Datasets:
knowrohit07
/
gita_dataset

Dataset card Data Studio Files
xet
Community
1
Chapter
stringclasses
18 values
sentence_range
stringlengths
3
9
Text
stringlengths
7
7.34k
1
5318-5321
20 © NCERT not to be republished MATHEMATICS 444 is called the projection vector, and its magnitude | pr | is simply called as the projection of the vector AB uuur on the directed line l For example, in each of the following figures (Fig 10 20(i) to (iv)), projection vector of AB uuur along the line l is vector AC uuur Observations 1
1
5319-5322
For example, in each of the following figures (Fig 10 20(i) to (iv)), projection vector of AB uuur along the line l is vector AC uuur Observations 1 If ˆp is the unit vector along a line l, then the projection of a vector ar on the line l is given by a pˆ r
1
5320-5323
20(i) to (iv)), projection vector of AB uuur along the line l is vector AC uuur Observations 1 If ˆp is the unit vector along a line l, then the projection of a vector ar on the line l is given by a pˆ r 2
1
5321-5324
Observations 1 If ˆp is the unit vector along a line l, then the projection of a vector ar on the line l is given by a pˆ r 2 Projection of a vector ar on other vector b r , is given by a bˆ, r⋅ or , or 1 ( ) | | | | b a a b b b ⎛ ⎞ ⋅ ⋅ ⎜ ⎟ ⎝ ⎠ r r r r r r 3
1
5322-5325
If ˆp is the unit vector along a line l, then the projection of a vector ar on the line l is given by a pˆ r 2 Projection of a vector ar on other vector b r , is given by a bˆ, r⋅ or , or 1 ( ) | | | | b a a b b b ⎛ ⎞ ⋅ ⋅ ⎜ ⎟ ⎝ ⎠ r r r r r r 3 If θ = 0, then the projection vector of AB uuur will be AB uuur itself and if θ = π, then the projection vector of AB uuur will be BA uuur
1
5323-5326
2 Projection of a vector ar on other vector b r , is given by a bˆ, r⋅ or , or 1 ( ) | | | | b a a b b b ⎛ ⎞ ⋅ ⋅ ⎜ ⎟ ⎝ ⎠ r r r r r r 3 If θ = 0, then the projection vector of AB uuur will be AB uuur itself and if θ = π, then the projection vector of AB uuur will be BA uuur 4
1
5324-5327
Projection of a vector ar on other vector b r , is given by a bˆ, r⋅ or , or 1 ( ) | | | | b a a b b b ⎛ ⎞ ⋅ ⋅ ⎜ ⎟ ⎝ ⎠ r r r r r r 3 If θ = 0, then the projection vector of AB uuur will be AB uuur itself and if θ = π, then the projection vector of AB uuur will be BA uuur 4 If = 2 π θ or 3 = 2 π θ , then the projection vector of AB uuur will be zero vector
1
5325-5328
If θ = 0, then the projection vector of AB uuur will be AB uuur itself and if θ = π, then the projection vector of AB uuur will be BA uuur 4 If = 2 π θ or 3 = 2 π θ , then the projection vector of AB uuur will be zero vector Remark If α, β and γ are the direction angles of vector 1 2 3 ˆ ˆ ˆ a a i a j a k = + + r , then its direction cosines may be given as 3 1 2 ˆ cos , cos , and cos ˆ | | | | | | | || | a a a a i a a a a i r r r r r Also, note that | a| cos , | |cos and | |cos a a α β γ r r r are respectively the projections of ar along OX, OY and OZ
1
5326-5329
4 If = 2 π θ or 3 = 2 π θ , then the projection vector of AB uuur will be zero vector Remark If α, β and γ are the direction angles of vector 1 2 3 ˆ ˆ ˆ a a i a j a k = + + r , then its direction cosines may be given as 3 1 2 ˆ cos , cos , and cos ˆ | | | | | | | || | a a a a i a a a a i r r r r r Also, note that | a| cos , | |cos and | |cos a a α β γ r r r are respectively the projections of ar along OX, OY and OZ i
1
5327-5330
If = 2 π θ or 3 = 2 π θ , then the projection vector of AB uuur will be zero vector Remark If α, β and γ are the direction angles of vector 1 2 3 ˆ ˆ ˆ a a i a j a k = + + r , then its direction cosines may be given as 3 1 2 ˆ cos , cos , and cos ˆ | | | | | | | || | a a a a i a a a a i r r r r r Also, note that | a| cos , | |cos and | |cos a a α β γ r r r are respectively the projections of ar along OX, OY and OZ i e
1
5328-5331
Remark If α, β and γ are the direction angles of vector 1 2 3 ˆ ˆ ˆ a a i a j a k = + + r , then its direction cosines may be given as 3 1 2 ˆ cos , cos , and cos ˆ | | | | | | | || | a a a a i a a a a i r r r r r Also, note that | a| cos , | |cos and | |cos a a α β γ r r r are respectively the projections of ar along OX, OY and OZ i e , the scalar components a1, a2 and a3 of the vector ar , are precisely the projections of ar along x-axis, y-axis and z-axis, respectively
1
5329-5332
i e , the scalar components a1, a2 and a3 of the vector ar , are precisely the projections of ar along x-axis, y-axis and z-axis, respectively Further, if ar is a unit vector, then it may be expressed in terms of its direction cosines as ˆ ˆ ˆ cos cos cos a i j k = α + β + γ r Example 13 Find the angle between two vectors and a rb r with magnitudes 1 and 2 respectively and when a b1 ⋅ = rr
