Formula:
1) The vector equation of a straight line passing through a fixed point with position vector a and parallel to a given vector b is
r= a + $ b, where $ is scalar.
Note:
a) r is the position vector of any point P(x,y,z) on the line. Therefore
r= xi+ yj + zk
2) The cartesian equations of a straight line passing through a fixed point (x', y', z') and having direction ratios proportional to a, b, c is
(x - x')/a = (y - y')/b = (z - z')/c.
Note:
a) The above form of a line is known as the symmetrical form of a line.
b) The parametric equations of the line (x - x')/a = (y - y')/b = (z - z')/c are
x= x'+ a$, y= y'+ b$, z= z'+ c$, where $ is the parameter.
c) The coordinates of any point on the line (x - x')/a = (y - y')/b = (z - z')/c are x'+ a$, y'+ b$, z'+ c$, where $ belongs to R.
d) Since the direction cosines of a line are also its direction ratios. Therefore, equations of a line passing through (x', y', z') and having direction cos6 l, m, n are
(x - x')/l = (y - y')/m = (z - z')/n
e) Since x, y and z axes pass through the origin and have direction cosines 1,0,0; 0, 1,0 and 0,0,1 respectively. Therefore, their equations are:
i) x-axis: (x - 0)/1 = (y - 0)/0 = (z - 0)/0 or, y= 0 and z=0
ii) y-axis: (x - 0)/0 = (y - 0)/1 = (z - 0)/0 or, x= 0 and z=0.
iii) z-axis: (x - 0)/0 = (y - 0)/0 = (z - 0)/1 or, x= 0 and y=0
3) The vector equation of a line passing through two points with position vectors a and b is
r= a + $(b - a).
4) The cartesian equations of a line passing through two given points (x', y', z') and (x", y", z") are
(x - x')/(x" - x') = (y - y')/(y" - y') = (z - z')/(z"- z').
** Reduction of Cartesian form of a line to vector form and vice versa:
Cartesian to vector:
(x - x')/a = (y - y')/b = (z - z')/c
These are the equation of a line passing through the point A(x', y', z') and its direction ratios are proportional to a, b, c. In vector form this means that the line passes through point having position vector a= x' i+ y' j + z, k and is parallel to the vector m= ai + bj + ck. So, the vector equation of the line is r= a+ $m,
OR
r= (x' i+ y' j + z' k)+ $(ai + bj+ ck), where $ is a parameter.
Exercise -1
1) Find the vector equation of a line which passes through through the point with position vector 2i- j +4k and is in the direction of i+ j - 2k. Also reduce it to cartesian form. r= (2i - j + 4k) + $(i + j -2k). (x-1)/1= (y+2)/1= (z -4)/-2
2) Find the vector equation of the line through A(3,4,-7) and B (1,-1,6). find also, its Cartesian equation. r= (3i +4 j -7k) + $(-2i - 5 j +13k), (x-3)/-2= (y-4)/-5= (z +7)/13
3) The points A(4,5,10), B(2,3,4) and C(1,2,-1) are three vertices of a parallelogram ABCD. Find vector and Cartesian equation for the side AB and BC and find the coordinates of D. r= (4i +5j + 10k) + $(i + j +3k) where $= -2¥, (x-4)/1= (y-5)/1= (z - 10)/3, (3,4,5)
4) Find the vector equation of a line passing through a point with the position 2i- j + k, and parallel to the line joining the points - 4j + k and i+ 2j + 2k. Also, find the Cartesian equivalent of this equation. r= (2i - j + k) + $(2i - 2j +k), (x-2)/2= (y+1)/-2= (z -1)/1
5) Find the cartesian equation of a line passing through the point A(2,-1,3) and B(4,2,1). Also, reduce it to vector form. (x-2)/2 = (y+1)/3 = (z - 3)/-2 , r= (2i - j + 3k) + $(2i + 3j -2k).
