Advanced mathematics and its application, second edition, second volume, detailed explanation of after-school exercises.
Experience Network 20 14 May 2 1
Core Tip: This set of answers is the answer to the usual homework questions and some exam questions when I study advanced mathematics, which is especially suitable for the postgraduate entrance examination or the review of the final exam, highlighting the filial piety exercise 5-13; Prove with vectors:
This set of answers is the answer to my homework and some exam questions when I study high mathematics, which is especially suitable for postgraduate entrance examination or final exam review.
filial piety/mourning
Exercise 5- 1
3; It is proved by vector that the straight line connecting the midpoints of two sides of a triangle is parallel to the third variable and equal to half of the third side.
Proved as follows:
In the triangle OAB, EF is the midpoint of OA and AB respectively, connecting EF.
Let the vector OA be a and the vector AB be b, then according to the law of vector addition,
Vector ob = a+b,
The vector ef = a/2+b/2 = (a+b)/2.
So ef = 1/2 * OB, that is, vector EF‖ vector OB,
And according to ef = 1/2 * ob, take the modulus of both sides and get /ef/= 1/2 */ob/
That is, the modulus of the vector EF is equal to half the modulus of the vector OB.
5-2
7; Try to determine the values of m and n, and the test vectors a=-2i+3j+nk and b=mi-6j+2k are parallel.
A and B are parallel, which must be related: a=tb, that is, (-2i+3j+nk)=t(mi-6j+2k), that is, tm=-2, -6t=3, 2t=n, that is, t=- 1/2, and m=-2.
8; Given points A (- 1, 2,4) and B (6 6,2,2) and |AB|=9, find the z value.
10; It is known that two points M 1(4, radical number 2, 1) and m2 (3,0,2) calculate the modulus of vector M 1M2. direction cosine
M 1m2 = (3，0，2)-(4，sqrt (2)， 1) = (- 1，-sqrt (2)， 1)， Therefore: | M 1m2 | = This is beneficial to the calculation of cosines in three directions: COSA = m1m2 (x)/| m1m2 | =-1/2, Therefore: A = 2π/3cosb Therefore: b = 3π/4cosc = m1m2 (z)/| m1m2 | =1/2 Therefore: c=π/3M 1M2(x).
1 1;
Known vector a? At an equal acute angle to each coordinate axis, if | a |a|=2 the root of 2 is 3? , ask for one? Coordinates of
Exercise 5-3
1, let a=3i-j-2k, b=i+2j-k, and find a, b, a * b；; (-2a) 3b and a * b; The angle between a and b
2. Let A, B and C be unit vectors and satisfy a+b+c=0. Find a * b+b * c+c * a.
∫(a+b+c)*(a+b+c)= a？ +b？ +c？ +2ac+2ab+2bc: a, b and c are unit vectors ∴a? = 1，b？ = 1，c？ = 1∴a？ +b？ +c？ +2ac+2ab+2bc=3+2(ab+bc+ca)
3 Find the known points A (1,-1, 2), B (5, -6, 2), C (1, 3,-1).
(1) The unit vector is perpendicular to both vectors AB and AC;
(2) The area of the triangle ABC.
AB:(4,-5,0) AC: (0,4, -3) The unit vector AB X AC = Ijk4-5004-3 =15i+12j+16k perpendicular to both vectors AB and AC is 3/5i.
4. Let A = (3 3,5, -2) and B = (2, 1, 4), and ask what is the relationship between λ and μ, so that λa+μb can be perpendicular to the Z axis.
λ a+μ b = (3λ+5λ-2λ)+(2μ+μ+4μ) = (3λ+2μ, 5λ+μ, 4μ-2λ) z = (0,0, n) vertical, so z (λ a+μ b) = (3λ+2μ).
5. Prove that the circumferential angle of the diameter is a right angle by vector.
Let the center of the circle be zero, the diameter be AB, and the point opposite to the diameter be C. It is proved that AC * BC = 0Ac = 0C-0A, BC = 0C-0B because the modules of vectors 0a, 0B and 0C are equal, AC * BC = (0C-0A) * (0C-0B) = | 0C | 2.
Exercise 5-4
2. Find the plane equation that passes through the point m (3,0,-1) and is parallel to the plane 3X-7y+5z- 12=0.
Let the plane equation be 3x-7y+5z+a = 0; 3 * 3-7 * 0+5 *( 1)+A = 0 because it is too late; So a =-4; So the plane equation is 3X-7y+5z-4=0.
4. Find the plane equation of three points (1, 1,-1) (-2, -2,2) (1,-1, 2).
Three points (1, 1,-1) (-2, -2,2) (1,-1, 2) get the vector (3,3,3) (0,2. Z+ 1) (X- 1, Y-65438+)2) plane equation (x- 1, y- 1, z+ 1) (- 1.
6. Find the distance from (1, 2, 1) to the plane X+2Y+2Z- 10=0.
D = |1*1+2 * 2 *1-10 |/(√ (the square of1+2)) = 1 The formula is: a (x.
9. Find the plane equation that satisfies the following conditions
(2) Passing through points (4,0, -2) and (5,0 1 7) and parallel to the X axis.
Parallel to the X-axis: so its normal vector N is perpendicular to the X-axis and the projection of N on X is 0, so its equation can be set as by+CZ+D = 0; Then if there is -2C+D=0 B+7C+D=0, then D=2C B=-9C, then if there is -9Cy+Cz+2C=0, then c is eliminated, and -9y+z+2=0 is obtained.
Exercise 5-5
1, the straight line {x-y+z = 0, 2x+y+z=4 is expressed by point equation and parameter equation.
The normal vector n 1 of x-y+z=0 is (1,-1, 1)2x+Y+Z = 4 and the normal vector n2 is (2, 1,1) n. Add these two formulas to get X=4/3. Substituting the previous formula, the direction equation with y=4/3 is obtained: [x-(4/3)]/(-2) = [y-(4/3)]/1= z/1parameter equation: X = (.
5、
A linear equation that crosses the point (2, 1, 0) and intersects with the straight line x-11= y-11= z/2.
The plane equation perpendicular to the straight line x-11= y-11= z/2 can be obtained, that is, x-(y- 1)+2(z-2)=0 is at the same time as the known straight line. 1/2, 1) passes through two intersection points (0, 1, 2) and (3/2, 1), and the straight line is x/3 = y-/kloc-0.
Exercise 5-6
2. Write the equation of the surface of revolution obtained by rotating the following curves around the specified coordinate axis.
3. Explain how the following surfaces of revolution are formed.
Solution: Ellipse on (1)xOy plane
? Rotate around the x axis; Or that ellipse on the xOz plane rotate around the x axis.
(2) 2) The hyperbola on the XOY plane rotates around the Y axis; Or a hyperbola on the yOz plane.
Yz rotates around the y axis.
(3) 3) The hyperbola 122yx on the XOY plane rotates around the X axis; Or the hyperbola on the xOz plane rotates around the X axis.
(4) 4) Is the straight line on the YOZ plane rotating around the Z axis or the straight line on the xOz plane? Rotate around the z axis
Exercise 5-6
4. Convert the general equations of the following curves into parametric equations.
5. Find the equation of the projection curve of the following curve on the xoy plane.