Objective Test Questions Used in Teaching the Lesson "Plane Equations"

Answer : A

Analysis : This question is intended to help students realize that the equation of a sphere cannot have the product xy or yz or xz in the equation. Options B , C , D are eliminated because they do not satisfy this condition and so of course A is the correct option.

Question 4 : (Recognize the coordinates of the center and radius of the sphere)

Given sphere ( S ) with equation: ( x – 2) 2 + ( y + 3) 2 + ( z – 1) 2 = 16. The coordinates of center I and radius R of sphere ( S ) are:

( A ) I (2 ; – 3 ; 1) and R = 16.

( B ) I ( 2 ; – 3 ; 1) and R = 4.

( C ) I (– 2 ; 3 ; – 1) and R = 16.

( D ) I (– 2 ; 3 ; – 1) and R = 4.

Answer : B

Analysis : Option A has the wrong radius, options C and D have the wrong center coordinates.

Question 5 : ( Understand coordinate expressions, distance formula between two points ) Given two points M (2; – 1; 4), N (– 3; 2; 0), I is the midpoint of MN . Which of the following results is correct?

( A ) MN = 50



( B ) MN = ( 5 ; – 3 ; 4)

( C ) MN = 5 2

( D ) I (– 1; 1; 4)

Answer : C

Analysis : Option A was given due to the student's mistake of missing the square root when applying the distance formula between two points. Option B is based on

The student incorrectly calculated the coordinates of the vector when knowing the two endpoints: subtract the coordinates of the point written first from the coordinates of the point written later. As for option D , it is based on the student only adding the corresponding coordinates of the two points M and N without taking the average of those coordinates.

Question 6 : (Understand the coordinates of a point)

In space with coordinate system Oxyz for rectangular box

ABCD . A ' B ' C ' D ' with A (0 ; 0 ; 0), B (4 ; 0 ; 0), D (0 ; 2 ; 0), A '(0 ; 0 ; 3).

Find the correct result in the following results:

( A ) C (4 ; 2 ; 3)

( B ) C ' ( 4 ; 2 ; 3)

( C ) B ' (4 ; 3 ; 0)

( D ) D '(2 ; 3 ; 0)

Answer : B

A ' 3 B '

D ' C '

O 4 x

Figure 2.1

Analysis : The question posed here requires students to base on the drawing in figure 2.1 to determine the coordinates of the remaining points of the given box with the intention that students are familiar with how to determine the coordinates of a point in a plane, so they may mistake option C or D as the correct option.

Question 7 : (Understanding the equation of the sphere)

Given the equation: ax 2 + bxy + y 2 + cz 2 + 2 x – 4 y + 6 z – 11 = 0(*). Equation (*) is a sphere equation when:

a = 1

( A ) b = 1

c = 1

( B )

a = 1

b = 0

c = 0

( C )

a = 0

b = 1

c = 1

( D )

a = 1

b = 0

c = 1

Answer : D

Analysis : Students must understand that in the spherical equation, there cannot be a term containing the product xy , so b = 0. From there, looking at all 4 options, only options D and B satisfy this condition. On the other hand, the coefficients of x 2 , y 2 , z 2 must be equal, but the coefficient of y 2 is 1, so a = c = 1, so option B is eliminated.

Question 8 : (Understanding the equation of a sphere knowing the coordinates of the center and radius) The sphere ( S ) with center I (1; – 2; – 3) and radius R = 4 has the equation:

( A ) ( x – 1) 2 + ( y + 2) 2 + ( z – 3) 2 = 4 .

( B ) ( x + 1) 2 + ( y – 2) 2 + ( z – 3) 2 = 16 .

( C ) ( x – 1) 2 + ( y + 2) 2 + ( z + 3) 2 = 4 .

( D ) ( x – 1) 2 + ( y + 2) 2 + ( z + 3) 2 = 16 .

Answer : D

Analysis : The similar options have the wrong sign of the center coordinate or the wrong square of the radius, only option D is correct.

Question 9 : (Understand coordinate expressions of vector operations)



Let i , j , k be three unit vectors on the three axes x'Ox , y'Oy , z'Oz and

yes ,





ab . At that time



a . b is which of the following results:

( A ) – 1 ( B ) 1 ( C ) 3 ( D ) (1 ; – 1 ; – 1) .

Answer : A

Analysis : This question is intended not to ask students to multiply two expressions.

jk )( ikj )





formula of type a . b = ( i



= … that students must understand is:



a = (1 ; – 1 ; 1) and b

= (1 ; 1 ; – 1) so a . b = – 1. The following options appear:

B, C , D are due to errors in the multiplication process.

or vector order error

 i ,

 j ,

in performance

lead to wrong results



or mistake

a . b = | a | . | b |

or just multiply the corresponding coordinates together when



The scalar product of two vectors a and b results in a . b is a vector, not a number (option D ).

Question 10 : (Understand coordinate expressions) Consider the problem:

Given A (0 ; 2 ; – 2), B (– 3 ; 1 ; – 1), C (4 ; 3 ; 0) and D (1 ; 2 ; m ). Find m so that

four points A , B , C , D are coplanar.

