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5418-5421
2 Let aand rb r be two nonzero vectors Then 0 a ×b r= r r if and only if a and rb r are parallel (or collinear) to each other, i e
1
5419-5422
Let aand rb r be two nonzero vectors Then 0 a ×b r= r r if and only if a and rb r are parallel (or collinear) to each other, i e , a b × r r = 0 ⇔a b r r r � Fig 10
1
5420-5423
Then 0 a ×b r= r r if and only if a and rb r are parallel (or collinear) to each other, i e , a b × r r = 0 ⇔a b r r r � Fig 10 23 © NCERT not to be republished MATHEMATICS 450 In particular, 0 a ×a = r r r and ( ) 0 a × −a = r r r , since in the first situation, θ = 0 and in the second one, θ = π, making the value of sin θ to be 0
1
5421-5424
e , a b × r r = 0 ⇔a b r r r � Fig 10 23 © NCERT not to be republished MATHEMATICS 450 In particular, 0 a ×a = r r r and ( ) 0 a × −a = r r r , since in the first situation, θ = 0 and in the second one, θ = π, making the value of sin θ to be 0 3
1
5422-5425
, a b × r r = 0 ⇔a b r r r � Fig 10 23 © NCERT not to be republished MATHEMATICS 450 In particular, 0 a ×a = r r r and ( ) 0 a × −a = r r r , since in the first situation, θ = 0 and in the second one, θ = π, making the value of sin θ to be 0 3 If 2 then | || | a b a b r r r r
1
5423-5426
23 © NCERT not to be republished MATHEMATICS 450 In particular, 0 a ×a = r r r and ( ) 0 a × −a = r r r , since in the first situation, θ = 0 and in the second one, θ = π, making the value of sin θ to be 0 3 If 2 then | || | a b a b r r r r 4
1
5424-5427
3 If 2 then | || | a b a b r r r r 4 In view of the Observations 2 and 3, for mutually perpendicular unit vectors ˆ ˆ ,ˆ and i j k (Fig 10
1
5425-5428
If 2 then | || | a b a b r r r r 4 In view of the Observations 2 and 3, for mutually perpendicular unit vectors ˆ ˆ ,ˆ and i j k (Fig 10 24), we have ˆ ˆ i i × = ˆ ˆ ˆ ˆ 0 j j k k × = × = r ˆ ˆ i ×j = ˆ ˆ ˆ ˆ ˆ ˆ ˆ , , k j k i k i j × = × = 5
1
5426-5429
4 In view of the Observations 2 and 3, for mutually perpendicular unit vectors ˆ ˆ ,ˆ and i j k (Fig 10 24), we have ˆ ˆ i i × = ˆ ˆ ˆ ˆ 0 j j k k × = × = r ˆ ˆ i ×j = ˆ ˆ ˆ ˆ ˆ ˆ ˆ , , k j k i k i j × = × = 5 In terms of vector product, the angle between two vectors and a rb r may be given as sin θ = | | | || | a b a b × r r r r 6
1
5427-5430
In view of the Observations 2 and 3, for mutually perpendicular unit vectors ˆ ˆ ,ˆ and i j k (Fig 10 24), we have ˆ ˆ i i × = ˆ ˆ ˆ ˆ 0 j j k k × = × = r ˆ ˆ i ×j = ˆ ˆ ˆ ˆ ˆ ˆ ˆ , , k j k i k i j × = × = 5 In terms of vector product, the angle between two vectors and a rb r may be given as sin θ = | | | || | a b a b × r r r r 6 It is always true that the vector product is not commutative, as a ×b r r = b a − r× r
1
5428-5431
24), we have ˆ ˆ i i × = ˆ ˆ ˆ ˆ 0 j j k k × = × = r ˆ ˆ i ×j = ˆ ˆ ˆ ˆ ˆ ˆ ˆ , , k j k i k i j × = × = 5 In terms of vector product, the angle between two vectors and a rb r may be given as sin θ = | | | || | a b a b × r r r r 6 It is always true that the vector product is not commutative, as a ×b r r = b a − r× r Indeed, ˆ | || | sin a b a b n × = θ r r r r , where ˆ , a b and n rr form a right handed system, i
1
5429-5432
In terms of vector product, the angle between two vectors and a rb r may be given as sin θ = | | | || | a b a b × r r r r 6 It is always true that the vector product is not commutative, as a ×b r r = b a − r× r Indeed, ˆ | || | sin a b a b n × = θ r r r r , where ˆ , a b and n rr form a right handed system, i e
1
5430-5433
It is always true that the vector product is not commutative, as a ×b r r = b a − r× r Indeed, ˆ | || | sin a b a b n × = θ r r r r , where ˆ , a b and n rr form a right handed system, i e , θ is traversed from ato rb r , Fig 10
1
5431-5434
Indeed, ˆ | || | sin a b a b n × = θ r r r r , where ˆ , a b and n rr form a right handed system, i e , θ is traversed from ato rb r , Fig 10 25 (i)
1
5432-5435
e , θ is traversed from ato rb r , Fig 10 25 (i) While, 1ˆ | || | sin b a a b n × = θ r r r r , where 1ˆ , b aand n r r form a right handed system i
1
5433-5436
, θ is traversed from ato rb r , Fig 10 25 (i) While, 1ˆ | || | sin b a a b n × = θ r r r r , where 1ˆ , b aand n r r form a right handed system i e
1
5434-5437
25 (i) While, 1ˆ | || | sin b a a b n × = θ r r r r , where 1ˆ , b aand n r r form a right handed system i e θ is traversed from bto a r r , Fig 10
1
5435-5438
While, 1ˆ | || | sin b a a b n × = θ r r r r , where 1ˆ , b aand n r r form a right handed system i e θ is traversed from bto a r r , Fig 10 25(ii)
1
5436-5439
e θ is traversed from bto a r r , Fig 10 25(ii) Fig 10
1
5437-5440
θ is traversed from bto a r r , Fig 10 25(ii) Fig 10 25 (i), (ii) Thus, if we assume aand rb r to lie in the plane of the paper, then 1 ˆ n and ˆ n both will be perpendicular to the plane of the paper
1
5438-5441
25(ii) Fig 10 25 (i), (ii) Thus, if we assume aand rb r to lie in the plane of the paper, then 1 ˆ n and ˆ n both will be perpendicular to the plane of the paper But, ˆn being directed above the paper while 1ˆn directed below the paper
