* ON FINDING THE VECTOR EQUATION OF A LINE SATISFYING THE GIVEN CONDITION AND REDUCING IT CARTESIAN FORM:
=> Formula to be used:
1) r= a + ¥b
2) r= a+ ¥(b - a)
3) The Cartesian form of a straight line passing through a fixed point (x₁, y₁,z₁) and having direction Ratios proportional a, b, c is :(x- x₁)/a = (y- y₁/b= (z- z₁)/c
4) The Cartesian equation of a line passing through two given points (x₁, y₁,z₁) and (x₂, y₂, z₂) is: (x- x₁)/(x₂ - x₁)= (y- y₁)/(y₂ - y₁)= (z- z₁)/((z₂ - z₁).
EXERCISE- 1
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1) Find the vector equation of a line which passes 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-2)/1= (y+1)/1= (z -4)/-2
2) Find the vector equation of the line A(3,4,7) and B(1,-1,6). Find also, its cartesian equations. r=(3i +4j -7k)+ ¥(-2i -5j +13k). (x-3)/-2= (y-4)/-5= (z +7)/13
3) The point A(4,5,10), B(2,3,4) and C(1,2,-1) are three vertices of a parallelogram ABCD. Find vector and cartesian equations for the side AB and BC and find the coordinates of D. r=(4i + 5j+10k)+ ¥(I+j +3k). (x-4)/1= (y-5)/1= (z -10)/3. And r=(2i +3j+4k)+ €(I+j+5k). (x-2)/1= (y-3)/1= (z -4)/5. And (3,4,5)
4) Find the vector equation of a line passing through a point with position vector 2i- j+ k, and parallel to the line joining the point - i + 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. r=(2i-j+3k)+ ¥(2i+ 3j- 2k). (x-2)/2= (y+1)/3= (z - 3)/-2
6) The cartesian equations of a lines are 6x - 2= 3y +1= 2z -2. Find its direction Ratios and also find vector equation of the line. r=(i/3 -j/3 +k)+ ¥(i+ 2j +3k). (x-1/3)/1= (y+1/3)/2= (z -4)/3.
7) Find the direction cosines of the line (x-2)/2= (2y -5)/-3, z= -1 Also, find the vector equation of the line. 4/5, -3/5, 0, r=(2i +5j/2 - k)+ ¥(2i- 3j/2 +0k).
8) show that the points whose position vectors 5i +5k, 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, m) are collinear, find the value of m. 10
10) Find the point on the line (x+2)/3= (y+1)/3= (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)
11) find the vector and 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
12) Find the vector equation of the line passing through the points (-1,0,2) and (3,4,6). r=(-i + 2k)+ ¥(4i+ 4j + 4k).
13) 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.
14) A line passes through the point with position vector 2i - 3j + 4k and is in the direction of 3i+ 4j - 5k. find the equation of the line in vector and cartesian form. r=(2i- 3j+4k)+ ¥(3i+ 4j- 5k). (x-2)/3= (y+3)/4= (z -4)/-5
15) ABCD is a parallelogram. The position vectors of the point A, B and C are respectively, r= 4i + 5j - 10k, 2i - 3j + 4k, and - i+2j+ k. Find the vector equation of the line BD. Also, reduced it to Cartesian form.
r=(2i- 3j+4k)+ ¥(i - 13j + 17k). (x-2)/1= (y+3)/-13= (z -4)/17
16) find in vector form as well as in cartesian 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-2)/1= (y-2)/-1= (z +1)/2
17) 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 in cartesian form. r=(i +2 j+3k)+ ¥(i - 2j + 3k). (x-1)/1= (y -2)/-2= (z - 3)/3
18) Find the vector equation of a line passing through (2,-1,1) and parallel to the line whose equations are (x-3)/2= (y+1)/7= (z -2)/-3. r=(2i-j+k)+ ¥(2i+ 7j- 3k).
19) The Cartesian equation of a lines are (x-5)/3= (y+4)/7= (z - 6)/2. Find a vector equation for the line. r=(5i-4j+6k)+ ¥(3i+7j + 2k).
