Class 12 Maths - Vectors by ednexa.com

1. Vectors
1.
ABCD is a parallelogram. P, Q are the midpoints of the sides AB and CD
respectively. Show that DP and BQ trisect AC and trisected by AC.
Sol:
Let DP and BQ intersect AC at M and N respectively.
Let a , b , c, d , p, q , m , n be the position vectors of the points A, B, C, D, P, Q, M, N
respectively with respect to some origin O.
□ ABCD is a parallelogram.
 AB  DC




bacd
...(i)
P is the midpoint of AB


a b
 p
2



 2p  a  b
...(ii)
Q is the midpoint of CD


cd
q 
2



 2q  c  d
...(iii)

For eliminating b from (i) and (ii), subtract (i) from (ii)
b  a  2p
b  a  c d
 
 
_________________
2 a  2p  c  d
 2p  d  2a  c

2p  d 2a  c

2 1
2 1
This means that M divides DP internally in the ratio 2 : 1 and M divides CA
internally in the ratio 2 : 1.
For eliminating d from (i) and (iii), add (i) and (iii),
…(iv)

2. c  d  2q
c d b a
____________
2c  2q  b  a
 2q  b  2c  a

2q  b 2c  a

2 1
2 1
This means that N divides BQ internally in the ratio 2: 1 and N divides AC internally
in the ratio 2 : 1.
From (iv) and (v), if follows that DP and BQ trisect AC and are trisected by AC.
2.
If in a tetrahedron, the two pairs of opposite edges are perpendicular, then show
that the edges in the third pair is also perpendicular.
Sol:
Let O-ABC be a tetrahedron. Then (OA, BC), (OB, CA) and (OC, AB) are the pair of
opposite edges.
Take O as the origin of reference and let a, b and c be the position vectors of the
vertices A, B and C respectively. Then
OA  a, OB  b, OC  c,
AB  b  a, BC  c  b and CA  a  c.
Now, suppose the pair (OA, BC) and (OB, CA) are perpendicular to each other.
Then

3. OA. BC  0, i.e., a. (c  b)  0
 a. c  a. b  0
...(i)
and OB. CA  0, i.e., b. (a  c)  0
 b. a  b. c  0
 a. b  b. c  0
Adding (i) and (2), we get,
...(ii)
a. c  b. c  0
 c. b  c. a  0
i.e., c. (b  a)  0
 OC. AB  0
∴ the third pair (OC, AB) is perpendicular.
3.
Show that the line segment joining the midpoints of two sides of a triangle is
parallel to the third side and half the length of that of the third side.
Sol:
Let ABC be a triangle and M and N be the midpoints of the AB and AC.
Then we have to show MN parallel to BC and
1
(MN)  . (BC).
2
Let a, b, c, m and n be the position
vectors of A, B, C, M and N respectively. Since M and N are the midpoints of AB and
AC respectively,

4. ab
a c
and n 
2
2
 MN  n  m
m
ac ab



 2   2 
1
 (a  c  a  b)
2
1
 (c  b)
2
1
 BC
2
Thus MN is non –zero scalar multiple of BC .
 MN is parallel to BC.
seg MN is parallel to seg BC.
1
Also, | MN | 
| BC |
2
1
 (MN)  . (BC).
2
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