1
5439-5442
Fig 10 25 (i), (ii) Thus, if we assume aand rb r to lie in the plane of the paper, then 1 ˆ n and ˆ n both will be perpendicular to the plane of the paper But, ˆn being directed above the paper while 1ˆn directed below the paper i
1
5440-5443
25 (i), (ii) Thus, if we assume aand rb r to lie in the plane of the paper, then 1 ˆ n and ˆ n both will be perpendicular to the plane of the paper But, ˆn being directed above the paper while 1ˆn directed below the paper i e
1
5441-5444
But, ˆn being directed above the paper while 1ˆn directed below the paper i e 1ˆ ˆ n = −n
1
5442-5445
i e 1ˆ ˆ n = −n Fig 10
1
5443-5446
e 1ˆ ˆ n = −n Fig 10 24 © NCERT not to be republished VECTOR ALGEBRA 451 Hence a b × r r = ˆ | || a b|sin n r r = 1ˆ | || a b|sin n − θ r r b a = − r× r 7
1
5444-5447
1ˆ ˆ n = −n Fig 10 24 © NCERT not to be republished VECTOR ALGEBRA 451 Hence a b × r r = ˆ | || a b|sin n r r = 1ˆ | || a b|sin n − θ r r b a = − r× r 7 In view of the Observations 4 and 6, we have ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ , and
1
5445-5448
Fig 10 24 © NCERT not to be republished VECTOR ALGEBRA 451 Hence a b × r r = ˆ | || a b|sin n r r = 1ˆ | || a b|sin n − θ r r b a = − r× r 7 In view of the Observations 4 and 6, we have ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ , and j i k k j i i k j × = − × = − × = − 8
1
5446-5449
24 © NCERT not to be republished VECTOR ALGEBRA 451 Hence a b × r r = ˆ | || a b|sin n r r = 1ˆ | || a b|sin n − θ r r b a = − r× r 7 In view of the Observations 4 and 6, we have ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ , and j i k k j i i k j × = − × = − × = − 8 If aand rb r represent the adjacent sides of a triangle then its area is given as 1 | | 2 a b r r
1
5447-5450
In view of the Observations 4 and 6, we have ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ , and j i k k j i i k j × = − × = − × = − 8 If aand rb r represent the adjacent sides of a triangle then its area is given as 1 | | 2 a b r r By definition of the area of a triangle, we have from Fig 10
1
5448-5451
j i k k j i i k j × = − × = − × = − 8 If aand rb r represent the adjacent sides of a triangle then its area is given as 1 | | 2 a b r r By definition of the area of a triangle, we have from Fig 10 26, Area of triangle ABC = 1 AB CD
1
5449-5452
If aand rb r represent the adjacent sides of a triangle then its area is given as 1 | | 2 a b r r By definition of the area of a triangle, we have from Fig 10 26, Area of triangle ABC = 1 AB CD 2 ⋅ But AB | b| = r (as given), and CD = | | ar sinθ
1
5450-5453
By definition of the area of a triangle, we have from Fig 10 26, Area of triangle ABC = 1 AB CD 2 ⋅ But AB | b| = r (as given), and CD = | | ar sinθ Thus, Area of triangle ABC = 1 | || | sin 2 b a θ r r 1 | |
1
5451-5454
26, Area of triangle ABC = 1 AB CD 2 ⋅ But AB | b| = r (as given), and CD = | | ar sinθ Thus, Area of triangle ABC = 1 | || | sin 2 b a θ r r 1 | | 2 a b = × r r 9
1
5452-5455
2 ⋅ But AB | b| = r (as given), and CD = | | ar sinθ Thus, Area of triangle ABC = 1 | || | sin 2 b a θ r r 1 | | 2 a b = × r r 9 If a and rb r represent the adjacent sides of a parallelogram, then its area is given by | | a b × r r
1
5453-5456
Thus, Area of triangle ABC = 1 | || | sin 2 b a θ r r 1 | | 2 a b = × r r 9 If a and rb r represent the adjacent sides of a parallelogram, then its area is given by | | a b × r r From Fig 10
1
5454-5457
2 a b = × r r 9 If a and rb r represent the adjacent sides of a parallelogram, then its area is given by | | a b × r r From Fig 10 27, we have Area of parallelogram ABCD = AB
1
5455-5458
If a and rb r represent the adjacent sides of a parallelogram, then its area is given by | | a b × r r From Fig 10 27, we have Area of parallelogram ABCD = AB DE
1
5456-5459
From Fig 10 27, we have Area of parallelogram ABCD = AB DE But AB | =b| r (as given), and DE | =a|sin θ r
1
5457-5460
27, we have Area of parallelogram ABCD = AB DE But AB | =b| r (as given), and DE | =a|sin θ r Thus, Area of parallelogram ABCD = | || | sin b a θ r r | |
1
5458-5461
DE But AB | =b| r (as given), and DE | =a|sin θ r Thus, Area of parallelogram ABCD = | || | sin b a θ r r | | a b = × r r We now state two important properties of vector product
1
5459-5462
But AB | =b| r (as given), and DE | =a|sin θ r Thus, Area of parallelogram ABCD = | || | sin b a θ r r | | a b = × r r We now state two important properties of vector product Property 3 (Distributivity of vector product over addition): If , a band c rr r are any three vectors and λ be a scalar, then (i) ( ) a b c × r+ r r = a b a c r r r r (ii) ( λa b) × r r = ( ) ( ) a b a b λ × = × λ r r r r Fig 10
1
5460-5463
Thus, Area of parallelogram ABCD = | || | sin b a θ r r | | a b = × r r We now state two important properties of vector product Property 3 (Distributivity of vector product over addition): If , a band c rr r are any three vectors and λ be a scalar, then (i) ( ) a b c × r+ r r = a b a c r r r r (ii) ( λa b) × r r = ( ) ( ) a b a b λ × = × λ r r r r Fig 10 26 Fig 10
