Example.1
If
,
and
are
vectors such that
=
, prove that
= 0
Solution
We
have ![](http://www.quizsolver.com/radix/dth/notif/VECTOR_4_SUB_files/image007.gif)
&⇒![](http://www.quizsolver.com/radix/dth/notif/VECTOR_4_SUB_files/image008.gif)
![](http://www.quizsolver.com/radix/dth/notif/VECTOR_4_SUB_files/image009.gif)
=![](http://www.quizsolver.com/radix/dth/notif/VECTOR_4_SUB_files/image010.gif)
&⇒ ![](http://www.quizsolver.com/radix/dth/notif/VECTOR_4_SUB_files/image011.gif)
=
= ![](http://www.quizsolver.com/radix/dth/notif/VECTOR_4_SUB_files/image013.gif)
= 0 (Given that
=
)
Example.2
Given three non-zero vectors
such
that
. Show that either the vectors
are parallel or
is normal to the plane containing
.
Solution
Given equation can be written as ![](http://www.quizsolver.com/radix/dth/notif/VECTOR_4_SUB_files/image020.gif)
&⇒ ![](http://www.quizsolver.com/radix/dth/notif/VECTOR_4_SUB_files/image021.gif)
&⇒ ![](http://www.quizsolver.com/radix/dth/notif/VECTOR_4_SUB_files/image022.gif)
&⇒ 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
.
Example.3
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 ![](http://www.quizsolver.com/radix/dth/notif/VECTOR_4_SUB_files/image032.gif)
Solution
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 ![](http://www.quizsolver.com/radix/dth/notif/VECTOR_4_SUB_files/image036.gif)
&⇒
![](http://www.quizsolver.com/radix/dth/notif/VECTOR_4_SUB_files/image037.gif)
&⇒
![](http://www.quizsolver.com/radix/dth/notif/VECTOR_4_SUB_files/image038.gif)
&⇒
![](http://www.quizsolver.com/radix/dth/notif/VECTOR_4_SUB_files/image039.gif)
&⇒
![](http://www.quizsolver.com/radix/dth/notif/VECTOR_4_SUB_files/image040.gif)
Comparing we get, ![](http://www.quizsolver.com/radix/dth/notif/VECTOR_4_SUB_files/image041.gif)
Putting these values in (1) we get
![](http://www.quizsolver.com/radix/dth/notif/VECTOR_4_SUB_files/image042.gif)
Example.4
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)
Solution
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
as the
x-axis and a line perpendicular to ![](http://www.quizsolver.com/radix/dth/notif/VECTOR_4_SUB_files/image044.gif)
it
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
![](http://www.quizsolver.com/radix/dth/notif/VECTOR_4_SUB_files/image047.gif)
,
.
since
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
=![](http://www.quizsolver.com/radix/dth/notif/VECTOR_4_SUB_files/image052.gif)
&⇒
&⇒ cosq = ![](http://www.quizsolver.com/radix/dth/notif/VECTOR_4_SUB_files/image054.gif)
&⇒ tanq =
.
Hence the result.
Alternate:
![](http://www.quizsolver.com/radix/dth/notif/VECTOR_4_SUB_files/image056.gif)
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.
Clearly
∠a¢h1a
= q = ∠b¢h2b.
Draw al ^ bh2
&⇒ h2l
= 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) =
tan2 qsin2a
&⇒ q =
tan–1![](http://www.quizsolver.com/radix/dth/notif/VECTOR_4_SUB_files/image058.gif)
Example.5
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
.
Solution
If
is the circumcentre, then
![](http://www.quizsolver.com/radix/dth/notif/VECTOR_4_SUB_files/image061.gif)
Consider ![](http://www.quizsolver.com/radix/dth/notif/VECTOR_4_SUB_files/image062.gif)
\ ![](http://www.quizsolver.com/radix/dth/notif/VECTOR_4_SUB_files/image063.gif)
Similarly,
, ![](http://www.quizsolver.com/radix/dth/notif/VECTOR_4_SUB_files/image065.gif)
we have, ![](http://www.quizsolver.com/radix/dth/notif/VECTOR_4_SUB_files/image066.gif)
&⇒ ![](http://www.quizsolver.com/radix/dth/notif/VECTOR_4_SUB_files/image067.gif)
\![](http://www.quizsolver.com/radix/dth/notif/VECTOR_4_SUB_files/image068.gif)
Similarly m =
and g = ![](http://www.quizsolver.com/radix/dth/notif/VECTOR_4_SUB_files/image070.gif)
\![](http://www.quizsolver.com/radix/dth/notif/VECTOR_4_SUB_files/image071.gif)