Chapter 23 Algebra of Vectors Exercise Ex. 23.1

Question 1(i)

Represent graphically a dispacement of 40 km, 30° east of north.Solution 1(i)

Question 1(ii)

Represent graphically a displacement of 50 km, south-eastSolution 1(ii)

Here, vector begin mathsize 12px style OP with rightwards arrow on top end style represents the displacement of 50 km, south-east.Question 1(iii)

Represent graphically a displacement of 70 km, 40° north of west.Solution 1(iii)

Here, vector begin mathsize 12px style OP with rightwards arrow on top end style represents the displacement of 70 km, 40° north of west.Question 2

Classify the following measures as scalars and vectors.

(i) 15 kg

(ii) 20 kg weight

(iii) 45°

(iv) 10 metres south-east

(v) 50 m/s2Solution 2

(i) 15 kg is a scalar quantity because it involves only

(ii) 20 kg weight is a vector quantity as it involves both magnitude and direction.

(iii) 45° is a scalar quantity as it involves only magnitude.

(iv) 10 metres south-east is a vector quantity as it involve direction.

(v) 50 m/s2 is a scalar quantity as it involves magnitude of acceleration.Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Chapter 23 Algebra of Vectors Exercise Ex. 23.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Chapter 23 Algebra of Vectors Exercise Ex. 23.3

Question 1

Find the position vector of a point R which divides the line joining the two points P and Q with position vectors

Solution 1

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

T h e space v e r t i c e s space A comma space B comma space C space o f space t r i a n g l e space A B C space h a v e space r e s p e c t i v e l y space p o s i t i o n space v e c t o r s space
a with rightwards arrow on top comma space b with rightwards arrow on top comma space c with rightwards arrow on top space w i t h space r e s p e c t space t o space a space g i v e n space o r i g i n space O. space S h o w space t h a t space t h e space p o i n t space D space w h e r e space t h e space b i s e c t o r space o f space
angle A space m e e t s space B C space h a s space p o s i t i o n space v e c t o r
d with rightwards arrow on top equals fraction numerator beta b with rightwards arrow on top plus gamma c with rightwards arrow on top over denominator beta plus gamma end fraction comma space w h e r e space beta equals open vertical bar c with rightwards arrow on top minus a with rightwards arrow on top close vertical bar equals gamma equals open vertical bar a with rightwards arrow on top minus b with rightwards arrow on top close vertical bar
H e n c e space d e d u c e space t h a t space t h e space i n c e n t r e space I space h a s space p o s i t i o n space v e c t o r space fraction numerator alpha a with rightwards arrow on top plus beta b with rightwards arrow on top plus gamma c with rightwards arrow on top over denominator alpha plus beta plus gamma end fraction comma space w h e r e
alpha equals open vertical bar b with rightwards arrow on top minus c with rightwards arrow on top close vertical bar

Solution 7

L e t space A B C space b e space a space t r i a n g l e.
L e t space t h e space p o s i t i o n space v e c t o r s space o f space A comma space B space a n d space C space w i t h space r e s p e c t space t o space s o m e space o r i g i n comma space O space b e
a with rightwards arrow on top comma space b with rightwards arrow on top space a n d space c with rightwards arrow on top space r e s p e c t i v e l y.
L e t space D space b e space t h e space p o i n t space o n space B C space w h e r e space t h e space b i s e c t o r space o f space angle A space m e e t s.
L e t space d with rightwards arrow on top space p o s i t i o n space v e c t o r space o f space D space w h i c h space d i v i d e s space B C space i n t e r n a l l y space i n space t h e space r a t i o space beta space a n d space gamma comma
w h e r e space beta equals open vertical bar stack A C with rightwards arrow on top close vertical bar space a n d space gamma equals open vertical bar stack A B with rightwards arrow on top close vertical bar
T h u s comma space beta equals open vertical bar c with rightwards arrow on top minus a with rightwards arrow on top close vertical bar space a n d space gamma equals open vertical bar b with rightwards arrow on top minus a with rightwards arrow on top close vertical bar
T h u s comma space b y space s e c t i o n space f o r m u l a comma space t h e space p o s i t i o n space v e c t o r space o f space D space i s space g i v e n space b y
stack O D with rightwards arrow on top equals fraction numerator beta b with rightwards arrow on top plus gamma c with rightwards arrow on top over denominator beta plus gamma end fraction
L e t space alpha equals open vertical bar b with rightwards arrow on top minus c with rightwards arrow on top close vertical bar
I n c e n t r e space i s space t h e space c o n c u r r e n t space p o i n t space o f space a n g l e space b i s e c t o r s.
T h u s comma space I n c e n t r e space d i v i d e s space t h e space l i n e space A D space i n space t h e space r a t i o space alpha : beta plus gamma
T h u s comma space t h e space p o s i t i o n space v e c t o r space o f space i n c e n t r e space i s
e q u a l space t o space fraction numerator alpha a with rightwards arrow on top plus fraction numerator beta b with rightwards arrow on top plus gamma c with rightwards arrow on top over denominator open parentheses beta plus gamma close parentheses end fraction cross times open parentheses beta plus gamma close parentheses over denominator alpha plus beta plus gamma end fraction equals fraction numerator alpha a with rightwards arrow on top plus beta b with rightwards arrow on top plus gamma c with rightwards arrow on top over denominator alpha plus beta plus gamma end fraction