20) find the Cartesian equation of a line passing through (1,-1,2) and parallel to the line whose equation are (x-4)/1= (y-1)/3= (z + 1)/-2. Also, reduce the equation obtained in a vector form. r=(i-j+2k)+ ¥(i+2j- 2k). (x-2)/1= (y+1)/2= (z -2)/-2
21) 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- 0j+k)+ ¥(2i+ 6j- 3k)
22) The Cartesian equation of a line are x= ay +b, z= cy + d. Find its direction Ratios and reduce it to vector form. a,1,c; r=(bi 0j+dk)+ ¥(ai + j + ck)
23) Find the vector equation of a line passing through the point with position vector i - 2j - 3k and parallel to the line joining the points with position vector i- j+4k) and 2i+j + 2k). and Also, find the Cartesian equivalent of this equation. r=(i- 2j -3k)+ ¥(i+2j- 2k). (x-1)/1= (y+2)/3= (z +3)/-2
24) Find the point on the line (x+2)/3= (y+1)/2= (z -3)/2 at a distance of 5 units from the point P(1,3,3). (4,3,7),(-2,-1,3)
25) Show that the points whose position vectors are 2i +3j+4k, i+ 2j + 3k) and 7i+ 9k are collinear.
26) 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. r=(i +2j+3k)+ ¥(-2i+ 14j +3k). (x-1)/-2= (y-2)/14= (z -3)/3
27) 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. (-1/3, 1/3, 1); 2, 1, -6; r= -i/3 + j/3 +k)+ ¥(2i +j- 6k)
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ANGLE BETWEEN TWO LINES
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₁₁₂₂
VECTOR FORM:
let the vector equations of the lines be r= a₁ + ¥b₁ and r= a₂ + ¥ 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 ¢ is the angle between the given lines, then
Cos ¢= b₁. b₂/(|b₁|.| b₂|)
* CONDITION of perpendicularity:
If the lines b₁ and b₂ are perpendicular. Than, b₁. b₂= 0
* CONDITION of parallelism:
If the lines are parallel, then b₁ and b₂ are parallel.
b₁ = K b₂ for some scalar K.
CARTESIAN FORM:
(x - x₁)/a₁ = (y- y₁)/b₁=(z- z₁)/c₁.. (1)
And
(x- x₂)/a₂= (y- y₂)/b₂ =(z - z₂)/c₂..(2)
*Direction Ratios of line (1) are proportional to a₁, b₁, c₁.
So m₁= Vector parallel to line (1)= a₁i + b₁j + c₁k
* Direction Ratios of line (2) are proportional to a₂, b₂, c₂.
So m₂= Vector parallel to line (2)= a₂i+ b₂j + c₂k
Let ¢ be the angle between (1) and (2). Then, ¢ is also the angle between m₁, m₂.
Cos ¢= m₁. m₂/(|m₁|. |m₂|) OR
cos ¢= (a₁a₂+ b₁b₂+ c₁c₂)/{√(a₁²b₁²+c₁²) √(a₂²+b₂²+ c₂²)}
* CONDITION of perpendicularity:
If the lines are perpendicular, then
m₁. m₂ = 0 => a₁a₂+ b₁b+ c₁c₂ = 0
* CONDITION of parallelism:
If the lines are parallel, then m₁ and m₂ are parallel.
Then m₁ = K m₂ for some scalar K.
=> a₁/a₂ = b₁/b₂= c₁/c₂.
ON FINDING THE ANGLE BETWEEN TWO LINES:
FORMULA TO BE USED:
cos ¢= b₁. b₂/(|b₁|.| b₂|) OR
cos ¢= (a₁a₂+ b₁b₂+ c₁c₂)/{√(a₁²b₁²+c₁²) √(a₂²+b₂²+ c₂²)}
ON FINDING THE EQUATION OF A LINE PARALLEL TO A GIVEN LINE AND PASSING THROUGH A GIVEN POINT:
Formula to be used:
1) r = a + Kb
2) (x - x₁)/a = (y - y₁)/b = (z - z₁)/c
ON FINDING THE EQUATION OF A LINE PASSING THROUGH A GIVEN POINT AND PERPENDICULAR TO TWO GIVEN LINES:
Result to be used:
A line passing through a point having position vector $ and perpendicular to the lins r= a₁+ Kb₁ and r= a₂+ M b₂ is parallel to the vector b₁ x b₂. So, its vector equation is r= $ + K(b₁ x b₂)
STEP- 1 Obtain the point through which the line passes. Let its position vector be $.
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 = $ + K(b₁ x b₂).
EXERCISE-2
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1) Find the angle between lines r= 3i+ 2j- 4k + K(i+ 2j + 2k) and r= 5j- 2k + M(3i+ 2j + 6k). cos ¢= (19/21)
2) Find the 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 and 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 proportional to 1, 1, 2 and (√3, -1),((- √3 -1), 4) π/3
5) Find the equation of a line passing through a point (2, -1, 3) and parallel to the line r= (i +j) + M(2i+ j - 2k). r= (2i - j +3k) + M(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.
