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A line with direction cosines proportional to 2, 1, 2 meets each of the lines x = y + a = z and x + a = 2y = 2z. The co-ordinates of each of the points of intersection are given by
- a)(3a, 3a, 3a), (a, a, a)
- b)(3a, 2a, 3a), (a, a, a)
- c)(3a, 2a, 3a), (a, a, 2a)
- d)(2a, 3a, 3a,), (2a, a, a)
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A line with direction cosines proportional to2, 1, 2 meets each of the...
The co-ordinates of any point on L1 in terms of parameter r1 are given by,x=y+a=z=r1
⇒x=r1,y=r1 - a,z=r1....(1)
Similarly the co-ordinates of any point on L2 in terms of parameter 2r2 are given by
⇒x+a=2y=2z=2r2
⇒ x=2r2−a,y=r2, z=r2...(2)
Let 'A' be a point on L1 and 'B' be a point on L2
Using (1) and (2), the direction ratios of AB are
2r2−a−r1,r2−r1+a,r2−r1
If the above line is same as the line whose direction cosines are proportional to (2,1,2) as given in the question, then
(2r2−r1−a)/2= (r2−r1+a)/1 = (r2−r1)/2
Solving the first two of the above equation, we get r1=3a
Again solving the last two, we get r2=a
Using these values in (1) and (2), we get the coordinates of points as
(3a,2a,3a) and (a,a,a).
⇒x=r1,y=r1 - a,z=r1....(1)
Similarly the co-ordinates of any point on L2 in terms of parameter 2r2 are given by
⇒x+a=2y=2z=2r2
⇒ x=2r2−a,y=r2, z=r2...(2)
Let 'A' be a point on L1 and 'B' be a point on L2
Using (1) and (2), the direction ratios of AB are
2r2−a−r1,r2−r1+a,r2−r1
If the above line is same as the line whose direction cosines are proportional to (2,1,2) as given in the question, then
(2r2−r1−a)/2= (r2−r1+a)/1 = (r2−r1)/2
Solving the first two of the above equation, we get r1=3a
Again solving the last two, we get r2=a
Using these values in (1) and (2), we get the coordinates of points as
(3a,2a,3a) and (a,a,a).
Most Upvoted Answer
A line with direction cosines proportional to2, 1, 2 meets each of the...
To find the coordinates of the points of intersection, we need to solve the given system of equations.
Equation 1: x = y
Equation 2: a = z
Equation 3: x = 2y = 2z
Let's solve these equations step by step:
Solving Equations 1 and 2:
Since x = y and a = z, we can substitute these values into Equation 3:
x = 2a
y = a
z = a
So, the coordinates of the point of intersection for Equations 1 and 2 are (a, a, a).
Solving Equations 1 and 3:
Substituting x = y into Equation 3:
y = 2y = 2z
Simplifying, we get:
2y = 2z
y = z
Substituting y = z into Equation 1, we get:
x = y = z
So, the coordinates of the point of intersection for Equations 1 and 3 are (a, a, a).
Thus, the coordinates of the points of intersection are (a, a, a) for both Equations 1 and 2, and Equations 1 and 3.
To verify if the direction cosines of the line are proportional to 2, 1, 2, we can calculate the direction cosines of the line passing through these points:
Direction cosines (l, m, n) are given by:
l = (x2 - x1) / sqrt((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2)
m = (y2 - y1) / sqrt((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2)
n = (z2 - z1) / sqrt((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2)
Using the coordinates of the points (a, a, a) and (3a, 2a, 3a), we can calculate the direction cosines:
l = (3a - a) / sqrt((3a - a)^2 + (2a - a)^2 + (3a - a)^2) = 2/3
m = (2a - a) / sqrt((3a - a)^2 + (2a - a)^2 + (3a - a)^2) = 1/3
n = (3a - a) / sqrt((3a - a)^2 + (2a - a)^2 + (3a - a)^2) = 2/3
Since the direction cosines are proportional to 2, 1, 2, the correct answer is option B: (3a, 2a, 3a), (a, a, a).
Equation 1: x = y
Equation 2: a = z
Equation 3: x = 2y = 2z
Let's solve these equations step by step:
Solving Equations 1 and 2:
Since x = y and a = z, we can substitute these values into Equation 3:
x = 2a
y = a
z = a
So, the coordinates of the point of intersection for Equations 1 and 2 are (a, a, a).
