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CBSE NCERT Solutions for Class 12 Maths Chapter 10
Back of Chapter Questions
Exercise: 10.1
1. Represent graphically a displacement of 40 km, 30o east of north. [2 Marks]
Solution:
[1 Mark]
Here, vector ππββββ β represents the displacement of 40 km, 30o East of North. [1 Mark]
2. Classify the following measures as scalars and vectors. [1 Mark each]
(i) 10 kg
(ii) 2 metres north-west
(iii) 40o
(iv) 40 watt
(v) 10β19 coulomb
(vi) 20 m s2β
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Solution:
(i) 10 kg is a scalar quantity because it involves only magnitude. [1 Mark]
(ii) 2 meters north-west is a vector quantity as it involves both magnitude and direction.
[1 Mark]
(iii) 40o is a scalar quantity as it involves only magnitude. [1 Mark]
(iv) 40 watts is a scalar quantity as it involves only magnitude. [1 Mark]
(v) 10β19 coulomb is a scalar quantity as it involves only magnitude. [1 Mark]
(vi) 20 m s2β is a vector quantity as it involves magnitude as well as direction. [1 Mark]
3. Classify the following as scalar and vector quantities. [1 Mark each]
(i) time period
(ii) distance
(iii) force
(iv) velocity
(v) work done
Solution:
(i) Time period is a scalar quantity as it involves only magnitude. [1 Mark]
(ii) Distance is a scalar quantity as it involves only magnitude. [1 Mark]
(iii) Force is a vector quantity as it involves both magnitude and direction. [1 Mark]
(iv) Velocity is a vector quantity as it involves both magnitude as well as direction. [1 Mark]
(v) Work done is a scalar quantity as it involves only magnitude. [1 Mark]
4. In Figure, identify the following vectors. [1 Mark each]
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(i) Coinitial
(ii) Equal
(iii) Collinear but not equal
Solution:
(i) Vectors π and π are coinitial because they have the same initial point. [1 Mark]
(ii) Vectors οΏ½βοΏ½ are π equal because they have the same magnitude and direction. [1 Mark]
(iii) Vectors π and οΏ½βοΏ½ are collinear but not equal. This is because although they are parallel, their directions are not the same. [1 Mark]
5. Answer the following as true or false. [1 Mark each]
(i) π and βπ are collinear.
(ii) Two collinear vectors are always equal in magnitude.
(iii) Two vectors having same magnitude are collinear.
(iv) Two collinear vectors having the same magnitude are equal.
Solution:
(i) True. Two vectors are collinear if they are parallel to line the same line.
Vectors π and βπ are parallel to the same line οΏ½ββοΏ½ . [1 Mark]
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So, π and βπ are collinear.
(ii) False.
Collinear vectors are those vectors that are parallel to the same line.
Here, π and οΏ½βοΏ½ are parallel to οΏ½ββοΏ½
Hence, collinear.
But, π and οΏ½βοΏ½ are not equal in magnitude. [1 Mark]
(iii) False. Two or more vectors are collinear if they are parallel to the same line.
πβ
and πβ
are equal in magnitude but not parallel to the same line
Hence, πβ
and πβ
are not collinear. [1 Mark]
(iv) False
Two or more vectors are equal if they have the same magnitude and same direction Collinear vectors may have the same magnitude but are not equal.
Hence false. [1 Mark]
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Exercise: 10.2
1. Compute the magnitude of the following vectors: [4 Marks]
π = πΜ + πΜ + οΏ½ΜοΏ½; οΏ½βοΏ½ = 2πΜ β 7πΜ β 3οΏ½ΜοΏ½; π =1
β3πΜ +
1
β3πΜ β
1
β3οΏ½ΜοΏ½
Solution:
Given vectors are π = πΜ + πΜ + οΏ½ΜοΏ½; οΏ½βοΏ½ = 2πΜ β 7πΜ β 3οΏ½ΜοΏ½; π =1
β3πΜ +
1
β3πΜ β
1
β3οΏ½ΜοΏ½ [1 Mark]
Magnitude of π is |π | = β(1)2 + (1)2 + (1)2 = β3
Hence, |π | = β3 [1 Mark]
Magnitude of οΏ½βοΏ½ is |οΏ½βοΏ½ | = β(2)2 + (β7)2 + (β3)2
= β4 + 49 + 9
= β62
Hence, |οΏ½βοΏ½ | = β62 [1 Mark]
Magnitude of π is |π | = β(1
β3)2+ (
1
β3)2+ (β
1
β3)2
= β1
3+
1
3+
1
3= 1
Hence, |π | = 1 [1 Mark]
2. Write two different vectors having same magnitude. [2 Marks]
Solution:
Let π = (βπΜ β 2πΜ + 3οΏ½ΜοΏ½) and οΏ½βοΏ½ = (2πΜ + πΜ β 3οΏ½ΜοΏ½)
Now, magnitude of π is |π | = β(β1)2 + (β2)2 + 32 = β1 + 4 + 9 = β14 [1 Mark]
|οΏ½βοΏ½ | = β22 + 12 + (β3)2 = β4 + 1 + 9 = β14 [1 Mark]
Hence, π and οΏ½βοΏ½ are two different vectors having the same magnitude.