1
5330-5333
e , the scalar components a1, a2 and a3 of the vector ar , are precisely the projections of ar along x-axis, y-axis and z-axis, respectively Further, if ar is a unit vector, then it may be expressed in terms of its direction cosines as ˆ ˆ ˆ cos cos cos a i j k = α + β + γ r Example 13 Find the angle between two vectors and a rb r with magnitudes 1 and 2 respectively and when a b1 ⋅ = rr Solution Given 1,| | 1and | | 2 a b a b r r r r
1
5331-5334
, the scalar components a1, a2 and a3 of the vector ar , are precisely the projections of ar along x-axis, y-axis and z-axis, respectively Further, if ar is a unit vector, then it may be expressed in terms of its direction cosines as ˆ ˆ ˆ cos cos cos a i j k = α + β + γ r Example 13 Find the angle between two vectors and a rb r with magnitudes 1 and 2 respectively and when a b1 ⋅ = rr Solution Given 1,| | 1and | | 2 a b a b r r r r We have 1 1 1 cos cos 2 3 | || a b| a b rr r r © NCERT not to be republished VECTOR ALGEBRA 445 Example 14 Find angle ‘θ’ between the vectors ˆ ˆ ˆ ˆ ˆ ˆ and a i j k b i j k = + − = − + r r
1
5332-5335
Further, if ar is a unit vector, then it may be expressed in terms of its direction cosines as ˆ ˆ ˆ cos cos cos a i j k = α + β + γ r Example 13 Find the angle between two vectors and a rb r with magnitudes 1 and 2 respectively and when a b1 ⋅ = rr Solution Given 1,| | 1and | | 2 a b a b r r r r We have 1 1 1 cos cos 2 3 | || a b| a b rr r r © NCERT not to be republished VECTOR ALGEBRA 445 Example 14 Find angle ‘θ’ between the vectors ˆ ˆ ˆ ˆ ˆ ˆ and a i j k b i j k = + − = − + r r Solution The angle θ between two vectors aand rb r is given by cosθ = | || a b| a b ⋅ r r r r Now a b ⋅ rr = ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) 1 1 1 1 i j k i j k + − ⋅ − + = − − = −
1
5333-5336
Solution Given 1,| | 1and | | 2 a b a b r r r r We have 1 1 1 cos cos 2 3 | || a b| a b rr r r © NCERT not to be republished VECTOR ALGEBRA 445 Example 14 Find angle ‘θ’ between the vectors ˆ ˆ ˆ ˆ ˆ ˆ and a i j k b i j k = + − = − + r r Solution The angle θ between two vectors aand rb r is given by cosθ = | || a b| a b ⋅ r r r r Now a b ⋅ rr = ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) 1 1 1 1 i j k i j k + − ⋅ − + = − − = − Therefore, we have cosθ = 31 − hence the required angle is θ = 1 1 cos 3 Example 15 If ˆ ˆ ˆ ˆ ˆ ˆ 5 3 and 3 5 a i j k b i j k = − − = + − r r , then show that the vectors and a b a b + − r r r r are perpendicular
1
5334-5337
We have 1 1 1 cos cos 2 3 | || a b| a b rr r r © NCERT not to be republished VECTOR ALGEBRA 445 Example 14 Find angle ‘θ’ between the vectors ˆ ˆ ˆ ˆ ˆ ˆ and a i j k b i j k = + − = − + r r Solution The angle θ between two vectors aand rb r is given by cosθ = | || a b| a b ⋅ r r r r Now a b ⋅ rr = ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) 1 1 1 1 i j k i j k + − ⋅ − + = − − = − Therefore, we have cosθ = 31 − hence the required angle is θ = 1 1 cos 3 Example 15 If ˆ ˆ ˆ ˆ ˆ ˆ 5 3 and 3 5 a i j k b i j k = − − = + − r r , then show that the vectors and a b a b + − r r r r are perpendicular Solution We know that two nonzero vectors are perpendicular if their scalar product is zero
1
5335-5338
Solution The angle θ between two vectors aand rb r is given by cosθ = | || a b| a b ⋅ r r r r Now a b ⋅ rr = ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) 1 1 1 1 i j k i j k + − ⋅ − + = − − = − Therefore, we have cosθ = 31 − hence the required angle is θ = 1 1 cos 3 Example 15 If ˆ ˆ ˆ ˆ ˆ ˆ 5 3 and 3 5 a i j k b i j k = − − = + − r r , then show that the vectors and a b a b + − r r r r are perpendicular Solution We know that two nonzero vectors are perpendicular if their scalar product is zero Here a b + r r = ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (5 3 ) ( 3 5 ) 6 2 8 i j k i j k i j k − − + + − = + − and a −b r r = ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (5 3 ) ( 3 5 ) 4 4 2 i j k i j k i j k − − − + − = − + So ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) (6 2 8 ) (4 4 2 ) 24 8 16 0
1
5336-5339
Therefore, we have cosθ = 31 − hence the required angle is θ = 1 1 cos 3 Example 15 If ˆ ˆ ˆ ˆ ˆ ˆ 5 3 and 3 5 a i j k b i j k = − − = + − r r , then show that the vectors and a b a b + − r r r r are perpendicular Solution We know that two nonzero vectors are perpendicular if their scalar product is zero Here a b + r r = ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (5 3 ) ( 3 5 ) 6 2 8 i j k i j k i j k − − + + − = + − and a −b r r = ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (5 3 ) ( 3 5 ) 4 4 2 i j k i j k i j k − − − + − = − + So ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) (6 2 8 ) (4 4 2 ) 24 8 16 0 a b a b i j k i j k + ⋅ − = + − ⋅ − + = − − = r r r r Hence and a b a b + − r r r r are perpendicular vectors