1
5461-5464
a b = × r r We now state two important properties of vector product Property 3 (Distributivity of vector product over addition): If , a band c rr r are any three vectors and λ be a scalar, then (i) ( ) a b c × r+ r r = a b a c r r r r (ii) ( λa b) × r r = ( ) ( ) a b a b λ × = × λ r r r r Fig 10 26 Fig 10 27 © NCERT not to be republished MATHEMATICS 452 Let aand rb r be two vectors given in component form as 1 2 3 ˆ ˆ ˆ a i a j a k + + and 1 2 3 ˆ ˆ ˆ b i b j b k + + , respectively
1
5462-5465
Property 3 (Distributivity of vector product over addition): If , a band c rr r are any three vectors and λ be a scalar, then (i) ( ) a b c × r+ r r = a b a c r r r r (ii) ( λa b) × r r = ( ) ( ) a b a b λ × = × λ r r r r Fig 10 26 Fig 10 27 © NCERT not to be republished MATHEMATICS 452 Let aand rb r be two vectors given in component form as 1 2 3 ˆ ˆ ˆ a i a j a k + + and 1 2 3 ˆ ˆ ˆ b i b j b k + + , respectively Then their cross product may be given by a b × r r = 1 2 3 1 2 3 ˆ ˆ ˆ i j k a a a b b b Explanation We have a b × r r = 1 2 3 1 2 3 ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) a i a j a k b i b j b k + + × + + = 1 1 1 2 1 3 2 1 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) ( ) ( ) a b i i a b i j a b i k a b j i × + × + × + × + 2 2 2 3 ˆ ˆ ˆ ˆ ( ) ( ) a b j j a b j k × + × + 3 1 3 2 3 3 ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) ( ) a b k i a b k j a b k k × + × + × (by Property 1) = 1 2 1 3 2 1 ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) ( ) a b i j a b k i a b i j × − × − × + 2 3 3 1 3 2 ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) ( ) a b j k a b k i a b j k × + × − × ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (as 0 and , and ) i i j j k k i k k i j i i j k j j k × = × = × = × = − × × = − × × = − × = 1 2 1 3 2 1 2 3 3 1 3 2 ˆ ˆ ˆ ˆ ˆ ˆ a b k a b j a b k a b i a b j a b i − − + + − ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (as , and ) i j k j k i k i j × = × = × = = 2 3 3 2 1 3 3 1 1 2 2 1 ˆ ˆ ˆ ( ) ( ) ( ) a b a b i a b a b j a b a b k − − − + − = 1 2 3 1 2 3 ˆ ˆ ˆ i j k a a a b b b Example 22 Find ˆ ˆ ˆ ˆ ˆ ˆ | |, if 2 3 and 3 5 2 a b a i j k b i j k × = + + = + − r r r r Solution We have a b × r r = ˆ ˆ ˆ 2 1 3 3 5 2 i j k − = ˆ ˆ ˆ ( 2 15) ( 4 9) (10 – 3) i j k − − − − − + ˆ ˆ ˆ 17 13 7 i j k = − + + Hence | | r ra b = 2 2 2 ( 17) (13) (7) 507 − + + = © NCERT not to be republished VECTOR ALGEBRA 453 Example 23 Find a unit vector perpendicular to each of the vectors ( ) a b + r r and ( ), a −b r r where ˆ ˆ ˆ ˆ ˆ ˆ , 2 3 a i j k b i j k = + + = + + r r
1
5463-5466
26 Fig 10 27 © NCERT not to be republished MATHEMATICS 452 Let aand rb r be two vectors given in component form as 1 2 3 ˆ ˆ ˆ a i a j a k + + and 1 2 3 ˆ ˆ ˆ b i b j b k + + , respectively Then their cross product may be given by a b × r r = 1 2 3 1 2 3 ˆ ˆ ˆ i j k a a a b b b Explanation We have a b × r r = 1 2 3 1 2 3 ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) a i a j a k b i b j b k + + × + + = 1 1 1 2 1 3 2 1 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) ( ) ( ) a b i i a b i j a b i k a b j i × + × + × + × + 2 2 2 3 ˆ ˆ ˆ ˆ ( ) ( ) a b j j a b j k × + × + 3 1 3 2 3 3 ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) ( ) a b k i a b k j a b k k × + × + × (by Property 1) = 1 2 1 3 2 1 ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) ( ) a b i j a b k i a b i j × − × − × + 2 3 3 1 3 2 ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) ( ) a b j k a b k i a b j k × + × − × ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (as 0 and , and ) i i j j k k i k k i j i i j k j j k × = × = × = × = − × × = − × × = − × = 1 2 1 3 2 1 2 3 3 1 3 2 ˆ ˆ ˆ ˆ ˆ ˆ a b k a b j a b k a b i a b j a b i − − + + − ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (as , and ) i j k j k i k i j × = × = × = = 2 3 3 2 1 3 3 1 1 2 2 1 ˆ ˆ ˆ ( ) ( ) ( ) a b a b i a b a b j a b a b k − − − + − = 1 2 3 1 2 3 ˆ ˆ ˆ i j k a a a b b b Example 22 Find ˆ ˆ ˆ ˆ ˆ ˆ | |, if 2 3 and 3 5 2 a b a i j k b i j k × = + + = + − r r r r Solution We have a b × r r = ˆ ˆ ˆ 2 1 3 3 5 2 i j k − = ˆ ˆ ˆ ( 2 15) ( 4 9) (10 – 3) i j k − − − − − + ˆ ˆ ˆ 17 13 7 i j k = − + + Hence | | r ra b = 2 2 2 ( 17) (13) (7) 507 − + + = © NCERT not to be republished VECTOR ALGEBRA 453 Example 23 Find a unit vector perpendicular to each of the vectors ( ) a b + r r and ( ), a −b r r where ˆ ˆ ˆ ˆ ˆ ˆ , 2 3 a i j k b i j k = + + = + + r r Solution We have ˆ ˆ ˆ ˆ ˆ 2 3 4 and 2 a b i j k a b j k + = + + − = − − r r r r A vector which is perpendicular to both and a b a b + − r r r r is given by ( ) ( ) a b a b + × − r r r r = ˆ ˆ ˆ ˆ ˆ ˆ 2 3 4 2 4 2 ( , say) 0 1 2 i j k i j k c = − + − = − − r Now | | cr = 4 16 4 24 2 6 + + = = Therefore, the required unit vector is | c| c r r = 1 2 1 ˆ ˆ ˆ 6 6 6 i j k − + − �Note There are two perpendicular directions to any plane
