is ·, then vector BD · vector AC1 = (- 1) * 1 +

The length of AC1 can be obtained from the known four prism abcd-a1b1c1d1, ab = 5, ad = 3, Aa1 = 7, B ∠ bad = 60, ° ∠ baa1 = ∠ daa1 = 45 °. (vector method)

If the quadrilateral ABCD is in the XY plane, and the quadrilateral a1b1c1d1 is in the positive direction of Z axis, a is the origin (0,0,0) and ab is in the positive direction of X axis, then the vector AB = (5,0,0)
Let d be in the positive direction of y-axis, because ∠ bad = 60 ° and | ad | = 3, then the vector ad = (3 / 2,3 √ 3 / 2,0)
Let Aa1 = (a, B, c)
|AA1|=√(a^2+b^2+c^2)=7；
Cos ∠ baa1 = (5 * a + 0 * B + 0 * c) / (| ab | Aa1 |) = A / 7 = √ 2 / 2, the solution is: a = 7 √ 2 / 2;
Cos ∠ daa1 = [(3 / 2) * a + (3 √ 3 / 2) * B + 0 * C] / (| ad | Aa1 |) = (a + √ 3b) / 14 = √ 2 / 2, the solution is: B = 7 √ 6 / 6
Then a = 7 √ 2 / 2, B = 7 √ 6 / 6 is brought back to √ (a ^ 2 + B ^ 2 + C ^ 2) = 7, and the solution is C = 7 √ 3 / 3
So, vector Aa1 = (7 √ 2 / 2, 7 √ 6 / 6, 7 √ 3 / 3)
Therefore, vector AC1 = vector Aa1 + vector A1B1 + vector b1c1 = vector Aa1 + vector AB + vector ad = (7 √ 2 / 2,7 √ 6 / 6,7 √ 3 / 3) + (5,0,0) + (3 / 2,3 √ 3 / 2,0) = ((13 + 7 √ 2) / 2, (7 √ 6 + 9 √ 3) / 2,7 √ 3 / 3)
Therefore, | AC1 | = √ (98 + 56 √ 2)

In the cube abcd-a1b1c1d1, m, N, e and F are the midpoint of edges B1C, a1d1, d1d and ab respectively
(1) Verification: a1e ⊥ plane abmn
(2) The angle between a1e and MF

①∵A1N=D1E
Yi Zheng RT △ aa1n ≌ RT △ a1d1e
Yizheng a1e ⊥ an
∵ ab ⊥ plane aa1d
∴AB⊥A1E
∩ ab ∩ an = plane abmn
{a1e ⊥ plane abmn
② ∵ a1e ⊥ plane abmn, FM ∈ plane abmn
∴A1E⊥FM
That is to say, the angle between a1e and FM is equal to 90 degrees

In the square abcd-a1b1c1d1 with edge length of 1, e and F are the midpoint of d1d and BD respectively, and CG = 1 / 4CD, h is the midpoint of C1G,
The space vector method is used to solve the following problems: (1) to prove that EF is perpendicular to B1C; (2) to find the cosine of the angle between EF and C1G; (3) to find the length of FH

Taking A1 as the origin, A1B1 as the x-axis, a1d1 as the y-axis, A1A as the z-axis, the space rectangular coordinate system is established. A1 XYZ has A1 (0,0,0), e (0,1,1 / 2), f (1 / 2,1 / 2,1), B1 (1,0,0), C (1,1,1), G (3 / 4,1,1), H (7 / 8,1,1 / 2) (1) from the known coordinates: vector EF = (1 / 2, - 1 / 2,1 / 2), vector B1C = (0,1,1)