1
5337-5340
Solution We know that two nonzero vectors are perpendicular if their scalar product is zero Here a b + r r = ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (5 3 ) ( 3 5 ) 6 2 8 i j k i j k i j k − − + + − = + − and a −b r r = ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (5 3 ) ( 3 5 ) 4 4 2 i j k i j k i j k − − − + − = − + So ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) (6 2 8 ) (4 4 2 ) 24 8 16 0 a b a b i j k i j k + ⋅ − = + − ⋅ − + = − − = r r r r Hence and a b a b + − r r r r are perpendicular vectors Example 16 Find the projection of the vector ˆ ˆ ˆ 2 3 2 a i j k = + + r on the vector ˆ ˆ 2ˆ b i j k = + + r
1
5338-5341
Here a b + r r = ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (5 3 ) ( 3 5 ) 6 2 8 i j k i j k i j k − − + + − = + − and a −b r r = ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (5 3 ) ( 3 5 ) 4 4 2 i j k i j k i j k − − − + − = − + So ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) (6 2 8 ) (4 4 2 ) 24 8 16 0 a b a b i j k i j k + ⋅ − = + − ⋅ − + = − − = r r r r Hence and a b a b + − r r r r are perpendicular vectors Example 16 Find the projection of the vector ˆ ˆ ˆ 2 3 2 a i j k = + + r on the vector ˆ ˆ 2ˆ b i j k = + + r Solution The projection of vector ar on the vector b r is given by 1 ( ) | | a b b ⋅ rr r = 2 2 2 (2 1 3 2 2 1) 10 5 6 3 6 (1) (2) (1) × + × + × = = + + Example 17 Find | | a b − r r , if two vectors aand rb r are such that | | 2, | | 3 a b r r and 4 a b ⋅ rr=
1
5339-5342
a b a b i j k i j k + ⋅ − = + − ⋅ − + = − − = r r r r Hence and a b a b + − r r r r are perpendicular vectors Example 16 Find the projection of the vector ˆ ˆ ˆ 2 3 2 a i j k = + + r on the vector ˆ ˆ 2ˆ b i j k = + + r Solution The projection of vector ar on the vector b r is given by 1 ( ) | | a b b ⋅ rr r = 2 2 2 (2 1 3 2 2 1) 10 5 6 3 6 (1) (2) (1) × + × + × = = + + Example 17 Find | | a b − r r , if two vectors aand rb r are such that | | 2, | | 3 a b r r and 4 a b ⋅ rr= Solution We have 2 | | a b r r = ( ) ( ) a b a b − ⋅ − r r r r =
1
5340-5343
Example 16 Find the projection of the vector ˆ ˆ ˆ 2 3 2 a i j k = + + r on the vector ˆ ˆ 2ˆ b i j k = + + r Solution The projection of vector ar on the vector b r is given by 1 ( ) | | a b b ⋅ rr r = 2 2 2 (2 1 3 2 2 1) 10 5 6 3 6 (1) (2) (1) × + × + × = = + + Example 17 Find | | a b − r r , if two vectors aand rb r are such that | | 2, | | 3 a b r r and 4 a b ⋅ rr= Solution We have 2 | | a b r r = ( ) ( ) a b a b − ⋅ − r r r r = a a a b b a b b − ⋅ − ⋅ + ⋅ r r r r r r r r © NCERT not to be republished MATHEMATICS 446 B C A a b + a b = 2 2 | | 2( ) | | a a b b − ⋅ + r r r r = 2 2 (2) 2(4) (3) − + Therefore | | a −b r r = 5 Example 18 If ar is a unit vector and ( ) ( ) 8 x a x a − ⋅ + = r r r r , then find | | xr
1
5341-5344
Solution The projection of vector ar on the vector b r is given by 1 ( ) | | a b b ⋅ rr r = 2 2 2 (2 1 3 2 2 1) 10 5 6 3 6 (1) (2) (1) × + × + × = = + + Example 17 Find | | a b − r r , if two vectors aand rb r are such that | | 2, | | 3 a b r r and 4 a b ⋅ rr= Solution We have 2 | | a b r r = ( ) ( ) a b a b − ⋅ − r r r r = a a a b b a b b − ⋅ − ⋅ + ⋅ r r r r r r r r © NCERT not to be republished MATHEMATICS 446 B C A a b + a b = 2 2 | | 2( ) | | a a b b − ⋅ + r r r r = 2 2 (2) 2(4) (3) − + Therefore | | a −b r r = 5 Example 18 If ar is a unit vector and ( ) ( ) 8 x a x a − ⋅ + = r r r r , then find | | xr Solution Since ar is a unit vector, | ra =| 1
1
5342-5345
Solution We have 2 | | a b r r = ( ) ( ) a b a b − ⋅ − r r r r = a a a b b a b b − ⋅ − ⋅ + ⋅ r r r r r r r r © NCERT not to be republished MATHEMATICS 446 B C A a b + a b = 2 2 | | 2( ) | | a a b b − ⋅ + r r r r = 2 2 (2) 2(4) (3) − + Therefore | | a −b r r = 5 Example 18 If ar is a unit vector and ( ) ( ) 8 x a x a − ⋅ + = r r r r , then find | | xr Solution Since ar is a unit vector, | ra =| 1 Also, ( ) ( ) x a x a − ⋅ + r r r r = 8 or x x x a a x a a ⋅ + ⋅ − ⋅ − ⋅ r r r r r r r r = 8 or 2 | | 1 rx = 8 i
1
5343-5346
a a a b b a b b − ⋅ − ⋅ + ⋅ r r r r r r r r © NCERT not to be republished MATHEMATICS 446 B C A a b + a b = 2 2 | | 2( ) | | a a b b − ⋅ + r r r r = 2 2 (2) 2(4) (3) − + Therefore | | a −b r r = 5 Example 18 If ar is a unit vector and ( ) ( ) 8 x a x a − ⋅ + = r r r r , then find | | xr Solution Since ar is a unit vector, | ra =| 1 Also, ( ) ( ) x a x a − ⋅ + r r r r = 8 or x x x a a x a a ⋅ + ⋅ − ⋅ − ⋅ r r r r r r r r = 8 or 2 | | 1 rx = 8 i e