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3. Write two different vectors having same direction. [2 Marks]
Solution:
Let π = (πΜ + πΜ + οΏ½ΜοΏ½) and οΏ½βοΏ½ = (2πΜ + 2πΜ + 2οΏ½ΜοΏ½).
The direction cosines of π are given by,
π =1
β12+12+12=
1
β3, π =
1
β12+12+12=
1
β3, and π =
1
β12+12+12=
1
β3 [1 Mark]
The direction cosines of οΏ½βοΏ½ are given by,
π =2
β22+22+22=
2
2β3=
1
β3, π =
2
β22+22+22=
2
2β3=
1
β3,
And π =2
β22+22+22=
2
2β3=
1
β3 [1 Mark]
The direction cosines of π and οΏ½βοΏ½ are the same. Hence, the two vectors have the same direction.
4. Find the values of π₯ and π¦ so that the vectors 2πΜ + 3πΜ and π₯πΜ + οΏ½ΜοΏ½ are equal [1 Mark]
Solution:
The two vectors 2πΜ + 3πΜ and π₯πΜ + οΏ½ΜοΏ½ will be equal if their corresponding components are equal.
Hence, the required values of π₯ and π¦ are 2 and 3 respectively. [1 Mark]
5. Find the scalar and vector components of the vector with initial point (2,1) and terminal point
(β 5,7). [1 Mark]
Solution:
Let the given points be π(2, 1) and π(β 5, 7)
The vector with the initial point π(2, 1) and terminal point π(β 5, 7) can be given by,
PQββββ β = (β5 β 2)πΜ + (7 β 1)πΜ [π
π Mark]
β PQββββ β = β7πΜ + 6πΜ
Hence, the required scalar components are β 7 and 6 while the vector components are β7π Μand
6πΜ. [π
π Mark]
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6. Find the sum of the vectors π = πΜ β 2πΜ + οΏ½ΜοΏ½, οΏ½βοΏ½ = β2οΏ½ΜοΏ½ + 4πΜ + 5οΏ½ΜοΏ½ and
π = πΜ β 6πΜ β 7οΏ½ΜοΏ½ [1 Mark]
Solution:
The given vectors are π = πΜ β 2πΜ + οΏ½ΜοΏ½, οΏ½βοΏ½ = β2πΜ + 4πΜ + 5οΏ½ΜοΏ½ and π = πΜ β 6πΜ β 7οΏ½ΜοΏ½
β΄ π + οΏ½βοΏ½ + π = (1 β 2 + 1)πΜ + (β2 + 4 β 6)πΜ + (1 + 5 β 7)οΏ½ΜοΏ½ [π
π Mark]
= 0 β πΜ β 4πΜ β 1 β οΏ½ΜοΏ½
= β4πΜ β οΏ½ΜοΏ½ [π
π Mark]
7. Find the unit vector in the direction of the vector π = πΜ + πΜ + 2οΏ½ΜοΏ½. [1 Mark]
Solution:
The unit vector οΏ½ΜοΏ½ in the direction of vector π = πΜ + πΜ + 2οΏ½ΜοΏ½.
|π | = β12 + 12 + 22 = β1 + 1 + 4 = β6 [π
π Mark]
β΄ οΏ½ΜοΏ½ =οΏ½βοΏ½
|οΏ½βοΏ½ |=
οΏ½ΜοΏ½+οΏ½ΜοΏ½+2οΏ½ΜοΏ½
β6=
1
β6πΜ +
1
β6πΜ +
2
β6οΏ½ΜοΏ½ [
π
π Mark]
8. Find the unit vector in the direction of vector,ππββββ β, where π and π are the points (1, 2, 3) and (4, 5, 6), respectively. [2 Marks]
Solution:
The given points are π(1, 2, 3) and π(4, 5, 6).
