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.
hence P, X and R are collinear.
Unit vectors are perpendicular to each other
and the unit vector is inclined at an angle
q to both . If are
real, prove that
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?
= = -
Equation of plane
passing through Q and perpendicular to PQ is
Hence distance from to the plane (1) is
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.
is perpendicular to
&⇒ x–y+2z =0 …(2)
Also it is given
that , (makes
equal angles with )
&⇒ xy – 2yz +3xz = 2xz +3xy– yz
&⇒ 2xy+yz– zx = 0 ….(3).
As the angle between
is given to be obtuse.
from (2) & (3)
we get, 2y (y – 2z) + yz – z(y – 2z) = 0
&⇒ (y – z)2 = 0,y= z and x
From (1) get z2
+ z2 + z2 = 12
z2 = 4
\ z = –2 (z = y < 0)
Show that midpoints of the
three diagonals of a complete quadrilateral are collinear . .
A complete quadrilateral
is a figure made by four straight lines no three of which are
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)
. . .
. . . (4)
where s, t , u and v
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 . .