1
5344-5347
Solution Since ar is a unit vector, | ra =| 1 Also, ( ) ( ) x a x a − ⋅ + r r r r = 8 or x x x a a x a a ⋅ + ⋅ − ⋅ − ⋅ r r r r r r r r = 8 or 2 | | 1 rx = 8 i e | rx |2 = 9 Therefore | | xr = 3 (as magnitude of a vector is non negative)
1
5345-5348
Also, ( ) ( ) x a x a − ⋅ + r r r r = 8 or x x x a a x a a ⋅ + ⋅ − ⋅ − ⋅ r r r r r r r r = 8 or 2 | | 1 rx = 8 i e | rx |2 = 9 Therefore | | xr = 3 (as magnitude of a vector is non negative) Example 19 For any two vectors aand rb r , we always have | | | || | a b a b ⋅ ≤ r r r r (Cauchy- Schwartz inequality)
1
5346-5349
e | rx |2 = 9 Therefore | | xr = 3 (as magnitude of a vector is non negative) Example 19 For any two vectors aand rb r , we always have | | | || | a b a b ⋅ ≤ r r r r (Cauchy- Schwartz inequality) Solution The inequality holds trivially when either 0 or 0 a b = r= r r r
1
5347-5350
| rx |2 = 9 Therefore | | xr = 3 (as magnitude of a vector is non negative) Example 19 For any two vectors aand rb r , we always have | | | || | a b a b ⋅ ≤ r r r r (Cauchy- Schwartz inequality) Solution The inequality holds trivially when either 0 or 0 a b = r= r r r Actually, in such a situation we have | | 0 | || | a b a b ⋅ = = r r r r
1
5348-5351
Example 19 For any two vectors aand rb r , we always have | | | || | a b a b ⋅ ≤ r r r r (Cauchy- Schwartz inequality) Solution The inequality holds trivially when either 0 or 0 a b = r= r r r Actually, in such a situation we have | | 0 | || | a b a b ⋅ = = r r r r So, let us assume that | | 0 | | a b ≠ ≠ r r
1
5349-5352
Solution The inequality holds trivially when either 0 or 0 a b = r= r r r Actually, in such a situation we have | | 0 | || | a b a b ⋅ = = r r r r So, let us assume that | | 0 | | a b ≠ ≠ r r Then, we have | | | || a b| a b ⋅ rr r r = | cos | 1 θ ≤ Therefore | a b| ⋅ rr ≤ | || | a rb r Example 20 For any two vectors aand rb r , we always have | | | | | | a b a b + ≤ + r r r r (triangle inequality)
1
5350-5353
Actually, in such a situation we have | | 0 | || | a b a b ⋅ = = r r r r So, let us assume that | | 0 | | a b ≠ ≠ r r Then, we have | | | || a b| a b ⋅ rr r r = | cos | 1 θ ≤ Therefore | a b| ⋅ rr ≤ | || | a rb r Example 20 For any two vectors aand rb r , we always have | | | | | | a b a b + ≤ + r r r r (triangle inequality) Solution The inequality holds trivially in case either 0 or 0 a b = r= r r r (How
1
5351-5354
So, let us assume that | | 0 | | a b ≠ ≠ r r Then, we have | | | || a b| a b ⋅ rr r r = | cos | 1 θ ≤ Therefore | a b| ⋅ rr ≤ | || | a rb r Example 20 For any two vectors aand rb r , we always have | | | | | | a b a b + ≤ + r r r r (triangle inequality) Solution The inequality holds trivially in case either 0 or 0 a b = r= r r r (How )
1
5352-5355
Then, we have | | | || a b| a b ⋅ rr r r = | cos | 1 θ ≤ Therefore | a b| ⋅ rr ≤ | || | a rb r Example 20 For any two vectors aand rb r , we always have | | | | | | a b a b + ≤ + r r r r (triangle inequality) Solution The inequality holds trivially in case either 0 or 0 a b = r= r r r (How ) So, let | | 0 | | a b r r r
1
5353-5356
Solution The inequality holds trivially in case either 0 or 0 a b = r= r r r (How ) So, let | | 0 | | a b r r r Then, 2 | | a +b r r = 2 ( ) ( ) ( ) a b a b a b + = + ⋅ + r r r r r r = a a a b b a b b ⋅ + ⋅ + ⋅ + ⋅ r r r r r r r r = 2 2 | | 2 | | a a b b + ⋅ + r r r r (scalar product is commutative) ≤ 2 2 | | 2| | | | a a b b + ⋅ r+ r r r (since | | x x x ≤ ∀ ∈ R ) ≤ 2 2 | | 2| || | | | a a b b + + r r r r (from Example 19) = 2 (| | | |) a b r r Fig 10
1
5354-5357
) So, let | | 0 | | a b r r r Then, 2 | | a +b r r = 2 ( ) ( ) ( ) a b a b a b + = + ⋅ + r r r r r r = a a a b b a b b ⋅ + ⋅ + ⋅ + ⋅ r r r r r r r r = 2 2 | | 2 | | a a b b + ⋅ + r r r r (scalar product is commutative) ≤ 2 2 | | 2| | | | a a b b + ⋅ r+ r r r (since | | x x x ≤ ∀ ∈ R ) ≤ 2 2 | | 2| || | | | a a b b + + r r r r (from Example 19) = 2 (| | | |) a b r r Fig 10 21 © NCERT not to be republished VECTOR ALGEBRA 447 Hence | | a b r r ≤ | | | | a b r r Remark If the equality holds in triangle inequality (in the above Example 20), i
1
5355-5358
So, let | | 0 | | a b r r r Then, 2 | | a +b r r = 2 ( ) ( ) ( ) a b a b a b + = + ⋅ + r r r r r r = a a a b b a b b ⋅ + ⋅ + ⋅ + ⋅ r r r r r r r r = 2 2 | | 2 | | a a b b + ⋅ + r r r r (scalar product is commutative) ≤ 2 2 | | 2| | | | a a b b + ⋅ r+ r r r (since | | x x x ≤ ∀ ∈ R ) ≤ 2 2 | | 2| || | | | a a b b + + r r r r (from Example 19) = 2 (| | | |) a b r r Fig 10 21 © NCERT not to be republished VECTOR ALGEBRA 447 Hence | | a b r r ≤ | | | | a b r r Remark If the equality holds in triangle inequality (in the above Example 20), i e
