vectors such that = , prove that
= 0 (Given that =
Given three non-zero vectors such
that . Show that either the vectors are parallel or is normal to the plane containing .
Given equation can be written as
&⇒ Either are parallel (i.e. )
or both are zero, which means is perpendicular to as well as, which is as saying is perpendicular to the plane
Show that the solution of the equation k where k is a non-zero scalar and
are two non–collinear vectors,
is of the form
The vectors are
non-collinear and therefore can be
expressed as ....(1)
where x, y, z are some scalars
Substituting this in the given equation
Comparing we get,
Putting these values in (1) we get
Three poles of height x, x + y and x + z are
posted at the vertices a, b and c of a triangular park of sides a, b and c
respectively. A plane sheet is mounted at the tops of the poles. If the plane
of the sheet is inclined at an angle q to the horizontal plane,
prove that q = . (By vector
Let A¢, B¢, C¢ be
the tops of the poles at a, b and c respectively. Through A¢,
draw a triangle A¢B1C1 congruent to DABC
and parallel to the horizontal plane of the park. Take A¢B1
x-axis and a line perpendicular to
as the y-axis (in the plane of DA¢B1C1),
and a line through A¢ and perpendicular to the plane A¢B1C1
as the z-axis. If are the unit vectors
along these axis, then
the planes A¢B¢C¢ is
inclined at an angle q to the plane A¢B1C1,
angle between the normals to the planes is p - q.
obviously the unit vector normal to the plane A¢B1C1
is and the normal vector to A¢B¢C¢ is
&⇒ cosq =
&⇒ tanq = .
Hence the result.
Let a¢, b¢, c¢ be
the tops of the poles posted at a, b, and c respectively
&⇒ aa¢ =
x, bb¢ =
x + y. Cc¢ = x + z extend ca and c¢a¢ to
meet at h1. From b, draw bh2 parallel to cah1
and from b¢, b¢h2 parallel to c¢a¢ h1.
= q = ∠b¢h2b.
Draw al ^ bh2
= h1a. .
in D a¢h1a, x = ah1tanq ...(1).
in D h1cc¢, x
+ z = (ah1 + b) tanq …(2)
in D h2bb¢, x
+ y = (h2b)tanq = (h2l + ccosa) tanq
&⇒ x + y = (ah1 + ccosa) tanq ...(3) .
from (1), (2) and (3), we get z= b tanq
and y = ccosa tanq. .
hence = cos2atan2q +
tan2q - 2cos2atan2q
= tan2q (1 – cos2a) =
&⇒ q =
OABC is a tetrahedron where O is the origin and A, B and C have respective
position vectors as , then prove that the
circumcentre of the tetrahedron is
If is the circumcentre, then
Similarly m = and g =