1
5464-5467
27 © NCERT not to be republished MATHEMATICS 452 Let aand rb r be two vectors given in component form as 1 2 3 ˆ ˆ ˆ a i a j a k + + and 1 2 3 ˆ ˆ ˆ b i b j b k + + , respectively Then their cross product may be given by a b × r r = 1 2 3 1 2 3 ˆ ˆ ˆ i j k a a a b b b Explanation We have a b × r r = 1 2 3 1 2 3 ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) a i a j a k b i b j b k + + × + + = 1 1 1 2 1 3 2 1 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) ( ) ( ) a b i i a b i j a b i k a b j i × + × + × + × + 2 2 2 3 ˆ ˆ ˆ ˆ ( ) ( ) a b j j a b j k × + × + 3 1 3 2 3 3 ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) ( ) a b k i a b k j a b k k × + × + × (by Property 1) = 1 2 1 3 2 1 ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) ( ) a b i j a b k i a b i j × − × − × + 2 3 3 1 3 2 ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) ( ) a b j k a b k i a b j k × + × − × ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (as 0 and , and ) i i j j k k i k k i j i i j k j j k × = × = × = × = − × × = − × × = − × = 1 2 1 3 2 1 2 3 3 1 3 2 ˆ ˆ ˆ ˆ ˆ ˆ a b k a b j a b k a b i a b j a b i − − + + − ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (as , and ) i j k j k i k i j × = × = × = = 2 3 3 2 1 3 3 1 1 2 2 1 ˆ ˆ ˆ ( ) ( ) ( ) a b a b i a b a b j a b a b k − − − + − = 1 2 3 1 2 3 ˆ ˆ ˆ i j k a a a b b b Example 22 Find ˆ ˆ ˆ ˆ ˆ ˆ | |, if 2 3 and 3 5 2 a b a i j k b i j k × = + + = + − r r r r Solution We have a b × r r = ˆ ˆ ˆ 2 1 3 3 5 2 i j k − = ˆ ˆ ˆ ( 2 15) ( 4 9) (10 – 3) i j k − − − − − + ˆ ˆ ˆ 17 13 7 i j k = − + + Hence | | r ra b = 2 2 2 ( 17) (13) (7) 507 − + + = © NCERT not to be republished VECTOR ALGEBRA 453 Example 23 Find a unit vector perpendicular to each of the vectors ( ) a b + r r and ( ), a −b r r where ˆ ˆ ˆ ˆ ˆ ˆ , 2 3 a i j k b i j k = + + = + + r r Solution We have ˆ ˆ ˆ ˆ ˆ 2 3 4 and 2 a b i j k a b j k + = + + − = − − r r r r A vector which is perpendicular to both and a b a b + − r r r r is given by ( ) ( ) a b a b + × − r r r r = ˆ ˆ ˆ ˆ ˆ ˆ 2 3 4 2 4 2 ( , say) 0 1 2 i j k i j k c = − + − = − − r Now | | cr = 4 16 4 24 2 6 + + = = Therefore, the required unit vector is | c| c r r = 1 2 1 ˆ ˆ ˆ 6 6 6 i j k − + − �Note There are two perpendicular directions to any plane Thus, another unit vector perpendicular to and a b a b + − r r r r will be 1 2 1 ˆ ˆ ˆ
1
5465-5468
Then their cross product may be given by a b × r r = 1 2 3 1 2 3 ˆ ˆ ˆ i j k a a a b b b Explanation We have a b × r r = 1 2 3 1 2 3 ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) a i a j a k b i b j b k + + × + + = 1 1 1 2 1 3 2 1 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) ( ) ( ) a b i i a b i j a b i k a b j i × + × + × + × + 2 2 2 3 ˆ ˆ ˆ ˆ ( ) ( ) a b j j a b j k × + × + 3 1 3 2 3 3 ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) ( ) a b k i a b k j a b k k × + × + × (by Property 1) = 1 2 1 3 2 1 ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) ( ) a b i j a b k i a b i j × − × − × + 2 3 3 1 3 2 ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) ( ) a b j k a b k i a b j k × + × − × ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (as 0 and , and ) i i j j k k i k k i j i i j k j j k × = × = × = × = − × × = − × × = − × = 1 2 1 3 2 1 2 3 3 1 3 2 ˆ ˆ ˆ ˆ ˆ ˆ a b k a b j a b k a b i a b j a b i − − + + − ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (as , and ) i j k j k i k i j × = × = × = = 2 3 3 2 1 3 3 1 1 2 2 1 ˆ ˆ ˆ ( ) ( ) ( ) a b a b i a b a b j a b a b k − − − + − = 1 2 3 1 2 3 ˆ ˆ ˆ i j k a a a b b b Example 22 Find ˆ ˆ ˆ ˆ ˆ ˆ | |, if 2 3 and 3 5 2 a b a i j k b i j k × = + + = + − r r r r Solution We have a b × r r = ˆ ˆ ˆ 2 1 3 3 5 2 i j k − = ˆ ˆ ˆ ( 2 15) ( 4 9) (10 – 3) i j k − − − − − + ˆ ˆ ˆ 17 13 7 i j k = − + + Hence | | r ra b = 2 2 2 ( 17) (13) (7) 507 − + + = © NCERT not to be republished VECTOR ALGEBRA 453 Example 23 Find a unit vector perpendicular to each of the vectors ( ) a b + r r and ( ), a −b r r where ˆ ˆ ˆ ˆ ˆ ˆ , 2 3 a i j k b i j k = + + = + + r r Solution We have ˆ ˆ ˆ ˆ ˆ 2 3 4 and 2 a b i j k a b j k + = + + − = − − r r r r A vector which is perpendicular to both and a b a b + − r r r r is given by ( ) ( ) a b a b + × − r r r r = ˆ ˆ ˆ ˆ ˆ ˆ 2 3 4 2 4 2 ( , say) 0 1 2 i j k i j k c = − + − = − − r Now | | cr = 4 16 4 24 2 6 + + = = Therefore, the required unit vector is | c| c r r = 1 2 1 ˆ ˆ ˆ 6 6 6 i j k − + − �Note There are two perpendicular directions to any plane Thus, another unit vector perpendicular to and a b a b + − r r r r will be 1 2 1 ˆ ˆ ˆ 6 6 6 i j k − + But that will be a consequence of ( ) ( ) a b a b − × + r r r r
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5466-5469
