The distance from point I to plane Ax + By + Cz+ D = 0 is equal to
3. B 2. CD
7
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B 2 C 2
B 2 C 2
2 2 3 2 49 7
A
A 2
A 2
6 2
. But =
A – 3 B + 2 C + D = 49. Substituting the coordinates of point I into the left sides of the plane equation in each option, only option C satisfies this.
Question 27 : (Apply the distance formula to determine the relative position of a plane with a sphere)
Given plane ( P ): 3 x + 4 z + 12 = 0 and sphere ( S ): x 2 + y 2 + ( z – 2) 2 = 1.
Then:
( A ) Plane ( P ) passes through the center of sphere ( S );
( B ) Plane ( P ) is tangent to sphere ( S );
( C ) Plane ( P ) intersects sphere ( S ) along a circle ;
( D ) Plane ( P ) does not intersect sphere ( S ) .
Answer : D
Analysis : This question is to check the relative position of plane (P) and sphere (S) so students need to calculate the distance from the center I (0; 0; 2) of sphere ( S ) to plane ( P ) and compare it with the radius R = 1 of sphere ( S ).
2.3 Objective multiple choice questions used in teaching the lesson "Equation of a straight line"
2.3.1 Content and requirements of the lesson
Required level of knowledge:
Know the parametric equation of a line, the conditions for two lines to intersect, cross, be parallel or perpendicular to each other.
For advanced geometry 12 textbooks, there is an additional canonical equation of a straight line.
Required skill level:
Know how to write parametric equations of lines, know how to use equations of two lines to determine the relative position of those two lines.
2.3.2 Manifestation of each level
A. Recognition
- Students know how to recognize parametric equations and canonical equations of a straight line and vice versa. When knowing parametric equations and canonical equations of a straight line, they must be able to "read" the coordinates of a parametric equation and the coordinates of a point on that line.
- Recognize two parallel and perpendicular lines.
B. Understanding
- Understand how to write the equation of a line passing through a point and having a given coordinate system and understand the nature that a line can pass through many different points, have many different coordinate systems (as long as they are in the same direction), leading to many different equations but representing the same line and vice versa, a line will have many different equations.
- Understand how to find the coordinate system, find points on a straight line when knowing the equation of the line.
C. Application
- Write parametric equations of straight lines.
- Solve problems about finding coordinates of intersection points, calculating distances; problems about relative positions of lines and planes; problems related to lines with planes and spheres.
2.3.3 Predict the mistakes that students may make when receiving knowledge in the lesson.
- When identifying or establishing parametric equations of a line, students often confuse the coordinates of a point and the coordinates of the vector with each other.
- When establishing the canonical equation of a straight line, the coordinates of the point passing through are often mistaken.
- The fact that a straight line can have many different equations depending on how the points it passes through are chosen or whether the equation is parametric or canonical causes confusion for students because they think that a straight line has only one equation.
- When solving problems about formulating straight line equations, we often do not check whether the straight line satisfies the requirements of the problem or not because there are cases where the problem has no solution.
- Calculation process, wrong formula application, wrong transformation.
2.3.4 Specific question system
Question 1 : (Recognize the parametric equation of a straight line)
In the following equations, which equation is a line equation?
(2; 3; 1)
straight line passing through point M (- 1 ; 5 ; 3) and having angular velocity u ?
x = 1 + 2 t
( A ) y = 5 + 3 tz = 3 + t
( B )
x = – 1 + 2 t
y = 5 + 3 tz = 3 + t
( C )
x = 1 + 2 t y = 5 – 3 t z = 3 – t
( D )
x = – 1 + 2 t
y = 5 – 3 t z = 3 – t
Answer : D
Analysis : The options are similar to option A and option C.
The coordinate sign of the point is wrong, option B has the wrong sign of the coordinates of vtcp.
Question 2 : (Recognize the vector of a line when knowing the equation of the line)
(that straight) Given line d :
(2; 3; 1)
u
( A )
(2; 1; 3)
u
( B )
(2 ; 1 ; 3)
u
( C )
x = 2 + 2 t y = – 3 + tz = – 1 – 3 t
.
.
.
. A vtcp of d has coordinates:
( 2 ; 1 ;
( D ) u 3) .
Answer : B.
x = x 0 + at
Analysis : From the parametric equation of the line:
y = y 0 + bt
z = z 0 + ct
we find
( a ; b ; c
its vtcp is: u ) (corresponding coefficient of t). Solution
A takes the coordinates of a point on line d, while options C and D have the wrong coordinate signs.
Question 3 : (Recognize parametric equations and canonical equations of
straight line)
2 yz 3
x
2
Given line d with parametric equation: which of the following is the canonical equation of d ?
3 5
( A )
2 yz 3
3 5
x
2
( B )
( C ) x – 2 = y = z + 3
( D ) x + 2 = y = z – 3
Answer : A
x = 2 + 2 t
y = – 3 t . Equation
z = – 3 + 5 t
Analysis : Because line d has the coordinates (2; - 3; 5), eliminate options C and D. Next, check the coordinates of point (2; 0; - 3) in options A and B to select option A.