β΄ PQββββ β = (4 β 1)πΜ + (5 β 2)πΜ + (6 β 3)οΏ½ΜοΏ½ = 3πΜ + 3πΜ + 3οΏ½ΜοΏ½
|PQββββ β| = β32 + 32 + 32 = β9 + 9 + 9 = β27 = 3β3 [1 Mark]
Hence, the unit vector in the direction of ππββββ β is
PQββββββ
|PQββββββ |=
3οΏ½ΜοΏ½+3οΏ½ΜοΏ½+3οΏ½ΜοΏ½
3β3=
1
β3πΜ +
1
β3πΜ +
1
β3οΏ½ΜοΏ½ [1 Mark]
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9. For given vectors, π = 2πΜ β πΜ + 2οΏ½ΜοΏ½ and οΏ½βοΏ½ = βπΜ + πΜ β οΏ½ΜοΏ½find the unit vector in the direction of the
vector π + οΏ½βοΏ½ [2 Marks]
Solution:
The given vectors are π = 2πΜ β πΜ + 2οΏ½ΜοΏ½ and οΏ½βοΏ½ = βπΜ + πΜ β οΏ½ΜοΏ½.
π = 2πΜ β πΜ + 2οΏ½ΜοΏ½
οΏ½βοΏ½ = βπΜ + πΜ β οΏ½ΜοΏ½
β΄ π + οΏ½βοΏ½ = (2 β 1)πΜ + (β1 + 1)πΜ + (2 β 1)οΏ½ΜοΏ½ = 1πΜ + 0πΜ + 1οΏ½ΜοΏ½ = πΜ + οΏ½ΜοΏ½ [1 Mark]
|π + οΏ½βοΏ½ | = β12 + 12 = β2
Hence, the unit vector in the direction of (π + οΏ½βοΏ½ ) is
(οΏ½βοΏ½ +οΏ½βοΏ½ )
|οΏ½βοΏ½ +οΏ½βοΏ½ |=
οΏ½ΜοΏ½+οΏ½ΜοΏ½
β2=
1
2πΜ +
1
β2οΏ½ΜοΏ½ [1 Mark]
10. Find a vector in the direction of vector 5πΜ β πΜ + 2οΏ½ΜοΏ½ which has magnitude 8 units.[2 Marks]
Solution:
Let π = 5πΜ β πΜ + 2οΏ½ΜοΏ½
β΄ |π | = β52 + (β1)2 + 22 = β25 + 1 + 4 = β30
β΄ οΏ½ΜοΏ½ =οΏ½βοΏ½
|οΏ½βοΏ½ |=
5οΏ½ΜοΏ½βοΏ½ΜοΏ½+2οΏ½ΜοΏ½
β30 [1 Mark]
Hence, the unit vector in the direction of vector 5πΜ β πΜ + 2οΏ½ΜοΏ½ which has magnitude 8 units is given by,
8οΏ½ΜοΏ½ = 8 (5οΏ½ΜοΏ½ β πΜ + 2οΏ½ΜοΏ½
β30) =
40
β30πΜ β
8
β30πΜ +
16
β30οΏ½ΜοΏ½
= 8(5π β π + 2οΏ½βοΏ½
β30)
=40
β30π β
8
β30π +
16
β30οΏ½βοΏ½ [1 Mark]
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11. Show that the vectors 2πΜ β 3πΜ + 4οΏ½ΜοΏ½ and β4οΏ½ΜοΏ½ + 6πΜ β 8οΏ½ΜοΏ½ are collinear. [2 Marks]
Solution:
Let π = 2πΜ β 3πΜ + 4οΏ½ΜοΏ½ and οΏ½βοΏ½ = β4πΜ + 6πΜ β 8οΏ½ΜοΏ½.
It is observed that οΏ½βοΏ½ = β4πΜ + 6πΜ β 8οΏ½ΜοΏ½ = β2(2πΜ β 3πΜ + 4οΏ½ΜοΏ½) = β2π [1 Mark]
β΄ οΏ½βοΏ½ = ππ
Where,
π = β2
Hence, the given vectors are collinear. [1 Mark]
12. Find the direction cosines of the vector πΜ + 2πΜ + 3οΏ½ΜοΏ½ [1 Mark]
Solution:
Let π = πΜ + 2πΜ + 3οΏ½ΜοΏ½
β΄ |π | = β12 + 22 + 32 = β1 + 4 + 9 = β14 [π
π Mark]
Hence, the direction cosines of π are (1
β14,
2
β14,
3
β14). [
π
π Mark]
13. Find the direction cosines of the vector joining the points π΄(1,2, β 3) and π΅(β 1, β 2,1) directed
from π΄ to π΅. [1 Mark]
Solution:
The given points are π΄(1,2, β 3) and π΅(β 1, β 2,1).