Solution We have ˆ ˆ ˆ ˆ ˆ 2 3 4 and 2 a b i j k a b j k + = + + − = − − r r r r A vector which is perpendicular to both and a b a b + − r r r r is given by ( ) ( ) a b a b + × − r r r r = ˆ ˆ ˆ ˆ ˆ ˆ 2 3 4 2 4 2 ( , say) 0 1 2 i j k i j k c = − + − = − − r Now | | cr = 4 16 4 24 2 6 + + = = Therefore, the required unit vector is | c| c r r = 1 2 1 ˆ ˆ ˆ 6 6 6 i j k − + − �Note There are two perpendicular directions to any plane Thus, another unit vector perpendicular to and a b a b + − r r r r will be 1 2 1 ˆ ˆ ˆ 6 6 6 i j k − + But that will be a consequence of ( ) ( ) a b a b − × + r r r r Example 24 Find the area of a triangle having the points A(1, 1, 1), B(1, 2, 3) and C(2, 3, 1) as its vertices
1
5467-5470
Thus, another unit vector perpendicular to and a b a b + − r r r r will be 1 2 1 ˆ ˆ ˆ 6 6 6 i j k − + But that will be a consequence of ( ) ( ) a b a b − × + r r r r Example 24 Find the area of a triangle having the points A(1, 1, 1), B(1, 2, 3) and C(2, 3, 1) as its vertices Solution We have ˆ ˆ ˆ ˆ AB 2 and AC 2 j k i j = + = + uuur uuur
1
5468-5471
6 6 6 i j k − + But that will be a consequence of ( ) ( ) a b a b − × + r r r r Example 24 Find the area of a triangle having the points A(1, 1, 1), B(1, 2, 3) and C(2, 3, 1) as its vertices Solution We have ˆ ˆ ˆ ˆ AB 2 and AC 2 j k i j = + = + uuur uuur The area of the given triangle is 1 | AB AC | 2 uuur× uuur
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5469-5472
Example 24 Find the area of a triangle having the points A(1, 1, 1), B(1, 2, 3) and C(2, 3, 1) as its vertices Solution We have ˆ ˆ ˆ ˆ AB 2 and AC 2 j k i j = + = + uuur uuur The area of the given triangle is 1 | AB AC | 2 uuur× uuur Now, AB AC × uuur uuur = ˆ ˆ ˆ ˆ ˆ ˆ 0 1 2 4 2 1 2 0 i j k i j k = − + − Therefore | AB AC| uuur× uuur = 16 4 1 21 + + = Thus, the required area is 1 21 2 © NCERT not to be republished MATHEMATICS 454 Example 25 Find the area of a parallelogram whose adjacent sides are given by the vectors ˆ ˆ ˆ ˆ ˆ ˆ 3 4 and a i j k b i j k = + + = − + r r Solution The area of a parallelogram with aand rb r as its adjacent sides is given by | | a ×b r r
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5470-5473
Solution We have ˆ ˆ ˆ ˆ AB 2 and AC 2 j k i j = + = + uuur uuur The area of the given triangle is 1 | AB AC | 2 uuur× uuur Now, AB AC × uuur uuur = ˆ ˆ ˆ ˆ ˆ ˆ 0 1 2 4 2 1 2 0 i j k i j k = − + − Therefore | AB AC| uuur× uuur = 16 4 1 21 + + = Thus, the required area is 1 21 2 © NCERT not to be republished MATHEMATICS 454 Example 25 Find the area of a parallelogram whose adjacent sides are given by the vectors ˆ ˆ ˆ ˆ ˆ ˆ 3 4 and a i j k b i j k = + + = − + r r Solution The area of a parallelogram with aand rb r as its adjacent sides is given by | | a ×b r r Now a b × r r = ˆ ˆ ˆ ˆ ˆ ˆ 3 1 4 5 4 1 1 1 i j k i j k = + − − Therefore | | a b × r r = 25 1 16 42 + + = and hence, the required area is 42
1
5471-5474
The area of the given triangle is 1 | AB AC | 2 uuur× uuur Now, AB AC × uuur uuur = ˆ ˆ ˆ ˆ ˆ ˆ 0 1 2 4 2 1 2 0 i j k i j k = − + − Therefore | AB AC| uuur× uuur = 16 4 1 21 + + = Thus, the required area is 1 21 2 © NCERT not to be republished MATHEMATICS 454 Example 25 Find the area of a parallelogram whose adjacent sides are given by the vectors ˆ ˆ ˆ ˆ ˆ ˆ 3 4 and a i j k b i j k = + + = − + r r Solution The area of a parallelogram with aand rb r as its adjacent sides is given by | | a ×b r r Now a b × r r = ˆ ˆ ˆ ˆ ˆ ˆ 3 1 4 5 4 1 1 1 i j k i j k = + − − Therefore | | a b × r r = 25 1 16 42 + + = and hence, the required area is 42 EXERCISE 10
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5472-5475
Now, AB AC × uuur uuur = ˆ ˆ ˆ ˆ ˆ ˆ 0 1 2 4 2 1 2 0 i j k i j k = − + − Therefore | AB AC| uuur× uuur = 16 4 1 21 + + = Thus, the required area is 1 21 2 © NCERT not to be republished MATHEMATICS 454 Example 25 Find the area of a parallelogram whose adjacent sides are given by the vectors ˆ ˆ ˆ ˆ ˆ ˆ 3 4 and a i j k b i j k = + + = − + r r Solution The area of a parallelogram with aand rb r as its adjacent sides is given by | | a ×b r r Now a b × r r = ˆ ˆ ˆ ˆ ˆ ˆ 3 1 4 5 4 1 1 1 i j k i j k = + − − Therefore | | a b × r r = 25 1 16 42 + + = and hence, the required area is 42 EXERCISE 10 4 1
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5473-5476
Now a b × r r = ˆ ˆ ˆ ˆ ˆ ˆ 3 1 4 5 4 1 1 1 i j k i j k = + − − Therefore | | a b × r r = 25 1 16 42 + + = and hence, the required area is 42 EXERCISE 10 4 1 Find ˆ ˆ ˆ ˆ ˆ ˆ | |, if 7 7 and 3 2 2 a b a i j k b i j k × = − + = − + r r r r
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5474-5477
EXERCISE 10 4 1 Find ˆ ˆ ˆ ˆ ˆ ˆ | |, if 7 7 and 3 2 2 a b a i j k b i j k × = − + = − + r r r r 2
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5475-5478
4 1 Find ˆ ˆ ˆ ˆ ˆ ˆ | |, if 7 7 and 3 2 2 a b a i j k b i j k × = − + = − + r r r r 2 Find a unit vector perpendicular to each of the vector and a b a b + − r r r r , where ˆ ˆ ˆ ˆ ˆ ˆ 3 2 2 and 2 2 a i j k b i j k = + + = + − r r
1
5476-5479
Find ˆ ˆ ˆ ˆ ˆ ˆ | |, if 7 7 and 3 2 2 a b a i j k b i j k × = − + = − + r r r r 2 Find a unit vector perpendicular to each of the vector and a b a b + − r r r r , where ˆ ˆ ˆ ˆ ˆ ˆ 3 2 2 and 2 2 a i j k b i j k = + + = + − r r 3
