Hints and answers

VECTOR ALGEBRA
of vector plots for potential differences and currents (they could all be on the
same plot if suitable scales were chosen), determine all unknown currents and
potential differences and find values for the inductance of L and the resistance
of R2 .
[ Scales of 1 cm = 0.1 V for potential differences and 1 cm = 1 mA for currents
are convenient. ]
7.11 Hints and answers
7.1
7.3
7.5
7.7
7.9
7.11
7.13
7.15
7.17
7.19
7.21
7.23
7.25
7.27
(c), (d) and (e).
(a) A sphere of radius k centred on the origin; (b) a plane with its normal in the
direction of u and at a distance l from the origin; (c) a cone with its axis parallel
to u and of semiangle cos−1 m; (d) a circular cylinder of radius n with its axis
parallel tou.
√
(a) cos−1 2/3; (b) z − x = 2; (c) 1/ 2; (d) 13 21 (c × g) · j = 13 .
Show that q × r is parallel to p; volume = 13 12 (q × r) · p = 53 .
Note that (a × b) · (c × d) = d · [(a × b) × c] and use the result for a triple vector
product to expand the expression in square brackets.
Show that the position vectors of the points are linearly dependent; r = a + λb
where a = i + k and b = −j + k.
Show that p must have the direction n̂ × m̂ and write a as xn̂ + y m̂. By obtaining a
pair of simultaneous equations for x and y, prove that x = (λ−µn̂· m̂)/[1−(n̂· m̂)2 ]
and that y = (µ − λn̂ · m̂)/[1 − (n̂ · m̂)2 ].
(a) Note that |a − g|2 = R 2 = |0 − g|2 , leading to a · a = 2a · g.
(b) Make p the new origin and solve the three simultaneous linear equations to
obtain √
λ = 5/18, µ = 10/18, ν = −3/18, giving g = 2i − k and a sphere of
radius 5 centred on (5, 1, −3).
(a) Find two points on both planes, say (0, 0, 0) and (1, −2, 1), and hence determine
the direction cosines of the line of intersection; (b) ( 23 )1/2 .
For (c) and (d), treat (c × a) × (a × b) as a triple vector product with c × a as one
of the three vectors.
(b) b = a−1 (−i + j + k), c = a−1 (i − j + k), d = a−1 (i + j − k); (c) a/2 for direction
√
k; successive planes through (0, 0, 0) and (a/2, 0, a/2) give a spacing of a/ 8 for
direction
√ i + j; successive planes through (−a/2, 0, 0) and (a/2, 0, 0) give a spacing
of a/ 3 for direction i + j + k.
Note that a2 − (n̂ · a)2 = a2⊥ .
p = −2a + 3b, q = 32 a − 32 b + 52 c and r = 2a − b − c. Remember that a · a = b · b =
c · c = 2 and a · b = a · c = b · c = 1.
With currents in mA and potential differences in volts:
I1 = (7.76, −23.2◦ ), I2 = (14.36, −50.8◦ ), I3 = (8.30, 103.4◦ );
V1 = (0.388, −23.2◦ ), V2 = (0.287, −50.8◦ ), V4 = (0.596, 39.2◦ );
L = 33 mH, R2 = 20 Ω.
240