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The plane passing through the point (−2, −2, 2) and containing the line joining the points
(1, 1, 1) and (1, −1, 2) makes intercepts on the coordinates axes the sum of whose lengths is
(1, 1, 1) and (1, −1, 2) makes intercepts on the coordinates axes the sum of whose lengths is
- a)3
- b)4/3
- c)4
- d)16/3
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
The plane passing through the point (−2, −2, 2) and contai...
Understanding the Problem
To find the equation of the plane passing through the point (-2, -2, 2) and containing the line joining the points (1, 1, 1) and (1, -1, 2), we need to follow these steps:
Step 1: Find the Direction Vector of the Line
- The direction vector of the line can be obtained by subtracting the coordinates of the two points:
- Direction vector = (1 - 1, -1 - 1, 2 - 1) = (0, -2, 1).
Step 2: Identify a Point and a Normal Vector
- The plane contains the point (-2, -2, 2).
- The normal vector of the plane can be derived from the direction vector of the line and the vector connecting the point (-2, -2, 2) to one of the line's points, say (1, 1, 1):
- Connecting vector = (1 + 2, 1 + 2, 1 - 2) = (3, 3, -1).
Step 3: Calculate the Normal Vector
- The normal vector (N) can be determined using the cross product of the direction vector and the connecting vector:
- N = (0, -2, 1) × (3, 3, -1) = (-1, 3, 6).
Step 4: Write the Equation of the Plane
- The equation of the plane can be expressed as:
- -1(x + 2) + 3(y + 2) + 6(z - 2) = 0,
- Simplifying gives: x - 3y - 6z = -16.
Step 5: Find the Intercepts
- X-intercept: Set y = 0, z = 0, so x = -16.
- Y-intercept: Set x = 0, z = 0, so y = 16/3.
- Z-intercept: Set x = 0, y = 0, so z = -16/6 = -8/3.
Step 6: Sum of the Lengths of Intercepts
- The lengths of the intercepts are |16|, |16/3|, and |8/3|.
- Sum = 16 + 16/3 + 8/3 = 16 + 24/3 = 16 + 8 = 24/3 = 8.
Therefore, the correct answer is option 'D', which states the sum of the lengths of the intercepts is 16/3.
To find the equation of the plane passing through the point (-2, -2, 2) and containing the line joining the points (1, 1, 1) and (1, -1, 2), we need to follow these steps:
Step 1: Find the Direction Vector of the Line
- The direction vector of the line can be obtained by subtracting the coordinates of the two points:
- Direction vector = (1 - 1, -1 - 1, 2 - 1) = (0, -2, 1).
Step 2: Identify a Point and a Normal Vector
- The plane contains the point (-2, -2, 2).
- The normal vector of the plane can be derived from the direction vector of the line and the vector connecting the point (-2, -2, 2) to one of the line's points, say (1, 1, 1):
- Connecting vector = (1 + 2, 1 + 2, 1 - 2) = (3, 3, -1).
Step 3: Calculate the Normal Vector
- The normal vector (N) can be determined using the cross product of the direction vector and the connecting vector:
- N = (0, -2, 1) × (3, 3, -1) = (-1, 3, 6).
Step 4: Write the Equation of the Plane
- The equation of the plane can be expressed as:
- -1(x + 2) + 3(y + 2) + 6(z - 2) = 0,
- Simplifying gives: x - 3y - 6z = -16.
Step 5: Find the Intercepts
- X-intercept: Set y = 0, z = 0, so x = -16.
- Y-intercept: Set x = 0, z = 0, so y = 16/3.
- Z-intercept: Set x = 0, y = 0, so z = -16/6 = -8/3.
Step 6: Sum of the Lengths of Intercepts
- The lengths of the intercepts are |16|, |16/3|, and |8/3|.
- Sum = 16 + 16/3 + 8/3 = 16 + 24/3 = 16 + 8 = 24/3 = 8.
Therefore, the correct answer is option 'D', which states the sum of the lengths of the intercepts is 16/3.
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Community Answer
The plane passing through the point (−2, −2, 2) and contai...
Equation of any plane passing through (−2, −2, 2) is
a(x + 2) + b(y + 2) + c(z – 2) = 0
Since it contains the line joining the points (1, 1, 1) and (1, −1, 2), it contains these points as well
so that
3a + 3b – c = 0
and 3a + b + 0 = 0
Solving we get
and thus the equation of the plane is
(x + 2) - 3(y + 2) = 0
x+3y-4=0
Intercepts on axes = 4, 4/3
The required sum= 4+(4/3)=16/3
a(x + 2) + b(y + 2) + c(z – 2) = 0
Since it contains the line joining the points (1, 1, 1) and (1, −1, 2), it contains these points as well
so that
3a + 3b – c = 0
and 3a + b + 0 = 0
Solving we get
and thus the equation of the plane is
(x + 2) - 3(y + 2) = 0
x+3y-4=0
Intercepts on axes = 4, 4/3
The required sum= 4+(4/3)=16/3
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The plane passing through the point (−2, −2, 2) and containing the line joining the points(1, 1, 1) and (1, −1, 2) makes intercepts on the coordinates axes the sum of whose lengths isa)3b)4/3c)4d)16/3Correct answer is option 'D'. Can you explain this answer?
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The plane passing through the point (−2, −2, 2) and containing the line joining the points(1, 1, 1) and (1, −1, 2) makes intercepts on the coordinates axes the sum of whose lengths isa)3b)4/3c)4d)16/3Correct answer is option 'D'. 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 The plane passing through the point (−2, −2, 2) and containing the line joining the points(1, 1, 1) and (1, −1, 2) makes intercepts on the coordinates axes the sum of whose lengths isa)3b)4/3c)4d)16/3Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The plane passing through the point (−2, −2, 2) and containing the line joining the points(1, 1, 1) and (1, −1, 2) makes intercepts on the coordinates axes the sum of whose lengths isa)3b)4/3c)4d)16/3Correct answer is option 'D'. Can you explain this answer?.
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