1
5477-5480
2 Find a unit vector perpendicular to each of the vector and a b a b + − r r r r , where ˆ ˆ ˆ ˆ ˆ ˆ 3 2 2 and 2 2 a i j k b i j k = + + = + − r r 3 If a unit vector ar makes angles ˆ ˆ with , with 3 i4 j π π and an acute angle θ with ˆk , then find θ and hence, the components of ar
1
5478-5481
Find a unit vector perpendicular to each of the vector and a b a b + − r r r r , where ˆ ˆ ˆ ˆ ˆ ˆ 3 2 2 and 2 2 a i j k b i j k = + + = + − r r 3 If a unit vector ar makes angles ˆ ˆ with , with 3 i4 j π π and an acute angle θ with ˆk , then find θ and hence, the components of ar 4
1
5479-5482
3 If a unit vector ar makes angles ˆ ˆ with , with 3 i4 j π π and an acute angle θ with ˆk , then find θ and hence, the components of ar 4 Show that ( ) ( ) a b a b − × + r r r r = 2( a b) × r r 5
1
5480-5483
If a unit vector ar makes angles ˆ ˆ with , with 3 i4 j π π and an acute angle θ with ˆk , then find θ and hence, the components of ar 4 Show that ( ) ( ) a b a b − × + r r r r = 2( a b) × r r 5 Find λ and μ if ˆ ˆ ˆ ˆ ˆ ˆ (2 6 27 ) ( ) 0 i j k i j k + + × + λ + μ = r
1
5481-5484
4 Show that ( ) ( ) a b a b − × + r r r r = 2( a b) × r r 5 Find λ and μ if ˆ ˆ ˆ ˆ ˆ ˆ (2 6 27 ) ( ) 0 i j k i j k + + × + λ + μ = r 6
1
5482-5485
Show that ( ) ( ) a b a b − × + r r r r = 2( a b) × r r 5 Find λ and μ if ˆ ˆ ˆ ˆ ˆ ˆ (2 6 27 ) ( ) 0 i j k i j k + + × + λ + μ = r 6 Given that 0 a b rr and 0 a ×b r= r r
1
5483-5486
Find λ and μ if ˆ ˆ ˆ ˆ ˆ ˆ (2 6 27 ) ( ) 0 i j k i j k + + × + λ + μ = r 6 Given that 0 a b rr and 0 a ×b r= r r What can you conclude about the vectors aand rb r
1
5484-5487
6 Given that 0 a b rr and 0 a ×b r= r r What can you conclude about the vectors aand rb r 7
1
5485-5488
Given that 0 a b rr and 0 a ×b r= r r What can you conclude about the vectors aand rb r 7 Let the vectors , , a b c rr r be given as 1 2 3 1 2 3 ˆ ˆ ˆ ˆ ˆ ˆ , , a i a j a k b i b j b k + + + + 1 2 3 ˆ ˆ ˆ c i c j c k + +
1
5486-5489
What can you conclude about the vectors aand rb r 7 Let the vectors , , a b c rr r be given as 1 2 3 1 2 3 ˆ ˆ ˆ ˆ ˆ ˆ , , a i a j a k b i b j b k + + + + 1 2 3 ˆ ˆ ˆ c i c j c k + + Then show that ( ) a b c a b a c × + = × + × r r r r r r r
1
5487-5490
7 Let the vectors , , a b c rr r be given as 1 2 3 1 2 3 ˆ ˆ ˆ ˆ ˆ ˆ , , a i a j a k b i b j b k + + + + 1 2 3 ˆ ˆ ˆ c i c j c k + + Then show that ( ) a b c a b a c × + = × + × r r r r r r r 8
1
5488-5491
Let the vectors , , a b c rr r be given as 1 2 3 1 2 3 ˆ ˆ ˆ ˆ ˆ ˆ , , a i a j a k b i b j b k + + + + 1 2 3 ˆ ˆ ˆ c i c j c k + + Then show that ( ) a b c a b a c × + = × + × r r r r r r r 8 If either 0 or 0, a b = r= r r r then 0 a ×b = r r r
1
5489-5492
Then show that ( ) a b c a b a c × + = × + × r r r r r r r 8 If either 0 or 0, a b = r= r r r then 0 a ×b = r r r Is the converse true
1
5490-5493
8 If either 0 or 0, a b = r= r r r then 0 a ×b = r r r Is the converse true Justify your answer with an example
1
5491-5494
If either 0 or 0, a b = r= r r r then 0 a ×b = r r r Is the converse true Justify your answer with an example 9
1
5492-5495
Is the converse true Justify your answer with an example 9 Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5)
1
5493-5496
Justify your answer with an example 9 Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5) © NCERT not to be republished VECTOR ALGEBRA 455 10
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5494-5497
9 Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5) © NCERT not to be republished VECTOR ALGEBRA 455 10 Find the area of the parallelogram whose adjacent sides are determined by the vectors ˆ ˆ ˆ 3 a i j k = − + r and ˆ ˆ ˆ 2 7 b i j k = − + r
1
5495-5498
Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5) © NCERT not to be republished VECTOR ALGEBRA 455 10 Find the area of the parallelogram whose adjacent sides are determined by the vectors ˆ ˆ ˆ 3 a i j k = − + r and ˆ ˆ ˆ 2 7 b i j k = − + r 11
1
5496-5499
© NCERT not to be republished VECTOR ALGEBRA 455 10 Find the area of the parallelogram whose adjacent sides are determined by the vectors ˆ ˆ ˆ 3 a i j k = − + r and ˆ ˆ ˆ 2 7 b i j k = − + r 11 Let the vectors a and rb r be such that 2 | | 3 and | | 3 a b = r= r , then a ×b r r is a unit vector, if the angle between and a rb r is (A) π/6 (B) π/4 (C) π/3 (D) π/2 12
1
5497-5500
Find the area of the parallelogram whose adjacent sides are determined by the vectors ˆ ˆ ˆ 3 a i j k = − + r and ˆ ˆ ˆ 2 7 b i j k = − + r 11 Let the vectors a and rb r be such that 2 | | 3 and | | 3 a b = r= r , then a ×b r r is a unit vector, if the angle between and a rb r is (A) π/6 (B) π/4 (C) π/3 (D) π/2 12 Area of a rectangle having vertices A, B, C and D with position vectors 1 1 ˆ ˆ ˆ ˆ ˆ ˆ – 4 , 4 2 2 i j k i j k + + + + , 1 ˆ ˆ ˆ 4 2 i j k − + and 1 ˆ ˆ ˆ – 4 2 i j k − + , respectively is (A) 1 2 (B) 1 (C) 2 (D) 4 Miscellaneous Examples Example 26 Write all the unit vectors in XY-plane
1
5498-5501
