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2.6 If H = 4a, – 3a, + 5a, at (1, 7/2, 0) the component of H parallel to surface p = 1 is (а) 4а, (d) -3a, + 5a, (e) 5a + 3a, (b) 5a, (c) - 3a

2.6 If H = 4a, – 3a, + 5a, at (1, 7/2, 0) the component of H parallel to surface p = 1 is (а) 4а, (d) -3a, + 5a, (e) 5a + 3a, (b) 5a, (c) - 3a

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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First pic question Second pic answer The problem is i did not understand stand the question so if you can clear it by sketch or explain for me further
2.6 If H = 4a, – 3a, + 5a, at (1, 7/2,0) the component of H parallel to surface p 1 is (а) — За, + 5а, (e) 5a, + 3a, (a) 4a (b) 5a, (c) -3a
Transcribed Image Text:2.6 If H = 4a, – 3a, + 5a, at (1, 7/2,0) the component of H parallel to surface p 1 is (а) — За, + 5а, (e) 5a, + 3a, (a) 4a (b) 5a, (c) -3a
Now, H = 4ap — За, + 5а,, In cartesian form: x = p cos o, y = p sin ø, z = z At p = 1, o 2, z = 0, we have x = 1 × cos ( 5 0, y = 1 × sin( (3)=1, z = z = 0 Therefore, P = 1 (along x and z-direction) means parallel to y i.e. given below, we have ap + az = cos pa, – sin øa, + az = COS ap sin (5) as + az (0)а, — (1)а, + а, - aø + az So, the component parallel to p = 1 is – 3a4 - Thus, the component is -3a6 + 5a,. Hence, the correct option will be (d) |-3ag + 5a, i.e.
Transcribed Image Text:Now, H = 4ap — За, + 5а,, In cartesian form: x = p cos o, y = p sin ø, z = z At p = 1, o 2, z = 0, we have x = 1 × cos ( 5 0, y = 1 × sin( (3)=1, z = z = 0 Therefore, P = 1 (along x and z-direction) means parallel to y i.e. given below, we have ap + az = cos pa, – sin øa, + az = COS ap sin (5) as + az (0)а, — (1)а, + а, - aø + az So, the component parallel to p = 1 is – 3a4 - Thus, the component is -3a6 + 5a,. Hence, the correct option will be (d) |-3ag + 5a, i.e.
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