7) Find the cartesian equation 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)/2= (y -3)/-7 = (z+2)/4.
8) A line passes through (2,-1,3) and is perpendicular to the line r= (i+ j - k)+ K(2i - 2j + k) and r= (2i- j - 3k)+ M(i + 2j + 2k). Obtain its equation. r= (2i- j +3k)+ M(2i +j - 2k), Where M= - 3K.
9) Find the value of P so that the lines. L ₁: (1-x)/3 =(7y- 14)/2P =(z-3)/2 and L₂: (7- 7x)/3P = (y -5)/1 = (6 - z)/5 are at right angle.
Also, find the equation of a line passing through the point (3,2, -4) and parallel to line L₁. 70/11, (x -3)/-3 = (y-2)/20/11 = (z+4)/2
10) Show that the three lines with direction cosines (12/13, -3/13, -4/13),(4/13, 12/13, 3/13),( 3/13, -4/13, 12)13) are mutually perpendicular.
11) 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).
12) Show that the line through the points (4,7,8) and (2,3,4) is parallel to the line through the points (-1,-2,1) and (1,2,5).
13) 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.
14) Show that the lines are (x -5)/7 = (y +2)/-5 = z/1 and x/1 = y/2 = z/3 are perpendicular to each other.
15) 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).
16) Find the equation of a line parallel to x-axis and passing through the origin. x/1 = y/0 = z/0
17) Find the angle between the following pairs of lines:
A) r= (4i- j)+ K(i + 2j - 2k) and r= (i- j +2k)+ M(2i + 4j - 4k). 0
B) r= (3i + 2j - 4k)+ K(i + 2j + 2k) and r= (5j - 2k)+ M(3i + 2j +6k). Cos ¢=(19/21)
C) r= K(i + j +2k) and r= 2j + M{(√3 -1)i - (√3+1)j + 4k). π/3
18) 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)/1 = (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 -5)/1 = (2y +6)/-2 = (z-3)/1. And (x -2)/3 = (y +1)/4 = (z - 6)/5. π/2
F) (-x +2)/-2 = (y -1)/7 = (z+3)/-3. And (x +2)/-1 = (2y -8)/4 = (z -5)/4. π/2
19) Find the angle between the pairs of the lines with direction Ratios proportional to
A) 5, -12,13 & -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
20) Find the angle between two lines, one of which has direction 2, 2,1 while the other one is obtained by joining the points (3,1,4) and (7,2,12). Cos ¢= (2/3)
21) Find the equation of the line passing through the point (1,2,-4) and parallel to the line (x-3)/4 = (y-5)/2 = (z+1)/3. (x-)/4 = (y-2)/2 = (z+4)/3.
22) find the equation 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.
23) Find the equation of the line passing through the point (2,-1,3) and parallel to the line r=(i - 2j + k) + K(2i + 3j - 5k). r=(2i - j + 3k) + K(2i + 3j - 5k).
24) Find the equation of the line passing through the point (2,1,3) and perpendicular to the lines (x-1)/1 = (y-2)/2= (z-3)/3 and x/-3 = y/2 = z/3. (x-2)/2 = (y-2)/-7 = (z-3)/4.
25) Find the equation of the line passing through the point i + j - 3k and perpendicular to the lines r=i + ¥(2i + j - 3k) and r=(2i + j -k) + M(i + j + k). r=(i +j +3 k) + ¥(4i - 5j +k)
26) 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
27) Determine the equations of the line passing through the point (1,2,-4) and perpendicular to the lines (x-8)/8 = (y +9)/-16 = (z- 10)/7 and (x- 15)/3 = (y-29)/8 = (z-5)/-5.
28) Show that the lines (x-5)/7 = (y+2)/-5 = z/1 and x/1 = y/2 =z/3 are perpendicular to each other.
29) Find the vector equation of the line passing through the point (2,-1,-1) which is parallel to the line 6x -2= 3y+ 1= 2z -2. r=(2i -j - k) + ¥(i + 2j + 3k).
30) If the lines (x-1)/-3 = (y-2)/2P = (z -3)/2 and (x-1)/3P = (y-1)/1 = (z - 6)/-5 are perpendicular, find the value of P. -10/7
31) 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 the angle between the lines AB and CD. 0
32) Find the value of P so that the following lines are perpendicular to each other. (x-5)/(5P+2) = (2-y)/5 = (1-z)/-1, x/1 = (2y +1)/4P = (1-z)/-3. 1
33) 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 (-1,2,3) and parallel to the given line. 2/7, 3/7,-6/7; (x+1)/2 = (y-2)/3 = (z -3)/-6
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