Solved Subjective Question on Vector Set 4

Example.1

If , and are vectors such that = , prove that

= 0

Solution

We have

&⇒

=
&⇒

= =

= 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

&⇒

&⇒ 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

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

&⇒

Comparing we get,

Putting these values in (1) we get

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

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

, .

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

&⇒

&⇒ cosq =

&⇒ tanq = . Hence the result.

Alternate:

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) = tan²qsin²a

&⇒ q = tan^–1

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

Consider

Similarly, ,

we have,
&⇒

Similarly m = and g =