Question 4 : (Recognize two parallel lines)
1 y 2 z 3
2 3
x
1
Given line d : . In the lines there are
3 y 4 z 5
4 6
1 y 2 z 3
x
2
x
1
2
3
In the following equation, which line is parallel to line d ? ( A )
( B )
1 y 2 z 3
4 6
x
2
( C )
3 y 4 z 5
3
x
1
2
( D )
Answer : A
Analysis : Option B and option D are eliminated because the numbers (1; 2; 3) and (1; 2; – 3) are not proportional to the number set (1; – 2; 3) which is the coordinate of line d , leaving only options A and C. It is easy to see that option C is also eliminated because the two lines coincide, so option A is chosen .
Question 5 : (Recognize two perpendicular lines)
Given line d :
x = 2 + t
y = – 3 + 2 tz = – 1
In the lines with the following equations, which line is perpendicular to line d ?
x = 1 + 2 t
( A ) y = 2 – t
z = 4 – 3 t
( B )
x = 1 + 2 t y = 2 + tz = 4 + 3 t
( C )
x = 1 – ty = 2 + 2 t z = 4 – 3 t
( D )
x = 1 + t
y = 2 – 2 tz = 4 + 3 t
Answer : A
Analysis : The options are similar, choose the equation with the coefficient of t corresponding to the rows of x and y being 2 and - 1, and the coefficient in the row of unknown z is not important, because the coordinates of line d are (1; 2; 0).
Question 6 : (Understand how to write the equation of a line passing through a point and having a given vector)
u
Equation of line d has perpendicular bisector = (1 ; – 3 ; 4) and passes through the point
Which of the following equations is M (1; 2; 3)?
x = 1 + t
( A ) y = 2 + 3 t
z = 3 + 4 t
( B )
x = 1 + t
y = – 3 + 2 t
z = 4 + 3 t
( C )
x = 1 + ty = 2 – 3 t z = 3 + 4 t
( D )
x = 1 + t
y = 2 + 3 t z = 3 – 4 t
Answer : C
Analysis : Students often make two mistakes when writing parametric equations of a line: they confuse the coordinates of the point and the coordinates of the point passing through each other due to mechanically understanding the order of how to give the line d at the beginning of the lesson. Option B confuses the coordinates of the point and the coordinates of the point, so it is eliminated. Options A , C , and D are similar, but only option C has the correct sign of the coordinates of the point.
Question 7 : (Understand how to find a point on a line with equation
Parameter) Line d : following coordinates:
x = 1 + 2 t y = 2 + tz = 3 – t
does not pass through any of the points
( A ) (1 ; 2 ; 3)
( B ) (3 ; 3 ; 2)
( C ) (– 1 ; 1 ; 4)
( D ) (– 1 ; 1 ; 2)
Answer : D
Analysis : Through the above question, students are required to "read" the coordinates of a point on a line with a given equation. Therefore, students must understand that the way to find the coordinates of a point on a line with a given equation is to simply substitute a value of the parameter t into the equation of the line to immediately get the coordinates of a point on that line.
Option A is immediately visible, option B takes t = 1, option C takes t = - 1, option D also takes t = - 1 like that but calculates wrongly.
Question 8 : (Understanding parametric equations of straight lines)
x = 1 + 2 t
Given line d : y = 2 + t
z = 3 – t
Which of the following equations is also
equation of line d ?
x = 1 + 2 t
( A ) y = 2 – tz = 3 + t
( B )
x = 3 – 2 t y = 3 – tz = 2 + t
( C )
x = 3 + 2 t y = 2 + tz = 2 – t
( D )
x = 3 – 2 t y = 3 – tz = 2 – t
Answer : B
Analysis : Students are often confused about the fact that the same line can have many equations because just by choosing different passing points or vectors with different coordinates (of course the vectors must be in the same direction as the vectors of the original line d ) the equations are already different.
From the given equation, students must "read" the point where the line d
passing through there is point M (1; 2; 3) and the vector of d has coordinates (2; 1; – 1). Solution
A and B have correct coordinates of the passing point but incorrect coordinates of the vtcp. Option C has correct coordinates of the vtcp but incorrect coordinates of the passing point. If not familiar, students will have difficulty finding the correct answer in this question.
Question 9 : (Understand parametric equations and canonical equations of straight lines)
Given point M (1; 2; 3) and point N (2; – 1; 4) and three equations:
x = 1 + t
y = 2 – 3 tz = 3 + t
( 1 )
x = 2 + t
y = – 1 – 3 t
z = 4 + t
( 2 )
2 y 1 z 4
1 3
x
1
( 3 )
Which of the following statements is true:
( A ) Only (1) is the equation of line MN
( B ) Only (3) is the equation of line MN
( C ) Only (2) and (3) are equations of line MN
( D ) Both (1), (2), (3) are equations of line MN. Answer : D
Analysis : This question is to test whether students have a deep understanding of the equation of a straight line? Because equations (1), (2) and (3) all represent the straight line MN, the only difference is that equation (1) chooses the point passing through point M , while equation (2) chooses the point passing through point N , and equation (3) is the canonical equation of the straight line MN .