6) The Cartesian equations of a line are 6x - 2 = 3y+ 1= 2z - 2. Find its direction ratios and also find vector equation of the line. 1,2,3, r= (i/3 - j/3 + k) + $(i + 2j +3k)
7) Find the direction cosine of the line (x-2)/2 = (2y -5)/-3, z= -1. also, Find the vector equation of the line. 2/5/2, -3/2/5/2,0 r= (2i +5j/2 -k) + $(2i -3j/2 +0k).
8) Show that the points whose position vectors are 5i +5j, 2i+ j+ 3k and -4i + 3j - k are collinear.
9) if the points A(-1,3,2), B(-4,2,-2) and C(5,5, k) are collinear. Find the value of k. 10.
10) Find the point on the line (x+2)/3 = (y+1)/2= (z - 3)/2 at a distance of 3√2 from the point (1, 2, 3). (-2,-1,-3) and (56/17,43/17, 111/17)
Exercise-A
1) find the vector and the Cartesian equations of the line through the point (5,2,-4) and which is parallel to the vector 3i + 2j - 8k. r= (5i +2j -4k) + $(3i + 2j -8k). (x-5)/3 = (y-2)/2= (z +4)/-8
2) Find the vector equation of the line passing through the points (-1,0,2) and (3,4,6). r= (-i + 3k) + $(4i + 4j +4k).
3) Find the vector equation of a line which is parallel to the vector 2i- j+ 3k and which passes through the point (5,-2,4). also reduce it to cartesian form. r= (5i - 2j + 4k) + $(2i - j +3k). (x-5)/2= (y+2)/-1= (z -4)/3
4) A line passes through the point with position vector 2i -3j +4k and is in the direction of 3i+4j -5k. Find the equations of the line in vector and Cartesian form. r= (2i - 3j + 4k) + $(3i + 4j -5k). (x-2)/3 = (y+3)/4 = (z -4)/-5
5) ABCD is a parallelogram. The position vectors of the position A, B and C are respectively. 4i +5j- 10k, 2i - 3j+ 4j and-i + 2j + k. Find the vector equation of the line BD. Also, reduce it to cartesian form. r= (2i - 3j + 4k) + $(i - 13j +17k). (x-2)/1= (y+3)/-13 = (z -4)/17
6) find in vector form as well as in the cartesium form, the equation of the line passing through the point a(1,2,-1) and B(2,1,1). r= (i +2j -k) + $(i - j +2k). (x-1)/1= (y-2)/-1= (z +1)/2
7) Find the vector equation for the line which passes through the point (1,2,3) and parallel to the vector i- 2j+ 3k. Reduce the corresponding equation to the Cartesian form. r= (i +2 j + 3k) + $(i -2j +3k). (x-1)/1 = (y-2)/-2= (z - 3)/3.
8) Find the vector equation of a line passing through (2,-1,1) and parallel to the line whose equations are (x-3)/2 = (y+2)/7 = (z -2)/-3. r= (2i - j + k) + $(2i +7 j -3k).
9) The Cartesian equation of a line (x-5)/3= (y+4)/7= (z - 6)/2. Find a vector equation for the line find the cartition equation for the line. r= (5i - 4j + 6k) + $(3i +7 j +2k).
10) Find the cartesian equation of a line passing through (1,-1,2) and parallel to the line whose equations are. (x-3)/1 = (y-1)/2= (z + 1)/-2. Also, reduce the equation obtained in vector form. (x-1)/1= (y+1)/2= (z - 2)/-2; r= (i - j + 2k) + $(i + 2j -2k)
11) Find the direction cosines of the line (4 -x)/2 = y/6= (1 -z)/3. Also, reduce it to vector form. -2/7,6/7,-3/7; r= (4i+ 0 j + k) + $(-2i +6 j - 3k)
12) The cartesian equations of a line x= ay +b, z= cy+ d. Find the direction ratios and reduce it to vector form. DRS: a, 1,c; r= (bi +0 j + dk) + $(ai + j +ck).
