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Solved Subjective Question on Vector Set 2

Posted on - 04-01-2017

Math

IIT JEE

Example.1

In a quadrilateral PQRS , , , M is the midpoint of and X is a point on SM such that SX = SM. Then prove that P, X and R are collinear.

Solution     


&⇒

also


&⇒
, hence P, X and R are collinear.

Example.2

Unit vectors are perpendicular to each other and the unit vector is inclined at an angle q to both . If are real, prove that

Solution

Example.3

Given that are the position vectors of points P and Q respectively. Find the equation for the plane passing through Q and perpendicular to the line PQ. What is the distance from the point (-1, 1, 1) to the plane?

Solution

= = -

Equation of plane passing through Q and perpendicular to PQ is


&⇒

&⇒


&⇒
…(1)

Let

Hence distance from to the plane (1) is

= == 5 units

Example.4

Find the vector which makes equal angles with the vectors and is perpendicular to the vector = (1,-1,2) with and the angle between and the unit vector is obtuse.

Solution

….(1)

is perpendicular to
&⇒


&⇒
x–y+2z =0 …(2)

Also it is given that , (makes equal angles with )

Since

We get
&⇒
xy – 2yz +3xz = 2xz +3xy– yz


&⇒
2xy+yz– zx = 0 …
.(3).

As the angle between is given to be obtuse.

….(4)

from (2) & (3) we get, 2y (y – 2z) + yz – z(y – 2z) = 0


&⇒
(y – z)2 = 0,y= z and x = –z

From (1) get z2 + z2 + z2 = 12
&⇒
z2 = 4

\ z = –2 (z = y < 0)

Hence

Example.5

Show that midpoints of the three diagonals of a complete quadrilateral are collinear . .

Solution

A complete quadrilateral is a figure made by four straight lines no three of which are concurrent. .

Let ABCD be a quadrilateral and let P, Q and R be the respective midpoints of the diagonals BD, CA and EF. .

Let be the position vectors of A, B, C, D, E, F, P, Q & R respectively.

then , and

since F lies on the two lines BC and AD and E lies on the two lines BA and CD respectively, we have

. . . (1)

. . . (2)

. . . (3)

. . . (4)

where s, t , u and v are scalars

adding (1) , (2), (3) and (4) , we get

(2 – s – u)() + (1 – t + u)() - 2() = 0

setting 1 – t + u = 1 + s – u and 2 – s – v =t + v

so that (2 – s – u) + (1 – t + u) –2 = 0

This shows that the points P, Q and R are collinear since the sum (2 – s - u) + ( 1 – t + u) – 2= 0 . .

 
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