1
5356-5359
Then, 2 | | a +b r r = 2 ( ) ( ) ( ) a b a b a b + = + ⋅ + r r r r r r = a a a b b a b b ⋅ + ⋅ + ⋅ + ⋅ r r r r r r r r = 2 2 | | 2 | | a a b b + ⋅ + r r r r (scalar product is commutative) ≤ 2 2 | | 2| | | | a a b b + ⋅ r+ r r r (since | | x x x ≤ ∀ ∈ R ) ≤ 2 2 | | 2| || | | | a a b b + + r r r r (from Example 19) = 2 (| | | |) a b r r Fig 10 21 © NCERT not to be republished VECTOR ALGEBRA 447 Hence | | a b r r ≤ | | | | a b r r Remark If the equality holds in triangle inequality (in the above Example 20), i e | | a b + r r = | | | | a +b r r , then | AC| uuur = | AB| uuur+| BC | uuur showing that the points A, B and C are collinear
1
5357-5360
21 © NCERT not to be republished VECTOR ALGEBRA 447 Hence | | a b r r ≤ | | | | a b r r Remark If the equality holds in triangle inequality (in the above Example 20), i e | | a b + r r = | | | | a +b r r , then | AC| uuur = | AB| uuur+| BC | uuur showing that the points A, B and C are collinear Example 21 Show that the points ˆ ˆ ˆ ˆ ˆ ˆ A( 2 3 5 ), B( 2 3 ) i j k i j k − + + + + and ˆ C(7ˆ ) i k − are collinear
1
5358-5361
e | | a b + r r = | | | | a +b r r , then | AC| uuur = | AB| uuur+| BC | uuur showing that the points A, B and C are collinear Example 21 Show that the points ˆ ˆ ˆ ˆ ˆ ˆ A( 2 3 5 ), B( 2 3 ) i j k i j k − + + + + and ˆ C(7ˆ ) i k − are collinear Solution We have AB uuur = ˆ ˆ ˆ ˆ ˆ ˆ (1 2) (2 3) (3 5) 3 2 i j k i j k , BC uuur = ˆ ˆ ˆ ˆ ˆ ˆ (7 1) (0 2) ( 1 3) 6 2 4 i j k i j k , AC uuur = ˆ ˆ ˆ ˆ ˆ ˆ (7 2) (0 3) ( 1 5) 9 3 6 i j k i j k | AB| uuur = 14, | BC| 2 14 and | AC| 3 14 uuur uuur Therefore AC uuur = | AB| uuur+| BC | uuur Hence the points A, B and C are collinear
1
5359-5362
| | a b + r r = | | | | a +b r r , then | AC| uuur = | AB| uuur+| BC | uuur showing that the points A, B and C are collinear Example 21 Show that the points ˆ ˆ ˆ ˆ ˆ ˆ A( 2 3 5 ), B( 2 3 ) i j k i j k − + + + + and ˆ C(7ˆ ) i k − are collinear Solution We have AB uuur = ˆ ˆ ˆ ˆ ˆ ˆ (1 2) (2 3) (3 5) 3 2 i j k i j k , BC uuur = ˆ ˆ ˆ ˆ ˆ ˆ (7 1) (0 2) ( 1 3) 6 2 4 i j k i j k , AC uuur = ˆ ˆ ˆ ˆ ˆ ˆ (7 2) (0 3) ( 1 5) 9 3 6 i j k i j k | AB| uuur = 14, | BC| 2 14 and | AC| 3 14 uuur uuur Therefore AC uuur = | AB| uuur+| BC | uuur Hence the points A, B and C are collinear �Note In Example 21, one may note that although AB BC CA 0 + + = uuur uuur uuur r but the points A, B and C do not form the vertices of a triangle
1
5360-5363
Example 21 Show that the points ˆ ˆ ˆ ˆ ˆ ˆ A( 2 3 5 ), B( 2 3 ) i j k i j k − + + + + and ˆ C(7ˆ ) i k − are collinear Solution We have AB uuur = ˆ ˆ ˆ ˆ ˆ ˆ (1 2) (2 3) (3 5) 3 2 i j k i j k , BC uuur = ˆ ˆ ˆ ˆ ˆ ˆ (7 1) (0 2) ( 1 3) 6 2 4 i j k i j k , AC uuur = ˆ ˆ ˆ ˆ ˆ ˆ (7 2) (0 3) ( 1 5) 9 3 6 i j k i j k | AB| uuur = 14, | BC| 2 14 and | AC| 3 14 uuur uuur Therefore AC uuur = | AB| uuur+| BC | uuur Hence the points A, B and C are collinear �Note In Example 21, one may note that although AB BC CA 0 + + = uuur uuur uuur r but the points A, B and C do not form the vertices of a triangle EXERCISE 10
1
5361-5364
Solution We have AB uuur = ˆ ˆ ˆ ˆ ˆ ˆ (1 2) (2 3) (3 5) 3 2 i j k i j k , BC uuur = ˆ ˆ ˆ ˆ ˆ ˆ (7 1) (0 2) ( 1 3) 6 2 4 i j k i j k , AC uuur = ˆ ˆ ˆ ˆ ˆ ˆ (7 2) (0 3) ( 1 5) 9 3 6 i j k i j k | AB| uuur = 14, | BC| 2 14 and | AC| 3 14 uuur uuur Therefore AC uuur = | AB| uuur+| BC | uuur Hence the points A, B and C are collinear �Note In Example 21, one may note that although AB BC CA 0 + + = uuur uuur uuur r but the points A, B and C do not form the vertices of a triangle EXERCISE 10 3 1
1
5362-5365
�Note In Example 21, one may note that although AB BC CA 0 + + = uuur uuur uuur r but the points A, B and C do not form the vertices of a triangle EXERCISE 10 3 1 Find the angle between two vectors aand rb r with magnitudes 3 and 2 , respectively having 6 a b ⋅ = rr
1
5363-5366
EXERCISE 10 3 1 Find the angle between two vectors aand rb r with magnitudes 3 and 2 , respectively having 6 a b ⋅ = rr 2
1
5364-5367
3 1 Find the angle between two vectors aand rb r with magnitudes 3 and 2 , respectively having 6 a b ⋅ = rr 2 Find the angle between the vectors ˆ ˆ 2ˆ 3 i j k − + and ˆ ˆ ˆ 3 2 i j k − + 3
1
5365-5368
Find the angle between two vectors aand rb r with magnitudes 3 and 2 , respectively having 6 a b ⋅ = rr 2 Find the angle between the vectors ˆ ˆ 2ˆ 3 i j k − + and ˆ ˆ ˆ 3 2 i j k − + 3 Find the projection of the vector ˆ ˆ i −j on the vector ˆ ˆ i +j
