9 maths vector

7/23/2019 9 Maths Vector

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MASTER CARD

VECTOR ALGEBRA

CONCEPT :

1.

Types of Vector :

2.

Vector Algebra

3.

Section Formula

4.

Product of Vectors

Q.1. Find unit vector which is perpendicular to each of the vectors a + b and a - b where a = i - j

and b = k - I .

Q.2. Find the position vector of a point R which divides the line joining the points P(i + 2j - k) and

Q(-I + j + k) in the ratio of 2 : 1.

i) Internally ii) Externally

Q.3. If a . b = a . c , a x b = a x c and a ≠ 0 the prove that b = c .

Q.4. Find the area of the triangle whose vertices are (1,1,1) , (2,1,3) and (0,0,1) using vectors.

Q.5. Show that the points A(2i – j + k), B(I -3j – 5k), C(3i -4j -4k) are the vertices of the right triangle.

Q.6. If a = I + j +7k and b= 5i - j + λk , find the value of λ so that (a + b) and (a - b) are

perpendicular vectors.

Q.7. Find the projection of the vectors a = 2i + 3j + 2k on the vector b = I + 2j + 2k.

Q.8. Three vectors a , b and c satisfy the condition a+ b +c =0. Evaluate the quantity μ = a . b + b .

c + c . a if |a| = 1 , |b| = 4 and |c| = 2.

Q.9. If I + j + k , 2i + j + k , 3i + 2j - 3k and I – 6j - 6k are the position vectors of the points A,B,C

and D respectively . Then find the angle between AB and CD . Deduce that AB and CD are

colinear.

Q.10. Find a vector of magnitude 5 units and parallel to resultant of the vectors a = 2i + 3j - k and

b = I – 2j + k .

NOTE : unit vectors are written without cap symbol.

BY G. K. Choudhary

PGT (MATHS)

KV Kankarbagh(SS)



SOLUTIONS

1.

Find a + b = -j + k

And a – b = 2i – j – k

You know that unit vector is perpendicular to both a + b and a – b

N = (a + b ) x (a – b)

=| (a + b ) x (a – b)|

Now (a + b ) x (a – b) = 1 12 1 1

= 2i + 2j + 2k

| (a + b ) x (a – b)| = √(22+22+22) = 2√3

n=++

√ 3

=

√ 3 i +

√ 3 j +

√ 3 k

2.

Let r divides PQ internally in the ratio 2:1. The position vector R

(i)

R = − + + + + −

+ =

− +4 + 3

(ii)

Let R divides PQ externally

R = − + + – ( + – )

−

= -3i + 3k

3.

Given that a . b = a . c

Or a.b=a.c

Or a.b-a.c = 0Or a(b-c) =0 a.b

Therefor b=c as a=≠0 or a perpendicular (b-c) (I)

Again a×b =0

Or a×b – c =0

Therefor b=c

Or a parallel b-c (II)

From (I) and (II) b=c

4.

Area of triangle = ½(AB×AC)

Given points are A(1,1,1), B(2,1,3) and C(0,0,1)

Therefore OA= I+J+K , OB= 2I+J+3K, OC=1.KTherefor AB=I+2K AC=I-J

Now AB×AC= (2I-2J-K)

AB×AC =3

Therefore Area of a traingle = 3/2 square unit.

5.

Here AB = -I – 2J – 6K

BC = 2I –J + K



CA =-I + 3J +5K

Therefore| AB|2 = 41 = 6+ 35 =| BC |2 + | CA |2

Hence points are vertices of a right triangle

6.

Here a + b = 6i -2j + (7 +λ) k

a-b = -4i + (7- λ) k

(a+b) is perpendicular to (a-b)

Therefore (a+b). (a-b) =0 this gives λ= ±5

7.

The projection of the vector a=2i+3j+2k on the vector b=i+2j+k=1 by |B|. (a.b)=

2.1+3.2+2.1/√(1)2 +(2)2+(1)2= 5/3√6

8.

Since a+b+c =0 we have

a.(a+b+c) =0

or a.a + a.b + a.c = 0

or a.b + a.c = - |a|2 = -1 (I)

again b.(a+b+c) = 0

or a.b + b.c = - |b|2 = -16 (ii)

similarly a.c + b.c = -4 (iii)

adding (i), (ii) and (iii) we have

2(a.b + b.c + a.c ) = -21

And therefore 2μ = -21 i.e μ =-21/2

9.

AB = position vector B – position vector A

=2+4j-k

Therefore |AB|= 3√2

Similarly CD= -2I -8J +2K AND |CD| =6√2.

Thus cosø = AB.CD/|AB| |CD| = -36/36 =-1

Since 0≤Q ≤λ . it follows that Q = hence AB and CD are collinear.

10.

Resultant vector a and b = a + b = 3i + j +0k any vector parallel to the vector 32+J+0K is of the

form λ.(32+J+0K)

As per question |λ.(32+J+0K)| = 5

i.e √(3λ)2 + (λ)2 + (0)2 = 5 i.e 10λ2 = 25 or λ = ± √5/2

hence required vector is either √5/2 (32+J) or - √5/2(32+J)

9 maths vector

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