13) Find the vector equation of a line passing through the point with position vector i- 2i - 3k and parallel to the line joining the points with position vector i- j + 4k and 2i + j + 2k. also, find the Cartesian equivalent of this equation. r= (i - 2j - 3k) + $(i + 2j -2k); (x-1)/1 = (y+2)/3= (z +3)/-2
14) Find the points on the line (x+2)/3= (y+1)/2= (z - 3)/2 at a distance of 5 units from the point (1,3,3). (4,3,7),(-2,-1,3)
15) Show that the points whose position vectors -2i +3j, i+ 2j +3k and 7i+ 9k are collinear.
16) Find the cartesian and vector equations of a line which passes through the point (1,2,3) and is parallel to the line (-x-2)/1= (y+3)/7= (2z -6)/3. (x-1)/-3= (y-2)/14 = (z -3)/3; r= (i +2j + 4k) + $(-2i + 14j +3k)
17) The Cartesian equations of a line are 3x+ 1= 6y - 2= 1- z. Find the fixed point through which it passes, its direction ratios and also its vector equation. (-2/3,1/3,1); 2,1, -6; r= (-i/3 +j/3 + k) + $(2i + j -6k)
ANGLE BETWEEN TWO LINES
Let the vector equation of the two lines be r= a₁+ λb₁
and r= a₂ +μ b₂.
These two lines are parallel to the vector b₁ and b₂ respectively. Therefore, angle between these two lines is equal to the angle between b₁ and b₂ . Thus, if θ is the angle between the given lines, then
cos θ= b₁. b₂ /(|b₁| |b₂|)
Condition of Perpendicularity
If the lines b₁ and b₂ are perpendicular. Then, b₁.b₂ = 0
Condition of Parallelism
If the lines are parallel, then b₁ and b₂ are parallel.
b₁ = λ b₂ for some scalar λ
CARTESIAN FORM
Let the Cartesian equation of the two lines be
(x - x₁)/a₁ = (y - y₁)/b₁ = (z - z₁)/c₁ ... (1)
And
(x- x₂)/a₂ = (x - x₂)/b₂ = (z - z₂)/c₂ .......(2)
* Direction ratios of the line (1) are proportional to a₁ , b₁ , c₁.
m₁ = Vector parallel to line (1) = a₁i + b₁j + c₁k
* Direction ratios of line (2) are proportional to a₂, b₂, c₂.
m₂ = Vector parallel to the line (2) = a₂i + b₂j + c₂k.
Let θ be the angle between (1) and (2). Then, θ is also the angle between m₁ and m₂.
cos θ = m₁.m₂/|m₁| |m₂|
=> cos θ = (a₁.a₂ + b₁b₂+ c₁.c₂)/{√(a₁²+ b₁²+ c₁²) √(a₂² + b₂²+ c₂²)}
Condition of Perpendicularity:
If the lines are perpendicular, then
m₁.m₂ = 0
=> a₁.a₂ + b₁.b₂ + c₁.c₂ = 0
Condition of Parallelism:
If the lines are parallel, then m₁ and m₂ are parallel.
So, m₁ = λm₂ for some scalar λ
=> a₁/a₂ = b₁/₂ = c₁/c₂ .
EXERCISE -2
TYPE - I
1) Find the angle between the lines
r= 3i+ 2j -4k+ λ(i + 2j + 2k) and
r= 5i- 2j + λ(3i + 2j + 6k). cos (19/21)
2) Find angle between the lines
(x-2)/3 = (y+1)/-2, z = 2 and (x-1)/1 = (2y+3)/3= (z+5)/2. π/2
3) Prove that the line x= ay+ b, z= cy + d, x= a'y + B' , z= c' y d' are perpendicular if AA'+ cc' + 1= 0.