1
5366-5369
2 Find the angle between the vectors ˆ ˆ 2ˆ 3 i j k − + and ˆ ˆ ˆ 3 2 i j k − + 3 Find the projection of the vector ˆ ˆ i −j on the vector ˆ ˆ i +j 4
1
5367-5370
Find the angle between the vectors ˆ ˆ 2ˆ 3 i j k − + and ˆ ˆ ˆ 3 2 i j k − + 3 Find the projection of the vector ˆ ˆ i −j on the vector ˆ ˆ i +j 4 Find the projection of the vector ˆ ˆ 3ˆ 7 i j k + + on the vector ˆ ˆ ˆ 7 8 i j k − +
1
5368-5371
Find the projection of the vector ˆ ˆ i −j on the vector ˆ ˆ i +j 4 Find the projection of the vector ˆ ˆ 3ˆ 7 i j k + + on the vector ˆ ˆ ˆ 7 8 i j k − + 5
1
5369-5372
4 Find the projection of the vector ˆ ˆ 3ˆ 7 i j k + + on the vector ˆ ˆ ˆ 7 8 i j k − + 5 Show that each of the given three vectors is a unit vector: 1 1 1 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (2 3 6 ), (3 6 2 ), (6 2 3 ) 7 7 7 i j k i j k i j k + + − + + − Also, show that they are mutually perpendicular to each other
1
5370-5373
Find the projection of the vector ˆ ˆ 3ˆ 7 i j k + + on the vector ˆ ˆ ˆ 7 8 i j k − + 5 Show that each of the given three vectors is a unit vector: 1 1 1 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (2 3 6 ), (3 6 2 ), (6 2 3 ) 7 7 7 i j k i j k i j k + + − + + − Also, show that they are mutually perpendicular to each other © NCERT not to be republished MATHEMATICS 448 6
1
5371-5374
5 Show that each of the given three vectors is a unit vector: 1 1 1 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (2 3 6 ), (3 6 2 ), (6 2 3 ) 7 7 7 i j k i j k i j k + + − + + − Also, show that they are mutually perpendicular to each other © NCERT not to be republished MATHEMATICS 448 6 Find | | and | | a rb r , if ( ) ( ) 8 and | | 8| | a b a b a b + ⋅ − = = r r r r r r
1
5372-5375
Show that each of the given three vectors is a unit vector: 1 1 1 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (2 3 6 ), (3 6 2 ), (6 2 3 ) 7 7 7 i j k i j k i j k + + − + + − Also, show that they are mutually perpendicular to each other © NCERT not to be republished MATHEMATICS 448 6 Find | | and | | a rb r , if ( ) ( ) 8 and | | 8| | a b a b a b + ⋅ − = = r r r r r r 7
1
5373-5376
© NCERT not to be republished MATHEMATICS 448 6 Find | | and | | a rb r , if ( ) ( ) 8 and | | 8| | a b a b a b + ⋅ − = = r r r r r r 7 Evaluate the product (3 5 ) (2 7 ) a b a b − ⋅ + r r r r
1
5374-5377
Find | | and | | a rb r , if ( ) ( ) 8 and | | 8| | a b a b a b + ⋅ − = = r r r r r r 7 Evaluate the product (3 5 ) (2 7 ) a b a b − ⋅ + r r r r 8
1
5375-5378
7 Evaluate the product (3 5 ) (2 7 ) a b a b − ⋅ + r r r r 8 Find the magnitude of two vectors aand rb r , having the same magnitude and such that the angle between them is 60o and their scalar product is 1 2
1
5376-5379
Evaluate the product (3 5 ) (2 7 ) a b a b − ⋅ + r r r r 8 Find the magnitude of two vectors aand rb r , having the same magnitude and such that the angle between them is 60o and their scalar product is 1 2 9
1
5377-5380
8 Find the magnitude of two vectors aand rb r , having the same magnitude and such that the angle between them is 60o and their scalar product is 1 2 9 Find | | xr , if for a unit vector ar , ( ) ( ) 12 x a x a − ⋅ + = r r r r
1
5378-5381
Find the magnitude of two vectors aand rb r , having the same magnitude and such that the angle between them is 60o and their scalar product is 1 2 9 Find | | xr , if for a unit vector ar , ( ) ( ) 12 x a x a − ⋅ + = r r r r 10
1
5379-5382
9 Find | | xr , if for a unit vector ar , ( ) ( ) 12 x a x a − ⋅ + = r r r r 10 If ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 2 2 3 , 2 and 3 a i j k b i j k c i j = + + = − + + = + r r r are such that a + λb r r is perpendicular to cr , then find the value of λ
1
5380-5383
Find | | xr , if for a unit vector ar , ( ) ( ) 12 x a x a − ⋅ + = r r r r 10 If ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 2 2 3 , 2 and 3 a i j k b i j k c i j = + + = − + + = + r r r are such that a + λb r r is perpendicular to cr , then find the value of λ 11
1
5381-5384
10 If ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 2 2 3 , 2 and 3 a i j k b i j k c i j = + + = − + + = + r r r are such that a + λb r r is perpendicular to cr , then find the value of λ 11 Show that | | | | a b r+b a r r r is perpendicular to | | | | a b r−b a r r r , for any two nonzero vectors aand rb r
1
5382-5385
If ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 2 2 3 , 2 and 3 a i j k b i j k c i j = + + = − + + = + r r r are such that a + λb r r is perpendicular to cr , then find the value of λ 11 Show that | | | | a b r+b a r r r is perpendicular to | | | | a b r−b a r r r , for any two nonzero vectors aand rb r 12