In the steps to solve the problem above, which of the following four steps did you start wrong?

( A ) Step 1:

( 3 ; 1; 1); AC (4; 1; 2) ; AD (1; 0; m 2).



( B ) Step 2:

AB , AC



| –1 1 | ; | 1 –3 | ; | –3 –1 | = (– 3 ; 10 ; 1);

1 2 2 4 4 1

AB , AC . AD



( C ) Step 3:

( D ) Step 4:

A , B , C , D are coplanar. Answer: m = – 5.

Answer : C

3 + m + 2 = m + 5

AB , AC . AD



0 + 5 = 0.

Analysis : For questions like the one above, students must follow each step to determine whether they are right or wrong. To do so, they must understand the coordinate expressions of the dot product and the directional product of two vectors in order to choose the correct option. Here, the student calculated the dot product incorrectly.

Question 11 : (Applying coordinate expressions of vector operations)

Given two points M (1; 3; – 2), N (7; 6; 4). Line MN intersects plane ( Oxz ) at point I. Point I divides line segment MN according to which ratio below?

( A ) 2. ( B ) 1. ( C ) 1

. ( D ) 2 .

Answer : C

Analysis : This is a question to practice creativity and flexibility in choosing the right option for students. This question does not require students to find the full coordinates of point I to find the result, but only requires students to write the correct vector formula expressing point I dividing line segment MN by ratio k , from there move on to

The coordinate expression will easily find k = 1 :

Because point I Oxz ) so I ( x I ; 0 ; z I )





and

have corresponding coordinates of 3 and 6.

k IN





3 1

Point I divides line segment MN according to ratio k M .

= 6k

6 2 choose option C.

If solving this exercise to choose the correct option will take a lot of time, so you need to know how to reason as above.

However, some students do not remember the exact definition of dividing a line segment by a given ratio, so they calculate the wrong result such as:

- Students write backwards:

needle



Point I divides line segment MN according to the ratio k 

= 3 k = 2 (option A ).

k MN





- Student wrote incorrectly: Point I divides line segment MN according to ratio k M

= 3 k = 1 (option B ).

k IN



Or point I divides line segment MN according to the ratio k 

3 = 6 k = (option D ).

Question 12 : (Applying to write the equation of a sphere) Choose the correct answer from the following answers:

The sphere with center I ( – 4 ; 1 ; 0), passing through point M (0 ; 1 ; 5), has the equation: ( A ) ( x – 4) 2 + ( y + 1) 2 + z 2 = 9.

( B ) ( x – 4) 2 + ( y + 1) 2 + z 2 = 41.

( C ) ( x + 4) 2 + ( y – 1) 2 + z 2 = 9.

( D ) ( x + 4) 2 + ( y – 1) 2 + z 2 = 41.

Answer : D

Analysis : Options A and B have incorrect center coordinates, options A and C calculate the distance IM (which is the radius length) incorrectly.

Question 13 : (Applying to write the equation of a sphere)

The sphere passes through the origin and three points M (2; 0; 0), N (0; – 4; 0), P (0; 0; – 6) with the equation:

( A ) ( x – 1) 2 + ( y + 2) 2 + ( z + 3) 2 = 14.

( B ) ( x + 1) 2 + ( y – 2) 2 + ( z – 3) 2 = 14.

( C ) ( x – 1) 2 + ( y – 2) 2 + ( z – 3) 2 = 14.

( D ) ( x + 1) 2 + ( y + 2) 2 + ( z + 3) 2 = 14.

Answer : A

Analysis : Option B, option C, and option D give incorrect center coordinates.

To find the correct solution, of course, students can substitute the coordinates of the 4 points O , M , N , P into the equations of the given solutions.

However, it is very time-consuming. If you pay attention, you can see that the 4 solutions are similar, the radius of the sphere is 14 so you only need to check the coordinates of the center of the sphere, but the center of the sphere is not given. How to find it? If students can see that this sphere is the sphere circumscribed around a rectangular box with four vertices O , M , N , P (Figure 2.2), then the center of the sphere must be the midpoint I of the line segment OK with K being the opposite vertex of vertex O in that rectangular box.

But K (2; – 4; – 6) so I (1; – 2; – 3) so option A is the correct option.

- 4 N

. I

Figure 2.2

Question 14 : (Apply the formula to calculate area)

The three vertices of a parallelogram have coordinates (1; 1; 1), (2; 3; 4), (6; 5; 2). The area of that parallelogram is:

( A ) . ( B ) 2

. C ) 2

. ( D ) 4.

Answer : B

AB , AC



Analysis : Calculate directly. Option A is calculated according to formula 1 .

The result of option C is based on a common mistake of students: If calculated correctly, we have the expression:

AB , AC

10) 2 14 2 ( 6) 2



(

S = = = 2 and the miscalculation is due to writing:

AB , AC

10 2 14 2 6 2



S = = = 2 .

Option D applies the correct formula but the calculation is incorrect.