4) Find the angle between two lines whose direction ratios are proportional to 1, 1,2 and (√3-1), (-√3 -1), 4. 1/2
TYPE-2
ON FINDING THE EQUATION OF A LINE PARALLEL TO A GIVEN LINE AND PASSING THROUGH A GIVEN POINT
Formula used
A) r= a+ λb
B) (x - x₁)/a = (y - y₁)/b = (z - z₁)/c
5) Find the equation of a line passing through a point (2, -1,3) and parallel to the line r= (i+ j) + λ(2i + j - 2k). r= (2i- j+ 3k) + μ(2i + j - 2k).
6) Find the equation of a line passing through (1,-1,0) and parallel to the line (x - 2)/3 = (2y +1)/2 = (5 -z)/1. (x - 1)/3 = (y +1)/1 = (z - 0)/-1
Type 3:
ON FINDING THE EQUATION OF A LINE PASSING THROUGH A GIVEN POINT AND PERPENDICULAR TO TWO GIVEN LINES
Result be used: A line passing through a point having position vectors γ and perpendicular to the lines r= a₁+ λb₁ and r= a₂+ μb₂ is parallel to the vector b₁x b₂. So, its vector equation is r= γ + λ(b₁ xb₂)
For Solution:
Step 1:
Obtain the point through which the line passes. Let its position vector be γ.
Step 2:
Obtain the vectors parallel to the two given lines. Let the vectors be b₁ and b₂.
Step 3:
Obtain b₁ x b₂
Step 4:
The vector equation of the required line is r= γ + λ(b₁ xb₂).
7) Find the cartesian equations of the line passing through the point (-1, 3, - 2) and perpendicular to the lines x /1 = y /2 = z/3 and (x +2)/-3 = (y -1)/2 = (z +1)/5. (x +1)/3 = (y -3)/-7 = (z+ 2)/4
8) A line passes through (2, -1, 3) and is perpendicular to the line r= (i + j - k)+ λ(2i - 2j +k) and r= (2i - j - 3k) + μ(i + 2j + 2k). Obtain its equation. r= (2i - j+ 3k)+ μ(2i + j - 2k), where. μ = - 3λ
TYPE 4
ON PERPENDICULARITY OF TWO LINES
9) Find the value of λ so that the lines. (1-x)/3 = (7y-14)/2λ = (z-3)/2 and (7 - 7x)/3λ = (y -5)/1 = (6 -z)/5 are at right angle. Also find the equations of a line passing through the point (3,2,-4) and parallel to line l₁. (x-3)/-3 = (y-2)/20/11 = (z+4)/2.
EXERCISE - B
1) Show that the three lines with direction cosines 12/13,-3/13, -4/13 ; 12/13, 3/13, 3/13 ; 3/23, -4/13, 12/ 13 are mutually perpendicular.
2) Show that the line through the points (1,-1,2) and (3,4,-2) is perpendicular to the through the points (0,3,2) and (3,5,6).
3) Show that the line through the point (4,7,8) and (2,3,4) is parallel to the line through point (-1,-2,1) and (1,2,5).
4) Find the cartesian equation of the line which passes through the point (-2,4,-5) and parallel to the line given by (x +3)/3 = (y-4)/5 = (z+8)/6. (x +2)/3 = (y-4)/5 = (z+5)/6
5) Show that the lines (x -5)/7 = (y+2)/-5 = z/1 and x/1 = y/2 = z/3 are perpendicular to each other.
6) Show that the line joining the origin to the point (2,1,1) is perpendicular to the line determined by the point (3,5,-1) and (4,3,-1).