Question 10 : (Understand how to check whether a point belongs to a line or plane)
12 y 9 z 1
x
4
3
1
Intersection M of line d :
( P ): 3 x + 5 y – z – 2 = 0 has which coordinate below?
and plane
( A ) (1 ; 0 ; 1)
( B ) (0 ; 0 ; – 2)
( C ) (1 ; 1 ; 6)
( D ) (12 ; 9 ; 1)
Answer : B
Analysis : Substitute the coordinates of point M in each option into the equation of line d (just substitute the first two fractions). If it is satisfied, continue to substitute it into the equation of plane ( P ). If it is not satisfied, immediately eliminate that option.
Question 11 : (Understand the relative position of a line and a plane)
x = 1 + t
Given line d :
y = 2 – tz = 1 + 2 t
and plane ( P ): x + 3 y + z +1 = 0.
Find the correct conclusion in the following conclusions:
( A ) d // ( P )
( B ) d cuts ( P )
( C ) d P )
1 ; 2)
( D ) d P ) Answer : A
Analysis : Line d has perpendicular bisector
u = (1 ;
, plane ( P ) has position
n = (1 ; 3 ; 1) . We have:
and option D.
u . n = 0
should d // ( P ) or d P ) in option B
On the other hand , point M (1; 2; 1) P ) so d // ( P ) chooses option A.
Question 12 : (Understand the relative position of a line and a plane)
x = 1
Given line d :
y = 1 + tz = – 1 + t
and two planes:
( P ): x – y + z + 1 = 0 , ( Q ): 2 x + y – z – 4 = 0.
Which of the following conclusions is correct?
( A ) d // ( P ) ( C ) d = ( P ) Q )
( B ) d // ( Q ) ( D ) d P )
Answer : C
Analysis : Line d has perpendicular bisector
u = (0 ; 1 ; 1) n P
= ( 1 
is
; 1
vtpt of ( P ) should be d // ( P ) or d P ) 
in option D .
On the other hand , point M (1; 1; – 1) 
and ( P ) should eliminate option A and get option C as the correct option because only option C has the condition dP ).
Question 13 : (Understand the relative position of two lines) Find the correct conclusion about the relative position of two lines:
x = 1 + t
d : y = 2 + tz = 3 – t
and d' :
x = 1 – 2 t '
y = 1 – 2 t '
z = – 1 + 2 t ' .
( A ) d cuts d'
( B ) d d'
( C ) d is diagonal to d'
( D ) d // d' Answer : D
Analysis : Line d has perpendicular bisector
u d
= (1 ; 1 ;
, line d' has
vtcp
1)
2; 2; 2)
u d ' = (
satisfy:
u d ' =
should eliminate option A and
2 u d
Option C.
On the other hand , point M (1; 2; 3) belongs to d but not to d' so d // d' . Choose option D.
Question 14 : (Applying to calculate the distance from a point to a straight line)
Radius R of the sphere with center I (1; 3; 5) is tangent to the line
x = t
y = – 1 – tz = 2 – t
Which of the following numbers is
( A )
Answer : A
. ( B ) 7 . ( C ) 14. ( D ) 42 .
14
3
Analysis : Calculating the distance from I to the line directly 
is time consuming):
1 ; 1)
IM
ó vtcp u = (1 ;
1; 4; 3)
= (
, n =
, point M (0 ; – 1 ; 2)
u , IM = ( 1 ;
4; 5)
14
42
| n |
| u | 3
R = d( I , 
= .
However, it can be quickly evaluated as follows (to train students' flexibility and creativity in choosing answer options):
4 2 3 2 26
R = d( I , IM = 1 2 choose option A .
Question 15 : (Applying the perpendicularity condition of two vectors)
Orthogonal projection coordinates of point M (2; 0; 1) on line
1 yz 2
x
1
2
1
Which of the following numbers is this?
( A ) (1 ; 0 ; 2) ( C ) (0 ; – 2 ; 1)
( B ) (2 ; 2 ; 3) ( D ) (– 1 ; – 4 ; 0)
Answer : A
Analysis : Solve directly (takes a lot of time). Or check if the points in each option are on the line, if so, then check the perpendicularity condition. Here, all four points in the four options are on the line, so students are required to check the perpendicularity condition, so this question will help teachers test their overall knowledge.
Question 16 : (Apply the method to calculate the distance between two parallel planes)
1 y 7 z 3
x
2
1
4
Given plane ( P ): 3 x – 2 y – z + 5 = 0 and straight line
. Let ( Q ) be the plane containing and parallel to ( P ).
The distance between ( P ) and ( Q ) is:
( A ) 9 14
14
( B ) 9
Answer : B
. ( C ) 9 .
2
. ( D ) 9 .
4
MP ).
Analysis : To solve this problem, students must understand that the distance between two parallel planes is equal to the distance from any point on one plane to the other plane (this is knowledge learned from grade 11).
We have point M (1; 7; 3)
2.7 1.3 5 | 9
( 2) 2 ( 1) 2
| 3.1
3 2
14
Since ( P ) // ( Q ) so d(( P ), ( Q )) = d( M , ( P )) = .