Question 15 : (Applying the directional product of two vectors)

Given A (2 ; – 1 ; 6 ), B (– 3 ; – 1 ; – 4), C (5 ; – 1 ; 0), D (1 ; 2 ; 1). Volume

of tetrahedron ABCD is equal to:

( A ) 30 ( B ) 60 ( C ) 90 ( D ) 180.

Answer : A



Analysis : The solutions given are because students do not remember the formula for calculating volume clearly.

AB , AC . AD

tetrahedron volume, option B gives the result of 1

AB , AC . AD



, option C

AB , AC . AD

get result as 1

, option D gives the result.

2.2 Objective multiple choice questions used in teaching the lesson "Plane equation"

2.2.1 Content and requirements of the lesson

Required level of knowledge:

Understand the concept of normal vectors of a plane, know the general equation of a plane, the conditions of perpendicularity or parallelism of two planes, and the formula for calculating the distance from a point to a plane.

Required skill level:

Determine the normal vector of a plane, know how to write the general equation of a plane and calculate the distance from a point to a plane.

2.2.2 Manifestation of each level

A. Recognition

In this lesson, students need to recognize each type of plane equation: general equation, plane equation according to intercept; recognize the equation of a plane when knowing its equation. In addition, students also need to recognize the relative position of two planes such as the conditions for two planes to be parallel, two planes to be perpendicular.

B. Understanding

With the above knowledge and skill requirements, the level of understanding in the lesson "Plane Equations" includes the following issues:

- Understand how to create a plane equation when knowing the coordinates of a point and a displacement of that plane.

- Understand special cases of plane equations

- Understand perpendicular planes and parallel planes.

C. Application

Students need to apply the knowledge of the lesson to be able to write the equation of a plane when knowing the conditions to determine it such as: a plane passing through a point and having a coordinate system, a plane passing through three non-collinear points, ...; apply the formula to calculate the distance from a point to a plane to do the exercise. Solve some problems related to spheres and planes.

2.2.3 Predict the mistakes that students may make when receiving knowledge in the lesson.

For the lesson "Plane Equations", students often make the following mistakes:

- Mistaking the equation of a plane (in case there are only two unknowns x , y ) for the equation of a straight line in the plane

- Write the plane equation mechanically, mistaking the coordinates of the plane for the coordinates of the plane.

coordinates of points on a plane, wrong coordinates of points on a plane especially when there are negative numbers...

- Special cases of plane equations often confuse students about the special position of the plane in relation to the coordinate axes or to the coordinate plane.

- When writing the equation of a plane according to intercepts, some students do not base on the positions of the points on the three coordinate axes to correctly determine a , b , c, but mechanically apply the order for the three points on the three coordinate axes.

- Wrong calculation and transformation process.

2.2.4 Specific question system

Question 1 : (Recognize the equation of a plane)

In the following equations, which equation is not a plane equation:

( A ) 2 x – 5 y + 1 = 0

( B ) x – 2 = 0

( C ) x + y + z = 1

( D ) 3 x + y – z 2 + 2 = 0

Answer : D

Analysis : The plane equation has the form: Ax + By + Cz + D = 0 (*) in which A 2 + B 2 + C 2 > 0 (that is, A , B , C are not simultaneously equal to 0, so the equation (*) can be missing at most two unknowns). Students often do not pay attention to this, so they will most likely choose option A or option B. as the correct option.

The equation in option C has the right side equal to 1, which can also confuse students (because they mechanically understand that the right side of the plane equation must be equal to 0).

Although option D has a complete equation with three unknowns x , y , z, it is not a first-degree equation so it is not a plane equation.

Question 2 : (Recognize the vector of a plane when knowing the equation of that plane) In the following vectors, which vector is the vector of the plane 2 x – y + 5 = 0?

(2 ; 1 ; 5)



( A ) n

(2 ; 1)



( B ) n

(2 ; 1 ; 0)

( C ) 

( 1 ; 2 ; 0



( D ) n ) .

Answer : C.

Analysis : Because a displacement of plane Ax + By + Cz + D = 0 is n  = ( A ; B ; C ) (corresponding coefficients of unknowns x , y , z ). Option A mistakenly uses the coefficient of z as 5, option B mistakenly uses the equation of the straight line in the plane, option D mistakenly uses the coefficient of

x and y .

Question 3 : (Recognize the equation of a plane according to the intercept)

Given M (0; 0; 1), N (2; 0; 0), P (0; 3; 0). Then the equation of plane ( MNP ) is:

yz 1

1 x	2	3
1 x	2	3
2 x	3	1
2	3	1

Maybe you are interested!

( A ) x .

yz 0

( B ) .

yz 1

( C ) .

yz 0

( D ) .

Answer : C

Analysis : Some students often write the equation of the plane according to the intercept segment arbitrarily in the order of the three points given in the question paper as M , N , P (option A, and option B ).

This question requires students to know the positions of points M , N , P on the coordinate axes: N (2; 0; 0) Ox , P (0; 3; 0) Oy , M (0; 0; 1) Oz to choose the correct option, which is option C.