7) Find the angle between the following pairs of line:
A) r= (4i - j)+ λ(i + 2j -2k) and r= (i - j +2k) - μ(2i + 4j -4k). 0°
B) r= (3i + 2j - 4k)+ λ(i + 2j + 2k) and r= (5j - 2k) + μ(3i + 2j + 6k). cos(19/21)
C) r= λ(i + j +2k) and r= 2i + μ{(√3 -2)i - (√3+1)j + 4k}. π/3
8) Find the angle between the following pairs of lines:
A) (x +4)/3 = (y-1)/5 = (z-3)/4 and (x+1)/1 = (y-4)/2 = (z-5)/2. Cos(8/5√3)
B) (x -1)/2 = (y-2)/3 = (z-3)/-3 and (x+3)/-1 = (y-5)/8 = (z-1)/4. Cos(10/9√22)
C) (5-x)/-2 = (y+3)/1 = (1 -z)/3 and x/3 = (1-y)/-2 = (z+5)/-1 Cos(11/14)
D) (x -2)/3 = (y+3)/-2 , z = 5 and (x+1)/1 = (2y-3)/3 = (z-5)/2. π/2
E) (-x +2)/-2 = (y-1)/7 = (z+3)/-3 and (x+2)/-1 = (2y-8)/4 = (z-5)/4. π/2
9) Fnd the angle between the pairs of lines with direction ratios proportional to:
A) 5,-12,13) and -3,4,5. Cos(1/65)
B) 2,2,1 and 4,1,8. Cos(2/3)
C) 1,2,-2 and -2,2,1. π/2
D) a,b,c and b- c, c- a, a- b. π/2
10) Find the angle between two lines, one of which has the direction ratios 2,2,1 while the other one is obtained by joining the points (3, 1,4) and (7, 2,12). Cos(2/3)
11) Find the equation of the line passing through the point (1,2,-4) and parallel to the line (x -3)/5 = (y-5)/2 = (z+1)/3. (x-1)/4 = (y-2)/2 = (z+4)/3
12) Find the equations of the line passing through the point (-1,2,1) and parallel to the line (2x -1)/4 = (3y+5)/2 = (2-z)/3. (x+1)/2 = (y-2)/2/3 = (z-1)/-3
13) Find the equation of the line passing through the point at (2,-1,3) and parallel to the line r= (i - 2j +k)+ λ(2i + 3j -5k). r= (2i - j +3k)+ λ(2i + 3j -5k).
14) Find the equations of the line passing through the point (2,1,3) and perpendicular to the lines.(x-1)/1 = (y-2)/2 = (z-3)/3. (x-3)/3 = (y-1)/-7 = (z-3)/4
15) Find the equation of the line passing through the point i+ j - 3k and perpendicular to the line r= i + λ(2i + j -3k) and r= (2i + j -6k)+ μ(i + j +k). r= (i +j -3k)+ λ(4i -5 j -k)
16) Find the equation of the line passing through the point (1,-1,1) and Perpendicular to the lines joining the points (4,3,2),(1,-1,0) and (1,2-1),(2,1,1). (x-1)/10 = (y+1)/-4 = (z-1)/-7
17) Determine the equation of the line passing through the point (1,2,-4) and perpendicular to the lines (x-8)/8 = (y+9)/-26 = (z-10)/7 and (x-15)/3 = (y-29)/8 = (z-5)/-5
18) Show that the lines (x-5)/7 = (y+2)/-5 = z/2 and x/2 = y/2 = z/3 are perpendicular to each other.
