Cross Product
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Refer to figure (2-E1). Find (a) the magnitude, (b) x and y components and (c) the angle with the X - axis of the resultant of →OA, →BC and →DE.
- 3.6×10−5 N
- 1.8×10−5 N
- 1.3×10−5 N
- 6.5×10−5 N
- √3W
- W
- √32W
- 2W
- →L=→P×→r
- →L×→P=0
- |→L×→P|=LP
- None of these
- −^i+^j−3^k
- −^i+^j+3^k
- ^i+^j+3^k
- ^i−^j−5^k
- 0 kg-m2/sec
- 60 kg-m2/sec
- 120 kg-m2/sec
- 20 kg-m2/sec
- −0.4 ^i kg-m2/sec
- −0.4 ^j kg-m2/sec
- −0.4 ^k kg-m2/sec
- 0.4 ^k kg-m2/sec
- zero
- 4.8×10−3 J
- 3.5×10−3 J
- 7.8×10−3 J
- 6.2×10−3 J
- 10 N-m, +20 J
- 20 N-m, −10 J
- 0 N-m, −10 J
- 0 N-m, +10 J
Vectors →C and →D have magnitudes of 3 and 4 units respectively. What is the angle between the directions of →C and →D if the magnitude of vector product →C×→D is x and y respectively?
(x) 0 (y) 12 units
(i) 0 (ii)π2 (iii)π3 (iv) π4
(x) - (iv); (y) - (iii)
(x) - (i); (y) - (ii)
(x) - (ii); (y) - (i)
(x) - (i); (y) - (iv)
If and are the two sides of a triangle, then the angle between them which gives the maximum area of the triangle is
- mu3 sin2 θ cos θ3g
- 3 mu3 sin2 θ cos θ2g
- mu3 sin2 θ cos θ2g
- None of these
- √2
- 2√2
- √32
- √23
and the angle between and is . The area of the triangle constructed on the vectors and is?
- 4√3×10−3 N
- 2√3×10−3 N
- 2×10−3 N
- 4×10−3 N
- μ0iaπr
- μ0ia2πr
- μ0ia2πr2
- μ0iaπr2
- √2×10−4 J
- (1−√2)×10−4 J
- 5×10−4 J
- 1√2×10−4 J
[Your answer must be an integer]
- θ = 30∘
- θ = 45∘
- θ = 60∘
- θ = 15∘
Calculate the area of the parallelogram when adjacent sides are given by the vectors →A = ^i + 2^j + 3^k and →B = 2^i − 3^j + ^k
195 sq unit
sq unit
0
sq unit
- The thermal power dissipated in the resistor is equal to the rate of work done by an external person pulling the rod
- If applied external force is doubled, then a part of the external power increases the velocity of the rod
- If resistance R is doubled, then power required to maintain the constant velocity v0 becomes half.
- Lenz's law is not satisfied if the rod is accelerated by an external force.
- k2a2
- zero
- −k4a2
- −3k2a2
- tan−1(1)
- tan−1(12)
- tan−1(14)
- tan−1(4)
- −0.4 ^i kg-m2/sec
- −0.4 ^k kg-m2/sec
- 0.4 ^k kg-m2/sec
- −0.4 ^j kg-m2/sec
Show that →a.(→b×→c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors, →a, →b and →c.
- 33 sq. unit
- 21 sq. unit
- 0 sq. unit
- 332 sq. unit