11 Let the vectors a and rb r be such that 2 | | 3 and | | 3 a b = r= r , then a ×b r r is a unit vector, if the angle between and a rb r is (A) π/6 (B) π/4 (C) π/3 (D) π/2 12 Area of a rectangle having vertices A, B, C and D with position vectors 1 1 ˆ ˆ ˆ ˆ ˆ ˆ – 4 , 4 2 2 i j k i j k + + + + , 1 ˆ ˆ ˆ 4 2 i j k − + and 1 ˆ ˆ ˆ – 4 2 i j k − + , respectively is (A) 1 2 (B) 1 (C) 2 (D) 4 Miscellaneous Examples Example 26 Write all the unit vectors in XY-plane Solution Let r x i y j ∧ ∧ = + r be a unit vector in XY-plane (Fig 10
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5499-5502
Let the vectors a and rb r be such that 2 | | 3 and | | 3 a b = r= r , then a ×b r r is a unit vector, if the angle between and a rb r is (A) π/6 (B) π/4 (C) π/3 (D) π/2 12 Area of a rectangle having vertices A, B, C and D with position vectors 1 1 ˆ ˆ ˆ ˆ ˆ ˆ – 4 , 4 2 2 i j k i j k + + + + , 1 ˆ ˆ ˆ 4 2 i j k − + and 1 ˆ ˆ ˆ – 4 2 i j k − + , respectively is (A) 1 2 (B) 1 (C) 2 (D) 4 Miscellaneous Examples Example 26 Write all the unit vectors in XY-plane Solution Let r x i y j ∧ ∧ = + r be a unit vector in XY-plane (Fig 10 28)
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5500-5503
Area of a rectangle having vertices A, B, C and D with position vectors 1 1 ˆ ˆ ˆ ˆ ˆ ˆ – 4 , 4 2 2 i j k i j k + + + + , 1 ˆ ˆ ˆ 4 2 i j k − + and 1 ˆ ˆ ˆ – 4 2 i j k − + , respectively is (A) 1 2 (B) 1 (C) 2 (D) 4 Miscellaneous Examples Example 26 Write all the unit vectors in XY-plane Solution Let r x i y j ∧ ∧ = + r be a unit vector in XY-plane (Fig 10 28) Then, from the figure, we have x = cos θ and y = sin θ (since | rr | = 1)
1
5501-5504
Solution Let r x i y j ∧ ∧ = + r be a unit vector in XY-plane (Fig 10 28) Then, from the figure, we have x = cos θ and y = sin θ (since | rr | = 1) So, we may write the vector rr as ( OP) r = uuur r = ˆ ˆ cos sin i j
1
5502-5505
28) Then, from the figure, we have x = cos θ and y = sin θ (since | rr | = 1) So, we may write the vector rr as ( OP) r = uuur r = ˆ ˆ cos sin i j (1) Clearly, | | rr = 2 2 cos sin 1 θ + θ = Fig 10
1
5503-5506
Then, from the figure, we have x = cos θ and y = sin θ (since | rr | = 1) So, we may write the vector rr as ( OP) r = uuur r = ˆ ˆ cos sin i j (1) Clearly, | | rr = 2 2 cos sin 1 θ + θ = Fig 10 28 Also, as θ varies from 0 to 2π, the point P (Fig 10
1
5504-5507
So, we may write the vector rr as ( OP) r = uuur r = ˆ ˆ cos sin i j (1) Clearly, | | rr = 2 2 cos sin 1 θ + θ = Fig 10 28 Also, as θ varies from 0 to 2π, the point P (Fig 10 28) traces the circle x2 + y2 = 1 counterclockwise, and this covers all possible directions
1
5505-5508
(1) Clearly, | | rr = 2 2 cos sin 1 θ + θ = Fig 10 28 Also, as θ varies from 0 to 2π, the point P (Fig 10 28) traces the circle x2 + y2 = 1 counterclockwise, and this covers all possible directions So, (1) gives every unit vector in the XY-plane
1
5506-5509
28 Also, as θ varies from 0 to 2π, the point P (Fig 10 28) traces the circle x2 + y2 = 1 counterclockwise, and this covers all possible directions So, (1) gives every unit vector in the XY-plane © NCERT not to be republished MATHEMATICS 456 Example 27 If ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ , 2 5 , 3 2 3 and 6 i j k i j i j k i j k are the position vectors of points A, B, C and D respectively, then find the angle between AB uuur and CD uuur
1
5507-5510
28) traces the circle x2 + y2 = 1 counterclockwise, and this covers all possible directions So, (1) gives every unit vector in the XY-plane © NCERT not to be republished MATHEMATICS 456 Example 27 If ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ , 2 5 , 3 2 3 and 6 i j k i j i j k i j k are the position vectors of points A, B, C and D respectively, then find the angle between AB uuur and CD uuur Deduce that AB uuur and CD uuur are collinear
1
5508-5511
So, (1) gives every unit vector in the XY-plane © NCERT not to be republished MATHEMATICS 456 Example 27 If ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ , 2 5 , 3 2 3 and 6 i j k i j i j k i j k are the position vectors of points A, B, C and D respectively, then find the angle between AB uuur and CD uuur Deduce that AB uuur and CD uuur are collinear Solution Note that if θ is the angle between AB and CD, then θ is also the angle between AB and CD uuur uuur
1
5509-5512
© NCERT not to be republished MATHEMATICS 456 Example 27 If ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ , 2 5 , 3 2 3 and 6 i j k i j i j k i j k are the position vectors of points A, B, C and D respectively, then find the angle between AB uuur and CD uuur Deduce that AB uuur and CD uuur are collinear Solution Note that if θ is the angle between AB and CD, then θ is also the angle between AB and CD uuur uuur Now AB uuur = Position vector of B – Position vector of A = ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (2 5 ) ( ) 4 i j i j k i j k + − + + = + − Therefore | AB| uuur = 2 2 2 (1) (4) ( 1) 3 2 + + − = Similarly CD uuur = ˆ ˆ ˆ 2 8 2 and |CD | 6 2 i j k − − + = uuur Thus cos θ = AB CD |AB||CD| uuur uuur uuur uuur = 1( 2) 4( 8) ( 1)(2) 36 1 36 (3 2)(6 2) − + − + − =− = − Since 0 ≤ θ ≤ π, it follows that θ = π
