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The angle between the line (x - 2)/2 = (y + 1)/-1 = (z - 3)/2 and the plane 3x + 6y - 2z + 5 = 0 is
- a)cos⁻1(4/21)
- b)sin⁻1-(4/21)
- c)sin⁻1(6/21)
- d)sin⁻1(4/21)
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
The angle between the line (x - 2)/2 = (y + 1)/-1 = (z - 3)/2 and the ...
To find the angle between the given line and the plane, we can use the dot product of the direction vector of the line and the normal vector of the plane. The dot product formula is given by:
a · b = |a| |b| cos θ
Where a and b are vectors, θ is the angle between them, and |a| and |b| are the magnitudes of vectors a and b, respectively.
Let's find the direction vector of the line first. From the given equation of the line, we can write the direction vector as:
V = (2, -1, 2)
Next, let's find the normal vector of the plane. The equation of the plane is given as:
3x - 6y - 2z + 5 = 0
Rewriting it in the form of ax + by + cz + d = 0, we get:
3x - 6y - 2z + 5 = 0
So, the normal vector of the plane is (3, -6, -2).
Now, let's calculate the dot product of the direction vector of the line and the normal vector of the plane:
V · N = (2, -1, 2) · (3, -6, -2)
= 2*3 + (-1)*(-6) + 2*(-2)
= 6 + 6 - 4
= 8
Next, let's calculate the magnitudes of the direction vector and the normal vector:
|V| = √(2^2 + (-1)^2 + 2^2)
= √(4 + 1 + 4)
= √9
= 3
|N| = √(3^2 + (-6)^2 + (-2)^2)
= √(9 + 36 + 4)
= √49
= 7
Now, we can substitute the values into the dot product formula to find the cosine of the angle:
cos θ = (V · N) / (|V| |N|)
= 8 / (3 * 7)
= 8 / 21
To find the sine of the angle, we can use the trigonometric identity:
sin^2 θ + cos^2 θ = 1
sin^2 θ + (8/21)^2 = 1
sin^2 θ = 1 - (64/441)
sin^2 θ = (441 - 64) / 441
sin^2 θ = 377 / 441
Taking the square root of both sides, we get:
sin θ = √(377/441)
= √377 / √441
= √377 / 21
Therefore, the answer is option (D) sin^(-1)(4/21)
a · b = |a| |b| cos θ
Where a and b are vectors, θ is the angle between them, and |a| and |b| are the magnitudes of vectors a and b, respectively.
Let's find the direction vector of the line first. From the given equation of the line, we can write the direction vector as:
V = (2, -1, 2)
Next, let's find the normal vector of the plane. The equation of the plane is given as:
3x - 6y - 2z + 5 = 0
Rewriting it in the form of ax + by + cz + d = 0, we get:
3x - 6y - 2z + 5 = 0
So, the normal vector of the plane is (3, -6, -2).
Now, let's calculate the dot product of the direction vector of the line and the normal vector of the plane:
V · N = (2, -1, 2) · (3, -6, -2)
= 2*3 + (-1)*(-6) + 2*(-2)
= 6 + 6 - 4
= 8
Next, let's calculate the magnitudes of the direction vector and the normal vector:
|V| = √(2^2 + (-1)^2 + 2^2)
= √(4 + 1 + 4)
= √9
= 3
|N| = √(3^2 + (-6)^2 + (-2)^2)
= √(9 + 36 + 4)
= √49
= 7
Now, we can substitute the values into the dot product formula to find the cosine of the angle:
cos θ = (V · N) / (|V| |N|)
= 8 / (3 * 7)
= 8 / 21
To find the sine of the angle, we can use the trigonometric identity:
sin^2 θ + cos^2 θ = 1
sin^2 θ + (8/21)^2 = 1
sin^2 θ = 1 - (64/441)
sin^2 θ = (441 - 64) / 441
sin^2 θ = 377 / 441
Taking the square root of both sides, we get:
sin θ = √(377/441)
= √377 / √441
= √377 / 21
Therefore, the answer is option (D) sin^(-1)(4/21)
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The angle between the line (x - 2)/2 = (y + 1)/-1 = (z - 3)/2 and the ...
D is the correct option.sin⁻1(4/21)
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The angle between the line (x - 2)/2 = (y + 1)/-1 = (z - 3)/2 and the plane 3x + 6y - 2z + 5 = 0 isa)cos⁻1(4/21)b)sin⁻1-(4/21)c)sin⁻1(6/21)d)sin⁻1(4/21)Correct answer is option 'D'. Can you explain this answer?
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The angle between the line (x - 2)/2 = (y + 1)/-1 = (z - 3)/2 and the plane 3x + 6y - 2z + 5 = 0 isa)cos⁻1(4/21)b)sin⁻1-(4/21)c)sin⁻1(6/21)d)sin⁻1(4/21)Correct answer is option 'D'. Can you explain this answer? for UPSC 2025 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about The angle between the line (x - 2)/2 = (y + 1)/-1 = (z - 3)/2 and the plane 3x + 6y - 2z + 5 = 0 isa)cos⁻1(4/21)b)sin⁻1-(4/21)c)sin⁻1(6/21)d)sin⁻1(4/21)Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for UPSC 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The angle between the line (x - 2)/2 = (y + 1)/-1 = (z - 3)/2 and the plane 3x + 6y - 2z + 5 = 0 isa)cos⁻1(4/21)b)sin⁻1-(4/21)c)sin⁻1(6/21)d)sin⁻1(4/21)Correct answer is option 'D'. Can you explain this answer?.
The angle between the line (x - 2)/2 = (y + 1)/-1 = (z - 3)/2 and the plane 3x + 6y - 2z + 5 = 0 isa)cos⁻1(4/21)b)sin⁻1-(4/21)c)sin⁻1(6/21)d)sin⁻1(4/21)Correct answer is option 'D'. Can you explain this answer? for UPSC 2025 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about The angle between the line (x - 2)/2 = (y + 1)/-1 = (z - 3)/2 and the plane 3x + 6y - 2z + 5 = 0 isa)cos⁻1(4/21)b)sin⁻1-(4/21)c)sin⁻1(6/21)d)sin⁻1(4/21)Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for UPSC 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The angle between the line (x - 2)/2 = (y + 1)/-1 = (z - 3)/2 and the plane 3x + 6y - 2z + 5 = 0 isa)cos⁻1(4/21)b)sin⁻1-(4/21)c)sin⁻1(6/21)d)sin⁻1(4/21)Correct answer is option 'D'. Can you explain this answer?.
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