Chapter 23 – Algebra of Vectors Exercise Ex. 23.4

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Prove by vector method that the internal bisectors of the angles of a triangle are concurrent.Solution 6

Chapter 23 – Algebra of Vectors Exercise Ex. 23.5

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Find |AB| in each case.Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

rightwards double arrow b with hat on top equals 1 half i with hat on top plus fraction numerator square root of 3 over denominator 2 end fraction j with hat on top

Question 12

Solution 12

Chapter 23 Algebra of Vectors Exercise Ex. 23.6

Question 1

Solution 1

Question 2

Solution 2

Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

Question 3

Solution 3

Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

F i n d space a space v e c t o r space o f space m a g n i t u d e space o f space 5 space u n i t s space p a r a l l e l space t o space t h e space r e s u l tan t space o f space t h e space v e c t o r s space
a with rightwards arrow on top equals 2 i with hat on top plus 3 j with hat on top minus k with hat on top space a n d space b with rightwards arrow on top equals i with hat on top minus 2 j with hat on top plus k with hat on top

Solution 18

G i v e n space t h a t space
a with rightwards arrow on top equals 2 stack i space with hat on top plus 3 j with hat on top minus k with hat on top
space a n d
b with rightwards arrow on top equals i with hat on top minus 2 j with hat on top plus k with hat on top
T h u s comma space F i n d space a space v e c t o r space o f space m a g n i t u d e space o f space 5 space u n i t s space p a r a l l e l space t o space t h e space r e s u l tan t space o f space t h e space v e c t o r s space
a with rightwards arrow on top plus b with rightwards arrow on top equals 2 i with hat on top plus 3 j with hat on top minus k with hat on top plus space i with hat on top minus 2 j with hat on top plus k with hat on top
rightwards double arrow a with rightwards arrow on top plus b with rightwards arrow on top equals 3 i with hat on top plus j with hat on top
rightwards double arrow open vertical bar a with rightwards arrow on top plus b with rightwards arrow on top close vertical bar equals square root of 9 plus 1 end root equals square root of 10
T h u s comma space t h e space u n i t space v e c t o r space a l o n g space t h e space r e s u l tan t space v e c t o r space a with rightwards arrow on top plus b with rightwards arrow on top space i s space
fraction numerator 3 i with hat on top plus j with hat on top over denominator square root of 10 end fraction
T h e space v e c t o r space o f space m a g n i t u d e space o f space 5 space u n i t s space p a r a l l e l space t o space t h e space r e s u l tan t
v e c t o r equals fraction numerator 3 i with hat on top plus j with hat on top over denominator square root of 10 end fraction cross times 5 equals square root of 5 over 2 end root open parentheses 3 i with hat on top plus j with hat on top close parentheses

Question 19

Solution 19

Chapter 23 Algebra of Vectors Exercise Ex. 23.7

Question 1

Solution 1

Question 2 (i)

Solution 2 (i)

Question 2 (ii)

Solution 2 (ii)

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

U sin g space v e c t o r space s h o w space t h a t space t h e space p o i n t s space A open parentheses minus 2 comma 3 comma 5 close parentheses comma space B open parentheses 7 comma 0 comma minus 1 close parentheses space a n d space C open parentheses minus 3 comma minus 2 comma minus 5 close parentheses space
a n d space D open parentheses 3 comma 4 comma 7 close parentheses space a r e space s u c h space t h a t space A B space a n d space C D space i n t e r s e c t space a t space t h e space p o i n t space P open parentheses 1 comma 2 comma 3 close parentheses