1
5510-5513
Deduce that AB uuur and CD uuur are collinear Solution Note that if θ is the angle between AB and CD, then θ is also the angle between AB and CD uuur uuur Now AB uuur = Position vector of B – Position vector of A = ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (2 5 ) ( ) 4 i j i j k i j k + − + + = + − Therefore | AB| uuur = 2 2 2 (1) (4) ( 1) 3 2 + + − = Similarly CD uuur = ˆ ˆ ˆ 2 8 2 and |CD | 6 2 i j k − − + = uuur Thus cos θ = AB CD |AB||CD| uuur uuur uuur uuur = 1( 2) 4( 8) ( 1)(2) 36 1 36 (3 2)(6 2) − + − + − =− = − Since 0 ≤ θ ≤ π, it follows that θ = π This shows that AB uuur and CD uuur are collinear
1
5511-5514
Solution Note that if θ is the angle between AB and CD, then θ is also the angle between AB and CD uuur uuur Now AB uuur = Position vector of B – Position vector of A = ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (2 5 ) ( ) 4 i j i j k i j k + − + + = + − Therefore | AB| uuur = 2 2 2 (1) (4) ( 1) 3 2 + + − = Similarly CD uuur = ˆ ˆ ˆ 2 8 2 and |CD | 6 2 i j k − − + = uuur Thus cos θ = AB CD |AB||CD| uuur uuur uuur uuur = 1( 2) 4( 8) ( 1)(2) 36 1 36 (3 2)(6 2) − + − + − =− = − Since 0 ≤ θ ≤ π, it follows that θ = π This shows that AB uuur and CD uuur are collinear Alternatively, 1 AB 2CD uuur uuur which implies that AB and CD uuur uuur are collinear vectors
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5512-5515
Now AB uuur = Position vector of B – Position vector of A = ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (2 5 ) ( ) 4 i j i j k i j k + − + + = + − Therefore | AB| uuur = 2 2 2 (1) (4) ( 1) 3 2 + + − = Similarly CD uuur = ˆ ˆ ˆ 2 8 2 and |CD | 6 2 i j k − − + = uuur Thus cos θ = AB CD |AB||CD| uuur uuur uuur uuur = 1( 2) 4( 8) ( 1)(2) 36 1 36 (3 2)(6 2) − + − + − =− = − Since 0 ≤ θ ≤ π, it follows that θ = π This shows that AB uuur and CD uuur are collinear Alternatively, 1 AB 2CD uuur uuur which implies that AB and CD uuur uuur are collinear vectors Example 28 Let , a band c rr r be three vectors such that | | 3, | | 4, | | 5 a b c = = = r r r and each one of them being perpendicular to the sum of the other two, find | a b c| + + r r r
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This shows that AB uuur and CD uuur are collinear Alternatively, 1 AB 2CD uuur uuur which implies that AB and CD uuur uuur are collinear vectors Example 28 Let , a band c rr r be three vectors such that | | 3, | | 4, | | 5 a b c = = = r r r and each one of them being perpendicular to the sum of the other two, find | a b c| + + r r r Solution Given ( ) a b c ⋅ r+ r r = 0, ( ) 0, ( ) 0
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Alternatively, 1 AB 2CD uuur uuur which implies that AB and CD uuur uuur are collinear vectors Example 28 Let , a band c rr r be three vectors such that | | 3, | | 4, | | 5 a b c = = = r r r and each one of them being perpendicular to the sum of the other two, find | a b c| + + r r r Solution Given ( ) a b c ⋅ r+ r r = 0, ( ) 0, ( ) 0 b c a c a b ⋅ + = ⋅ + = r r r r r r Now 2 | | a b c + + r r r = 2 ( ) ( ) ( ) a b c a b c a b c + + = + + ⋅ + + r r r r r r r r r = ( ) ( ) a a a b c b b b a c ⋅ + ⋅ + + ⋅ + ⋅ + r r r r r r r r r r +
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Example 28 Let , a band c rr r be three vectors such that | | 3, | | 4, | | 5 a b c = = = r r r and each one of them being perpendicular to the sum of the other two, find | a b c| + + r r r Solution Given ( ) a b c ⋅ r+ r r = 0, ( ) 0, ( ) 0 b c a c a b ⋅ + = ⋅ + = r r r r r r Now 2 | | a b c + + r r r = 2 ( ) ( ) ( ) a b c a b c a b c + + = + + ⋅ + + r r r r r r r r r = ( ) ( ) a a a b c b b b a c ⋅ + ⋅ + + ⋅ + ⋅ + r r r r r r r r r r + ( )
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Solution Given ( ) a b c ⋅ r+ r r = 0, ( ) 0, ( ) 0 b c a c a b ⋅ + = ⋅ + = r r r r r r Now 2 | | a b c + + r r r = 2 ( ) ( ) ( ) a b c a b c a b c + + = + + ⋅ + + r r r r r r r r r = ( ) ( ) a a a b c b b b a c ⋅ + ⋅ + + ⋅ + ⋅ + r r r r r r r r r r + ( ) c a b c c + r+ r r r r = 2 2 2 | | | | | | a b c + + r r r = 9 + 16 + 25 = 50 Therefore | | a b c + r+ r r = 50 5 2 = © NCERT not to be republished VECTOR ALGEBRA 457 Example 29 Three vectors , and a b c r r r satisfy the condition 0 a b c + + = r r r r
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b c a c a b ⋅ + = ⋅ + = r r r r r r Now 2 | | a b c + + r r r = 2 ( ) ( ) ( ) a b c a b c a b c + + = + + ⋅ + + r r r r r r r r r = ( ) ( ) a a a b c b b b a c ⋅ + ⋅ + + ⋅ + ⋅ + r r r r r r r r r r + ( ) c a b c c + r+ r r r r = 2 2 2 | | | | | | a b c + + r r r = 9 + 16 + 25 = 50 Therefore | | a b c + r+ r r = 50 5 2 = © NCERT not to be republished VECTOR ALGEBRA 457 Example 29 Three vectors , and a b c r r r satisfy the condition 0 a b c + + = r r r r Evaluate the quantity , if | | 1, | | 4 and | | 2 a b b c c a a b c μ = ⋅ + ⋅ + ⋅ = = = r r r r r r r r r