Solving Equations 1 and 3:
Substituting x = y into Equation 3:
y = 2y = 2z
Simplifying, we get:
2y = 2z
y = z
Substituting y = z into Equation 1, we get:
x = y = z
So, the coordinates of the point of intersection for Equations 1 and 3 are (a, a, a).
Thus, the coordinates of the points of intersection are (a, a, a) for both Equations 1 and 2, and Equations 1 and 3.
To verify if the direction cosines of the line are proportional to 2, 1, 2, we can calculate the direction cosines of the line passing through these points:
Direction cosines (l, m, n) are given by:
l = (x2 - x1) / sqrt((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2)
m = (y2 - y1) / sqrt((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2)
n = (z2 - z1) / sqrt((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2)
Using the coordinates of the points (a, a, a) and (3a, 2a, 3a), we can calculate the direction cosines:
l = (3a - a) / sqrt((3a - a)^2 + (2a - a)^2 + (3a - a)^2) = 2/3
m = (2a - a) / sqrt((3a - a)^2 + (2a - a)^2 + (3a - a)^2) = 1/3
n = (3a - a) / sqrt((3a - a)^2 + (2a - a)^2 + (3a - a)^2) = 2/3
Since the direction cosines are proportional to 2, 1, 2, the correct answer is option B: (3a, 2a, 3a), (a, a, a).
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A line with direction cosines proportional to2, 1, 2 meets each of the lines x = y + a = z and x + a = 2y = 2z. The co-ordinates of each of thepoints of intersection are given bya)(3a, 3a, 3a), (a, a, a)b)(3a, 2a, 3a), (a, a, a)c)(3a, 2a, 3a), (a, a, 2a)d)(2a, 3a, 3a,), (2a, a, a)Correct answer is option 'B'. Can you explain this answer?
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A line with direction cosines proportional to2, 1, 2 meets each of the lines x = y + a = z and x + a = 2y = 2z. The co-ordinates of each of thepoints of intersection are given bya)(3a, 3a, 3a), (a, a, a)b)(3a, 2a, 3a), (a, a, a)c)(3a, 2a, 3a), (a, a, 2a)d)(2a, 3a, 3a,), (2a, a, a)Correct answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A line with direction cosines proportional to2, 1, 2 meets each of the lines x = y + a = z and x + a = 2y = 2z. The co-ordinates of each of thepoints of intersection are given bya)(3a, 3a, 3a), (a, a, a)b)(3a, 2a, 3a), (a, a, a)c)(3a, 2a, 3a), (a, a, 2a)d)(2a, 3a, 3a,), (2a, a, a)Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A line with direction cosines proportional to2, 1, 2 meets each of the lines x = y + a = z and x + a = 2y = 2z. The co-ordinates of each of thepoints of intersection are given bya)(3a, 3a, 3a), (a, a, a)b)(3a, 2a, 3a), (a, a, a)c)(3a, 2a, 3a), (a, a, 2a)d)(2a, 3a, 3a,), (2a, a, a)Correct answer is option 'B'. Can you explain this answer?.
A line with direction cosines proportional to2, 1, 2 meets each of the lines x = y + a = z and x + a = 2y = 2z. The co-ordinates of each of thepoints of intersection are given bya)(3a, 3a, 3a), (a, a, a)b)(3a, 2a, 3a), (a, a, a)c)(3a, 2a, 3a), (a, a, 2a)d)(2a, 3a, 3a,), (2a, a, a)Correct answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A line with direction cosines proportional to2, 1, 2 meets each of the lines x = y + a = z and x + a = 2y = 2z. The co-ordinates of each of thepoints of intersection are given bya)(3a, 3a, 3a), (a, a, a)b)(3a, 2a, 3a), (a, a, a)c)(3a, 2a, 3a), (a, a, 2a)d)(2a, 3a, 3a,), (2a, a, a)Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A line with direction cosines proportional to2, 1, 2 meets each of the lines x = y + a = z and x + a = 2y = 2z. The co-ordinates of each of thepoints of intersection are given bya)(3a, 3a, 3a), (a, a, a)b)(3a, 2a, 3a), (a, a, a)c)(3a, 2a, 3a), (a, a, 2a)d)(2a, 3a, 3a,), (2a, a, a)Correct answer is option 'B'. Can you explain this answer?.
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