1.if \vec{a} \times \vec{b} = \vec{c} \times \vec{d} and\vec{a} \times \vec{c} = \vec{b} \times \vec{d} then prove that (\vec{a} -\vec{d})\left| \right| (\vec{b}-\vec{c}).
2.A line makes \alpha ,\beta ,\gamma and \delta with the diagnols of a cube.Prove that
cos^2\alpha + cos^2\beta +cos^2\gamma +cos^2\delta = 4/3
3.If A,B,C and D are four points in space, prove that
ar(ABC) = 1/4 \left|\vec{AB}\times \vec{CD} + \vec{BC}\times \vec{AD}+\vec{CA}\times \vec{BD}\right|
4. What is vectorial area?
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4 Answers
23for 1st
take cross product of (a - d) and ( b-c )
if it turns out to be zero, then the 2 vectors are parallel
12) Consider a unit length cube.
dcs of OP = (1/√3, 1/√3, 1/√3)
dcs of AR = (-1/√3, 1/√3, 1/√3)
dcs of BS = (1/√3, -1/√3, 1/√3)
dcs of CQ = (1/√3, 1/√3, -1/√3)
if l,m,n be the dcs of the required line,
cos2α + cos2β + cos2γ + cos2δ = 13{ (l+m+n)2 + (l-m+n)2 + (l+m-n)2)}
after solving
cos2α + cos2β + cos2γ + cos2δ = 43
6In geometry, for a finite planar surface of scalar area S, the vector area
is defined as a vector whose magnitude is S and whose direction is perpendicular to the plane, as determined by the right hand rule on the rim (moving one's right hand counterclockwise around the rim, when the palm of the hand is "touching" the surface, and the straight thumb indicates the direction).
For an orientable surface S composed of a set Si of flat facet areas, the vector area of the surface is given by
where is the unit normal vector to the area Si.
For bounded, oriented curved surfaces that are sufficiently well-behaved, we can still define vector area. First, we split the surface into infinitesimal elements, each of which is effectively flat. For each infinitesimal element of area, we have an area vector, also infinitesimal.
where is the local unit vector perpendicular to dS. Integrating gives the vector area for the surface.
For a curved or faceted surface, the vector area is smaller in magnitude than the area. As an extreme example, a closed surface can possess arbitrarily large area, but its vector area is necessarily zero.[1] Surfaces that share a boundary may have very different areas, but they must have the same vector area---the vector area is entirely determined by the boundary. These are consequences of Stokes theorem.
The concept of an area vector simplifies the equation for determining the flux through the surface. Consider a planar surface in a uniform field. The flux can be written as the dot product of the field and area vector. This is much simpler than multiplying the field strength by the surface area and the cosine of the angle between the field and the surface normal.
