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Determine the direction cosines of the normal to the plane and the distance from the origin. Plane z = 2
- a)0, 1, 0; 2
- b)1, 0, 0; 3
- c)1, 0, 1; 3
- d)0, 0, 1; 2
Correct answer is option 'D'. Can you explain this answer?
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Determine the direction cosines of the normal to the plane and the dis...
We have z = 2 . , it can be written as : 0x+0y+1z = 2. Compare it with lx+my+nz = d , we get ; l = 0 , m = 0 , n = 1 and d = 2 . therefore , D.C.’s of normal to the plane are 0 , 0 , 1 and distance from the origin = 2.
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Determine the direction cosines of the normal to the plane and the dis...
Understanding the Plane Equation
The equation of the plane given is z = 2. This indicates that the plane is parallel to the x-y plane and is located at a height of 2 units above it.
Normal Vector to the Plane
- The normal vector to a plane defined by the equation Ax + By + Cz + D = 0 can be derived from the coefficients A, B, and C.
- For the plane z = 2, we can rewrite it as 0x + 0y + 1z - 2 = 0.
- Here, A = 0, B = 0, C = 1.
Direction Cosines
- The direction cosines of a normal vector (n_x, n_y, n_z) are given by the ratios of the components of the normal vector to its magnitude.
- For our normal vector (0, 0, 1), the magnitude is √(0² + 0² + 1²) = 1.
- Therefore, the direction cosines are:
- n_x = 0/1 = 0
- n_y = 0/1 = 0
- n_z = 1/1 = 1
Distance from the Origin
- The distance d from the origin (0, 0, 0) to the plane z = 2 can be calculated as the absolute value of the constant term divided by the magnitude of the normal vector.
- Here, d = |D| / √(A² + B² + C²) = |2| / 1 = 2.
Conclusion
- The direction cosines of the normal to the plane z = 2 are (0, 0, 1), and the distance from the origin to the plane is 2.
- Hence, the correct answer is option 'D': (0, 0, 1).
The equation of the plane given is z = 2. This indicates that the plane is parallel to the x-y plane and is located at a height of 2 units above it.
Normal Vector to the Plane
- The normal vector to a plane defined by the equation Ax + By + Cz + D = 0 can be derived from the coefficients A, B, and C.
- For the plane z = 2, we can rewrite it as 0x + 0y + 1z - 2 = 0.
- Here, A = 0, B = 0, C = 1.
Direction Cosines
- The direction cosines of a normal vector (n_x, n_y, n_z) are given by the ratios of the components of the normal vector to its magnitude.
- For our normal vector (0, 0, 1), the magnitude is √(0² + 0² + 1²) = 1.
- Therefore, the direction cosines are:
- n_x = 0/1 = 0
- n_y = 0/1 = 0
- n_z = 1/1 = 1
Distance from the Origin
- The distance d from the origin (0, 0, 0) to the plane z = 2 can be calculated as the absolute value of the constant term divided by the magnitude of the normal vector.
- Here, d = |D| / √(A² + B² + C²) = |2| / 1 = 2.
Conclusion
- The direction cosines of the normal to the plane z = 2 are (0, 0, 1), and the distance from the origin to the plane is 2.
- Hence, the correct answer is option 'D': (0, 0, 1).
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Determine the direction cosines of the normal to the plane and the distance from the origin. Plane z = 2a)0, 1, 0; 2b)1, 0, 0; 3c)1, 0, 1; 3d)0, 0, 1; 2Correct answer is option 'D'. Can you explain this answer?
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Determine the direction cosines of the normal to the plane and the distance from the origin. Plane z = 2a)0, 1, 0; 2b)1, 0, 0; 3c)1, 0, 1; 3d)0, 0, 1; 2Correct 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 Determine the direction cosines of the normal to the plane and the distance from the origin. Plane z = 2a)0, 1, 0; 2b)1, 0, 0; 3c)1, 0, 1; 3d)0, 0, 1; 2Correct 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 Determine the direction cosines of the normal to the plane and the distance from the origin. Plane z = 2a)0, 1, 0; 2b)1, 0, 0; 3c)1, 0, 1; 3d)0, 0, 1; 2Correct answer is option 'D'. Can you explain this answer?.
Determine the direction cosines of the normal to the plane and the distance from the origin. Plane z = 2a)0, 1, 0; 2b)1, 0, 0; 3c)1, 0, 1; 3d)0, 0, 1; 2Correct 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 Determine the direction cosines of the normal to the plane and the distance from the origin. Plane z = 2a)0, 1, 0; 2b)1, 0, 0; 3c)1, 0, 1; 3d)0, 0, 1; 2Correct 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 Determine the direction cosines of the normal to the plane and the distance from the origin. Plane z = 2a)0, 1, 0; 2b)1, 0, 0; 3c)1, 0, 1; 3d)0, 0, 1; 2Correct answer is option 'D'. Can you explain this answer?.
Solutions for Determine the direction cosines of the normal to the plane and the distance from the origin. Plane z = 2a)0, 1, 0; 2b)1, 0, 0; 3c)1, 0, 1; 3d)0, 0, 1; 2Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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