1
5383-5386
11 Show that | | | | a b r+b a r r r is perpendicular to | | | | a b r−b a r r r , for any two nonzero vectors aand rb r 12 If 0 and 0 a a a b ⋅ = ⋅ = r r r r , then what can be concluded about the vector b r
1
5384-5387
Show that | | | | a b r+b a r r r is perpendicular to | | | | a b r−b a r r r , for any two nonzero vectors aand rb r 12 If 0 and 0 a a a b ⋅ = ⋅ = r r r r , then what can be concluded about the vector b r 13
1
5385-5388
12 If 0 and 0 a a a b ⋅ = ⋅ = r r r r , then what can be concluded about the vector b r 13 If , , a b c rr r are unit vectors such that 0 a b c + + = r r r r , find the value of a b b c c a ⋅ + ⋅ + ⋅ r r r r r r
1
5386-5389
If 0 and 0 a a a b ⋅ = ⋅ = r r r r , then what can be concluded about the vector b r 13 If , , a b c rr r are unit vectors such that 0 a b c + + = r r r r , find the value of a b b c c a ⋅ + ⋅ + ⋅ r r r r r r 14
1
5387-5390
13 If , , a b c rr r are unit vectors such that 0 a b c + + = r r r r , find the value of a b b c c a ⋅ + ⋅ + ⋅ r r r r r r 14 If either vector 0 or 0, then 0 a b a b = = ⋅ = r r r r r r
1
5388-5391
If , , a b c rr r are unit vectors such that 0 a b c + + = r r r r , find the value of a b b c c a ⋅ + ⋅ + ⋅ r r r r r r 14 If either vector 0 or 0, then 0 a b a b = = ⋅ = r r r r r r But the converse need not be true
1
5389-5392
14 If either vector 0 or 0, then 0 a b a b = = ⋅ = r r r r r r But the converse need not be true Justify your answer with an example
1
5390-5393
If either vector 0 or 0, then 0 a b a b = = ⋅ = r r r r r r But the converse need not be true Justify your answer with an example 15
1
5391-5394
But the converse need not be true Justify your answer with an example 15 If the vertices A, B, C of a triangle ABC are (1, 2, 3), (–1, 0, 0), (0, 1, 2), respectively, then find ∠ABC
1
5392-5395
Justify your answer with an example 15 If the vertices A, B, C of a triangle ABC are (1, 2, 3), (–1, 0, 0), (0, 1, 2), respectively, then find ∠ABC [∠ABC is the angle between the vectors BA uuur and BC uuur ]
1
5393-5396
15 If the vertices A, B, C of a triangle ABC are (1, 2, 3), (–1, 0, 0), (0, 1, 2), respectively, then find ∠ABC [∠ABC is the angle between the vectors BA uuur and BC uuur ] 16
1
5394-5397
If the vertices A, B, C of a triangle ABC are (1, 2, 3), (–1, 0, 0), (0, 1, 2), respectively, then find ∠ABC [∠ABC is the angle between the vectors BA uuur and BC uuur ] 16 Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, –1) are collinear
1
5395-5398
[∠ABC is the angle between the vectors BA uuur and BC uuur ] 16 Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, –1) are collinear 17
1
5396-5399
16 Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, –1) are collinear 17 Show that the vectors ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 2 , 3 5 and 3 4 4 i j k i j k i j k − + − − − − form the vertices of a right angled triangle
1
5397-5400
Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, –1) are collinear 17 Show that the vectors ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 2 , 3 5 and 3 4 4 i j k i j k i j k − + − − − − form the vertices of a right angled triangle 18
1
5398-5401
17 Show that the vectors ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 2 , 3 5 and 3 4 4 i j k i j k i j k − + − − − − form the vertices of a right angled triangle 18 If ar is a nonzero vector of magnitude ‘a’ and λ a nonzero scalar, then λ ar is unit vector if (A) λ = 1 (B) λ = – 1 (C) a = |λ| (D) a = 1/|λ| 10
1
5399-5402
Show that the vectors ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 2 , 3 5 and 3 4 4 i j k i j k i j k − + − − − − form the vertices of a right angled triangle 18 If ar is a nonzero vector of magnitude ‘a’ and λ a nonzero scalar, then λ ar is unit vector if (A) λ = 1 (B) λ = – 1 (C) a = |λ| (D) a = 1/|λ| 10 6
1
5400-5403
18 If ar is a nonzero vector of magnitude ‘a’ and λ a nonzero scalar, then λ ar is unit vector if (A) λ = 1 (B) λ = – 1 (C) a = |λ| (D) a = 1/|λ| 10 6 3 Vector (or cross) product of two vectors In Section 10
1
5401-5404
If ar is a nonzero vector of magnitude ‘a’ and λ a nonzero scalar, then λ ar is unit vector if (A) λ = 1 (B) λ = – 1 (C) a = |λ| (D) a = 1/|λ| 10 6 3 Vector (or cross) product of two vectors In Section 10 2, we have discussed on the three dimensional right handed rectangular coordinate system
1
5402-5405