19) Find the vector equation the line passing through (2, -1,-1) which is parallel to the lines 6x -2, 3y +1= 2z -2. r= (2i -j -k)+ λ(i + 2j +3k)
20)If the lines (x-1)/-3= (y-2)/2λ = (z-3)/2 and (x-1)/3λ = (y-1)/1 = (z-6)/-5 are perpendicular, find the value of λ. -10/7
21) If the coordinates of the points A, B ,C, D be (1,2,3), (4,5,7),(-4,3,-6) and (2,9,2) respectively, then find angle between the lines AB and CD. 0
22) Find the value λ so that the following lines are perpendicular to each other. (x-5)/(5λ+2) = (2- y)/5 = (1- z)/-1, x/1 = (2y+1)/4λ = (1-z)/-3. 1
23) Find the direction cosines of the line (x+2)/2 = (2y -7)/6 = (5- z)/6. Also, find the vector equation of the line through the point A(-1,2,3) and parallel to the given line. 2/7,3/7,-6/7; (x+2)/2 = ( y -2)/3 = (z -3)/-6
INTERSECTION OF TWO LINES
If the lines be in CARTESIAN form
Let the two lines be
(x - x₁)/a₁ = (y - y₁)/b₁ = (z - z₁)/c₁ ........(1) and
(x - x₂)/a₂ = (y - y₂)/b₂ = (z - z₂)/c₂.........(2)
Step 1: Write the co-ordinates of general points on (1) and (2). The coordinates of general points on (1) and (2) are given by
(x - x₁)/a₁ = (y - y₁)/b₁ = (z - z₁)/c₁ = λ and
(x - x₂)/a₂ = (y - y₂)/b₂ = (z - z₂)/c₂ = μ respectively.
i.e., (a₁λ+ x₁ , b₁λ+ y₁ , c₁λ + z₁) and
(a₂μ+ x₂ , b₂μ+ y₂ , c₂μ + z₂)
Step 2:
If the line (1) and (2) intersect, then they have a common point.
So, a₁λ+ x₁ = a₂μ+ x₂, b₁λ+ y₁ = b₂μ+ y₂ and c₁λ + z₁ = c₂μ + z₂
Step 3: Solve any two of the equations in λ and μ obtaining in step 2. If the values of λ and μ satisfy the third equation, then the line (1) and (2) intersect. Otherwise they do not interesect .
Step 4: To obtain the co-ordinates of the point of intersection, substitute the value λ and μ in the coordinates of general point/s obtain in step 1.
If the lines in VECTOR FORM
Let the two lines be
r= (a₁i+ a₂j + a₃k)+ λ(b₁i+ b₂j + b₃k)........(1) And
r= (a'₁i+ a'₂j + a'₃k)+ μ(b'₁i+ b'₂j + b'₃k)........(2)
Step 1: Since r in the equation of a line denotes the position vector of an arbitrary point on it.
Therefore , position vectors of arbitrary points on (1) and (2) are given by
(a₁i+ a₂j + a₃k)+ λ(b₁i+ b₂j + b₃k) and
(a'₁i+ a'₂j + a'₃k)+ μ(b'₁i+ b'₂j + b'₃k) respectively
Step 2: If the lines (1) and (2) intersect, than they have a common point. So,
(a₁i+ a₂j + a₃k)+ λ(b₁i+ b₂j + b₃k) = (a'₁i+ a'₂j + a'₃k)+ μ(b'₁i+ b'₂j + b'₃k)
=> (a₁ + λb₁)i+ (a₂ + λb₂)j + (a₃ + λb₃) = (a'₁ + μb'₁)i+ (a'₂ + μb'₂)j + (a'₃ + μ b'₃)k
= a₁ + λb₁ = a'₁ + μb'₁, a₂ + λb₂ = a'₂ + μb'₂ and a₃ + λb₃ = a'₃ + μ b'₃.
Step 3: Solve any two of the equation in λ and μ obtained in step 2. if the values of λ and μ satisfy the third equation, then the two lines interesect . Otherwise they do not.
Step 4: To obtain the position vector of the point of intersection, substitute the value of λ (or μ) in (1) (or (2)).
EXERCISE -C
1) Show that the lines intersect. Find their point of intersection.
(x-1)/2 = (y-2)/3 = (z-3)/4 and (x-4)/5 = (y-1)/2 = z. (-1,-1,-1)
2) Show that the lines do not intersect.
(x-1)/3 = (y +1)/2 = (z- 1)/5 and (x +2)/4 = (y-1)/3 = (z+1)/-2.
