If , and are vectors such that = , prove that

= 0

We have

&⇒

=

&⇒

= =

= 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 containing .

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 method)

Let A¢, B¢, C¢ be
the tops of the poles at a, b and c respectively. Through A¢,
draw a triangle A¢B_{1}C_{1} congruent to DABC
and parallel to the horizontal plane of the park. Take A¢B_{1}
as the

x-axis and a line perpendicular to

it
as the y-axis (in the plane of DA¢B_{1}C_{1}),
and a line through A¢ and perpendicular to the plane A¢B_{1}C_{1}
as the z-axis. If are the unit vectors
along these axis, then

, .

since
the planes A¢B¢C¢ is
inclined at an angle q to the plane A¢B_{1}C_{1},
angle between the normals to the planes is p - q.

obviously the unit vector normal to the plane A¢B_{1}C_{1}
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 h_{1}. From b, draw bh_{2} parallel to cah_{1}
and from b¢, b¢h_{2} parallel to c¢a¢ h_{1}.

Clearly

∠a¢h_{1}a
= q = ∠b¢h_{2}b.
Draw al ^ bh_{2
}

&⇒ h_{2}l
= h_{1}a. .

in D a¢h_{1}a, x = ah_{1}tanq ...(1).

in D h_{1}cc¢, x
+ z = (ah_{1} + b) tanq …(2)

in D h_{2}bb¢, x
+ y = (h_{2}b)tanq = (h_{2}l + ccosa) tanq

&⇒ x + y = (ah_{1} + ccosa) tanq ...(3) .

from (1), (2) and (3), we get z= b tanq and y = ccosa tanq. .

hence = cos^{2}atan^{2}q +
tan^{2}q - 2cos^{2}atan^{2}q

= tan^{2}q (1 – cos^{2}a) =
tan^{2 }qsin^{2}a

&⇒ q =
tan^{–1}

If 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

Consider

\

Similarly, ,

we have,

&⇒

\

Similarly m = and g =

\