6 3 Vector (or cross) product of two vectors In Section 10 2, we have discussed on the three dimensional right handed rectangular coordinate system In this system, when the positive x-axis is rotated counterclockwise © NCERT not to be republished VECTOR ALGEBRA 449 into the positive y-axis, a right handed (standard) screw would advance in the direction of the positive z-axis (Fig 10
1
5403-5406
3 Vector (or cross) product of two vectors In Section 10 2, we have discussed on the three dimensional right handed rectangular coordinate system In this system, when the positive x-axis is rotated counterclockwise © NCERT not to be republished VECTOR ALGEBRA 449 into the positive y-axis, a right handed (standard) screw would advance in the direction of the positive z-axis (Fig 10 22(i))
1
5404-5407
2, we have discussed on the three dimensional right handed rectangular coordinate system In this system, when the positive x-axis is rotated counterclockwise © NCERT not to be republished VECTOR ALGEBRA 449 into the positive y-axis, a right handed (standard) screw would advance in the direction of the positive z-axis (Fig 10 22(i)) In a right handed coordinate system, the thumb of the right hand points in the direction of the positive z-axis when the fingers are curled in the direction away from the positive x-axis toward the positive y-axis (Fig 10
1
5405-5408
In this system, when the positive x-axis is rotated counterclockwise © NCERT not to be republished VECTOR ALGEBRA 449 into the positive y-axis, a right handed (standard) screw would advance in the direction of the positive z-axis (Fig 10 22(i)) In a right handed coordinate system, the thumb of the right hand points in the direction of the positive z-axis when the fingers are curled in the direction away from the positive x-axis toward the positive y-axis (Fig 10 22(ii))
1
5406-5409
22(i)) In a right handed coordinate system, the thumb of the right hand points in the direction of the positive z-axis when the fingers are curled in the direction away from the positive x-axis toward the positive y-axis (Fig 10 22(ii)) Fig 10
1
5407-5410
In a right handed coordinate system, the thumb of the right hand points in the direction of the positive z-axis when the fingers are curled in the direction away from the positive x-axis toward the positive y-axis (Fig 10 22(ii)) Fig 10 22 (i), (ii) Definition 3 The vector product of two nonzero vectors aand rb r , is denoted by a b r r and defined as a b × r r = ˆ | || a b|sin n θ r r , where, θ is the angle between aand rb r , 0 ≤ θ ≤ π and ˆn is a unit vector perpendicular to both a and rb r , such that ˆ , a b and n rr form a right handed system (Fig 10
1
5408-5411
22(ii)) Fig 10 22 (i), (ii) Definition 3 The vector product of two nonzero vectors aand rb r , is denoted by a b r r and defined as a b × r r = ˆ | || a b|sin n θ r r , where, θ is the angle between aand rb r , 0 ≤ θ ≤ π and ˆn is a unit vector perpendicular to both a and rb r , such that ˆ , a b and n rr form a right handed system (Fig 10 23)
1
5409-5412
Fig 10 22 (i), (ii) Definition 3 The vector product of two nonzero vectors aand rb r , is denoted by a b r r and defined as a b × r r = ˆ | || a b|sin n θ r r , where, θ is the angle between aand rb r , 0 ≤ θ ≤ π and ˆn is a unit vector perpendicular to both a and rb r , such that ˆ , a b and n rr form a right handed system (Fig 10 23) i
1
5410-5413
22 (i), (ii) Definition 3 The vector product of two nonzero vectors aand rb r , is denoted by a b r r and defined as a b × r r = ˆ | || a b|sin n θ r r , where, θ is the angle between aand rb r , 0 ≤ θ ≤ π and ˆn is a unit vector perpendicular to both a and rb r , such that ˆ , a b and n rr form a right handed system (Fig 10 23) i e
1
5411-5414
23) i e , the right handed system rotated from ato rb r moves in the direction of ˆn
1
5412-5415
i e , the right handed system rotated from ato rb r moves in the direction of ˆn If either 0 or 0 a b = r= r r r , then θ is not defined and in this case, we define 0 a ×b r= r r
1
5413-5416
e , the right handed system rotated from ato rb r moves in the direction of ˆn If either 0 or 0 a b = r= r r r , then θ is not defined and in this case, we define 0 a ×b r= r r Observations 1
1
5414-5417
, the right handed system rotated from ato rb r moves in the direction of ˆn If either 0 or 0 a b = r= r r r , then θ is not defined and in this case, we define 0 a ×b r= r r Observations 1 a ×b r r is a vector
1
5415-5418
If either 0 or 0 a b = r= r r r , then θ is not defined and in this case, we define 0 a ×b r= r r Observations 1 a ×b r r is a vector 2
1
5416-5419
Observations 1 a ×b r r is a vector 2 Let aand rb r be two nonzero vectors
1
5417-5420
a ×b r r is a vector 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