3) Show that the lines r= (i + j -k) +λ(3i -j) and r= (4i -k)+ μ(2i + 4k) interesect. Find their point of intersection. (4,0,-1)
4) Find the equations of the two lines through the origin which intersect the line (x-3)/2 = (y-3)/1 = z/1 at angle of π/3 each. x/1 = y/2 = z/-1 and x/-1 = y/1= z/-2.
MISCELLANEOUS - 3
1) Show that the lines interesect and find their interesection.
a) x/1 = (y-2)/2 = (z +3)/3 and (x-2)/2 = (y- 6)/3 = (z-3)/4. (2,6,3)
b) (x +1)/3= (y+ 3)/5 = (z +)/7 and (x- 2)/1 = (y-4)/3 = (z-6)/5. (1/2, -1/2,-3/2)
c) A(0,-1,-1) and B(4,5,1) ; C(3,9,4) and D(-4,4,4). (10,14,4)
d) r= (i + j -k) +λ(3i -j) and r= (4i -k)+ μ(2i + 3k). (4,0,-1)
e) r= (3i + 2j - 4k) +λ(i +2j+ 2k) and r= (5i -2j)+ μ(3i +2j+ 6k). (-1,-6,-12)
2) Determine whether the following pair of lines intersect or not.
a) (x-1)/2 = (y+1)/3 = z and (x+1)/5 = (y-2)/1, 2 = z. No
b) (x-1)/3 = (y-1)/-1 = (z+1)/0 and (x-4)/2 = (y-0)/0 = (z+1)/3. Yes
c) (x-5)/4 = (y-7)/4 = (z+3)/-5 and (x-8)/7 = (y-4)/1 = (z-5)/3. Yes
D) r= (i - j) +λ(2i +k) and r= (2i - j)+ μ(i + j - k). No
PERPENDICULAR DISTANCE OF A LINE FROM A POINT:
CARTESIAN FORM:
Let P(α, β,γ) be a given point and let (x-x₁)/a = (y- y₁)/b = (z - z₁)/c be a given line.
Step.1: Write the coordinates of a general point on the given line. The coordinates of general point the line are (x₁ + aλ , y₁ + bλ , z₁ + cλ), where λ is a parameter. Assume that point L. is the foot of the perpendicular drawn from P on the given line .
Step.2: Write direction ratios of PL.
Step. 3: Apply the condition of Perpendicularity of the given line and PL.
Step. 4: Obtain the value of λ from the Step 3.
Step. 5: Substitute the δ in (x₁ + aλ , (y₁ + bλ , z₁ + cλ) to obtain the coordinates of L.
Step 6: Obtain PL by using distance formula
** VECTOR FORM:
Let P(α) be the given point, and let r= a + λb be the given line.
Step.1: Write the position vector of a general point on the given line. The position vector of a general point on r= a + λb is a + λb, where λ is a parameter. Assume that this point L is required foot of the Perpendicular from P on the given line.
Step.2: Obtain PL= Position vector of L - Position vector of P = a+ λb - α.
Step. 3: Put PL . b = 0 i.e., (a + λb - α). b = 0 to obtain the value of λ.
Step.4: Substitute the value of λ in r= a + λb to obtain the position vector of L.
Step. 5: Find | PL | to obtain the required length of the perpendicular.
