Syllabus Vector Differentiation - Velocity and Acceleration - Gradient ...
Syllabus Vector Differentiation - Velocity and Acceleration - Gradient ...
Syllabus Vector Differentiation - Velocity and Acceleration - Gradient ...
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Engineering Mathematics - II - <strong>Vector</strong> Calculus - 2007<br />
: 4x<br />
2<br />
y z<br />
3<br />
2 4 0<br />
...(2)<br />
Given (1, -1, 2) is a point on (1) <strong>and</strong> (2)<br />
1 2<br />
at (1, 1,2)<br />
i.e., a + 2b - (a + 2) -4 + 8 - 4<br />
2b - 2 = 0<br />
b 1<br />
1<br />
<br />
(2ax (a 2)) i<br />
bz j<br />
by k<br />
<br />
1 at (1, 1,2) (a 2) i 2 j<br />
by k<br />
2<br />
<br />
8xy i 4x<br />
2<br />
j<br />
3z<br />
2<br />
k<br />
<br />
2<br />
at (1, 1,2) 8<br />
i 4 j<br />
12 k<br />
since 1 <strong>and</strong> 2<br />
are orthogonal<br />
1. 2<br />
0<br />
-8(a - 2) - 8 + 12 = 0<br />
i.e., a = 5/2<br />
a 5 / 2<br />
<strong>and</strong> b 1<br />
37. Find the constant a, b, c so that the vector<br />
<br />
<br />
<br />
<br />
f (x 2y az) i (bx 3y z) j<br />
(4x cy 2z)k<br />
is irrotational.<br />
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