Solution 12

W e space h a v e
stack A P with rightwards arrow on top equals P o s i t i o n space v e c t o r space o f space P minus P o s i t i o n space v e c t o r space o f space A
rightwards double arrow stack A P with rightwards arrow on top equals i with hat on top plus 2 j with hat on top plus 3 k with hat on top minus open parentheses minus 2 i with hat on top plus 3 j with hat on top plus 5 k with hat on top close parentheses equals 3 i with hat on top minus j with hat on top minus 2 k with hat on top
stack P B with rightwards arrow on top equals P o s i t i o n space v e c t o r space o f space B minus P o s i t i o n space v e c t o r space o f space P
rightwards double arrow stack P B with rightwards arrow on top equals 7 i with hat on top minus k with hat on top minus open parentheses i with hat on top plus 2 j with hat on top plus 3 k with hat on top close parentheses equals 6 i with hat on top minus 2 j with hat on top minus 4 k with hat on top
C l e a r l y comma space stack P B with rightwards arrow on top equals 2 stack A P with rightwards arrow on top
s o space v e c t o r s space stack A P with rightwards arrow on top space a n d space stack P B with rightwards arrow on top space a r e space c o l l i n e a r.
B u t space P space i s space a space p o i n t space c o m m o n space t o space stack A P with rightwards arrow on top space a n d space stack P B with rightwards arrow on top. space
H e n c e space P comma space A comma space B space a r e space c o l l i n e a r space p o i n t s.
S i m i l a r l y comma space stack C P with rightwards arrow on top equals i with hat on top plus 2 j with hat on top plus 3 k with hat on top minus open parentheses minus 3 i with hat on top minus 2 j with hat on top minus 5 k with hat on top close parentheses equals 4 i with hat on top plus 4 j with hat on top plus 8 k with hat on top
a n d space stack P D with rightwards arrow on top equals 3 i with hat on top plus 4 j with hat on top plus 7 k with hat on top minus open parentheses i with hat on top plus 2 j with hat on top plus 3 k with hat on top close parentheses equals 2 i with hat on top plus 2 j with hat on top plus 4 k with hat on top
S o space v e c t o r s space stack C P with rightwards arrow on top space a n d space stack P D with rightwards arrow on top space a r e space c o l l i n e a r.
B u t space P space i s space a space c o m m o n space p o i n t space t o space stack C P with rightwards arrow on top space a n d space stack C D with rightwards arrow on top.
H e n c e comma space C comma space P comma space D space a r e space c o l l i n e a r space p o i n t s.
T h u s comma space A comma space B comma space C comma space D space a n d space P space a r e space p o i n t s space s u c h space t h a t space A comma space P comma space B space a n d space C comma space P comma space D space
a r e space t w o space s e t s space o f space c o l l i n e a r space p o i n t s. space H e n c e space A B space a n d space C D space i n t e r s e c t space a t space t h e
p o i n t space P

Question 13

Using vectors, find the value of λ such that the points

 (λ, – 10, 3), (1 -1, 3) and (3, 5, 3) are collinear.Solution 13

Chapter 23 Algebra of Vectors Exercise Ex. 23.8

Question 1

Solution 1

Question 2 (i)

Solution 2 (i)

Question 2 (ii)

Solution 2 (ii)

Question 2 (iii)

Solution 2 (iii)

Question 2 (iv)

Solution 2 (iv)

Question 3 (i)

Solution 3 (i)

Question 3 (ii)

Solution 3 (ii)

Question 4

Solution 4

Question 5 (i)

Solution 5 (i)

Question 5 (ii)

Solution 5 (ii)

Question 6 (i)

Solution 6 (i)

Question 6 (ii)

Solution 6 (ii)

Question 7 (i)

Solution 7 (i)

Question 7 (ii)

Solution 7 (ii)

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Chapter 23 Algebra of Vectors Exercise Ex. 23.9

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7 (i)

Solution 7 (i)

Question 7 (ii)

Solution 7 (ii)

Question 7 (iii)

Solution 7 (iii)

Question 8

Solution 8

Question 9

Solution 9

Question 10

If a unit vector begin mathsize 12px style straight a with rightwards arrow on top end style makes and angles begin mathsize 12px style straight pi over 3 space with space straight i with hat on top comma space straight pi over 4 space with space straight j with hat on top end style and an acute angle θ with begin mathsize 12px style straight k with hat on top end style, then find θ and hence, the components of begin mathsize 12px style straight a with rightwards arrow on top end style.Solution 10

Question 11

Solution 11

Question 12

Solution 12


Discover more from EduGrown School

Subscribe to get the latest posts sent to your email.