EXERCISE - D
1) Find the foot of the perpendicular from the point (0,2,3) on the line (x+ 3)/5 = (y- 1)/2 = (z+ 4)/3. √21
2) Find the length of the perpendicular to the point (1,2,3) to the line (x-6)/3 = (y- 7)/2 = (z- 7)/-2. 7
3) Find the foot of the perpendicular drawn from the point 2i - j + 5k to the line r= (11i - 2j - 8k)+ λ(10 i - 4j - 11k). Also, find the length of the perpendicular. √14
4) Find the image of the point (1,6,3) in the line x/1 = (y-1)/2 = (z -2)/3. Also, write the equation of the line joining the given point and its image and find the length of the segment joining the given point and its image. (1,3,5), 2√13
5) Find the distance from the point P(3, -8, 1) to the line (x -3)/3 = (y+7)/-1 = (z +2)/5. √(94/35)
6) Vertices B and C of ∆ ABC lie along the (x+2)/2 = (y -1)/1 = (z -0)/4. Find the area of the triangle given that A has coordinates (1, -1,2) and line segment BC has length 5. √(1775/28)
MISCELLANEOUS - 2
1) Find the perpendicular distance of the point (3,-1,11) from the line x/2 = (y- 2)/-3 = (z-3)/4. √53
2) Find the perpendicular distance of the point (1,0,0) from the line (x -1)/2 = (y+1)/-3 = (z+10)/8. Also, find the co-ordinates of the foot of the perpendicular and the equation of the perpendicular. 2√6, (3, -4, -2), r= i + β(i - 2j- k)
3) Find the foot of the perpendicular drawn from the point A(1,0,3) to the joint of the points B( 4, 7,1) and C(3,5,3). (5/3, 7/3, 17/3)
4) If A(1,0,4), B(0, -11,3), C(2,-3,1) are three points and D is the foot of perpendicular from A on BC. Find the coordinates of D. (22/9, -11/9, 5/9)
5) Find the foot of perpendicular from the point (2,3,4) to the line (4 -x)/2 = y/6 = (1 - z)/3. Also, find the perpendicular distance from the given to the line. (170/49, 78/49, 10/49), 3√(101/49)
6) Find the equation of the perpendicular drawn from the point P(2,4,-1) to the line (x +5)/1 = (y+3)/4 = (z-6)/-9. Also, write down the coordinates of the foot of the Perpendicular from P. (-4,1,-3); (x -2)/-6 = (y-4)/-3 = (z+1)/-2
7) Find the length of the perpendicular drawn from the point (5,4,-1) to the line r= i+ λ(2i+ 9j +5k). √(2109/110)
8) Find the foot of the perpendicular drawn from the point i + 6j + 3k to the line r= j+ 2k+ λ(i+ 2j +3k). Also, find the length of the perpendicular. i+ 3j +5k; √13
9) Find the equation of the perpendicular drawn from the point P(-1,3,2) to the line r= (2j + 3k) + λ(2i+ j +3k). Also, find the coordinates of the foot of the perpendicular from P. r= (-i + 3j +2k)+ λ(3i- 9j +k), (-4/7,12/7,15/7)
10) Find the foot of the perpendicular from (0,2,7) on the line (x -2)/-1 = (y-1)/3 = (z-3)/-2. (-3/2,-1/2,4)
11) Find the foot of the perpendicular from (1,2,-3) to the line (x +1)/2 = (y-3)/-2 = z/-1. (1,1,-1)
12) Find the equation of a line passing through the point A(0,6,-9) and (-3,-6,3). If D is the foot of perpendicular drawn from the point C(7,4,-1) on the line AB, then find the coordinates of the point D and the equation of the line CD. x/1 = (y-6)/4 = (z+9)/-4; (-1,2,-5); (x -7)/4 = (y-4)/1 = (z+1)/2
13) Find the distance of the point (2,4,-1) from the line (x +5)/1 = (y+3)/4 = (z-6)/-9. 7
14) Find the coordinates of the foot of perpendicular drawn from the point (1,8,4) to the line joining the points B(0,-1,3) and C(2,-3,-1). (-5/3,2/3,19/3)
SHORTEST DISTANCE BETWEEN TWO TWO LINES
Shortest distance between two skew lines(Vector Form)
is condition per to given lines to intersect if the line intersect with the sorted distance between the energy shortest distance between two skew lines curtain form
Unsatisfied between the lines by computer distance determine whether the following pair supplies intercept or not by the term between the lines equations are
₁₁₁₁₁₁₂₂₂₂₂₂₁₂₃ ₓ ₐ ᵢ ₙ μ γ ς λ θ
δβγ ₁ ₂ ₃
₃₃₂₃₂₃₂₃
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