β΄ ABββββ β = (β1 β 1)πΜ + (β2 β 2)πΜ + {1 β (β3)}οΏ½ΜοΏ½
β ABββββ β = β2πΜ β 4πΜ + 4οΏ½ΜοΏ½ [π
π Mark]
β΄ |ABββββ β| = β(β2)2 + (β4)2 + 42 = β4 + 16 + 16 = β36 = 6
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Hence, the direction cosines of π΄π΅ββββ β are (β2
6, β
4
6,4
6) = (β
1
3, β
2
3,2
3). [
π
π Mark]
14. Show that the vector πΜ + πΜ + οΏ½ΜοΏ½ is equally inclined to the axes ππ,ππ, and ππ.[1 Mark]
Solution:
Let π = πΜ + πΜ + οΏ½ΜοΏ½
Then,
|π | = β12 + 12 + 12 = β3
Therefore, the direction cosines of π are (1
β3,
1
β3,
1
β3). [
π
π Mark]
Now, let πΌ, π½, and πΎ be the angles formed by π with the positive directions of π₯, π¦, and π§ axes.
Then, we have
cosπΌ =1
β3, cosπ½ =
1
β3, cosπΎ =
1
β3
Hence, the given vector is equally inclined to axes ππ,ππ, and ππ. [π
π Mark]
15. Find the position vector of a point π which divides the line joining two points π and π whose
position vectors are πΜ + 2πΜ β οΏ½ΜοΏ½ and βπΜ + πΜ + οΏ½ΜοΏ½ respectively, in the ration 2: 1
(i) internally
(ii) externally
Solution:
The position vector of point π dividing the line segment joining two points π and π in the ratio m:n is given by:
(i) Internally: ποΏ½βοΏ½ +ποΏ½βοΏ½
π+π [1 Mark]
(ii) Externally: ποΏ½βοΏ½ βποΏ½βοΏ½
πβπ [1 Mark]
Position vectors of π and π are given as:
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OPββββ β = πΜ + 2πΜ β οΏ½ΜοΏ½ and OQββββ β = βπΜ + πΜ + οΏ½ΜοΏ½
(i) The position vector of point π which divides the line joining two points π and π internally in the ratio 2: 1 is given by,
ORββββ β =2(βπΜ + πΜ + οΏ½ΜοΏ½) + 1(πΜ + 2πΜ β οΏ½ΜοΏ½)
2 + 1=
(β2πΜ + 2πΜ + 2οΏ½ΜοΏ½) + (πΜ + 2πΜ β οΏ½ΜοΏ½)
3
=βοΏ½ΜοΏ½+4οΏ½ΜοΏ½+οΏ½ΜοΏ½
3= β
1
3πΜ +
4
3πΜ +
1
3οΏ½ΜοΏ½ [1 Mark]
(ii) The position vector of point π which divides the line joining two points π and π externally in the ratio 2: 1 is given by,
ORββββ β =2(βπΜ + πΜ + οΏ½ΜοΏ½) β 1(πΜ + 2πΜ β οΏ½ΜοΏ½)
2 β 1= (β2πΜ + 2πΜ + 2οΏ½ΜοΏ½) β (πΜ + 2πΜ β οΏ½ΜοΏ½)
= β3πΜ + 3οΏ½ΜοΏ½ [1 Mark]
16. Find the position vector of the mid-point of the vector joining the points π(2, 3, 4) and
π(4, 1, β 2). [1 Mark]
Solution:
The position vector of mid-point π of the vector joining points π(2, 3, 4) and π(4, 1, β 2) is given
by,
ORββββ β =(2οΏ½ΜοΏ½+3οΏ½ΜοΏ½+4οΏ½ΜοΏ½)+(4οΏ½ΜοΏ½+οΏ½ΜοΏ½β2οΏ½ΜοΏ½)
2=
(2+4)οΏ½ΜοΏ½+(3+1)οΏ½ΜοΏ½+(4β2)οΏ½ΜοΏ½
2 [
π
π Mark]
=6οΏ½ΜοΏ½+4οΏ½ΜοΏ½+2οΏ½ΜοΏ½
2= 3πΜ + 2πΜ + οΏ½ΜοΏ½ [
π
π Mark]
17. Show that the points π΄, π΅ and πΆ with position vectors,
π = 3πΜ β 4πΜ β 4οΏ½ΜοΏ½, οΏ½βοΏ½ = 2πΜ β πΜ + οΏ½ΜοΏ½ and π = πΜ β 3πΜ β 5οΏ½ΜοΏ½ respectively form the vertices of a right-angled triangle. [2 Marks]
Solution:
Position vectors of points π΄, π΅ and πΆ are respectively given as:
π = 3πΜ β 4πΜ β 4οΏ½ΜοΏ½, οΏ½βοΏ½ = 2πΜ β πΜ + οΏ½ΜοΏ½ and π = πΜ β 3πΜ β 5οΏ½ΜοΏ½
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β΄ ABββββ β = οΏ½βοΏ½ β π = (2 β 3)πΜ + (β1 + 4)πΜ + (1 + 4)οΏ½ΜοΏ½ = βπΜ + 3πΜ + 5οΏ½ΜοΏ½
BCββββ β = π β οΏ½βοΏ½ = (1 β 2)πΜ + (β3 + 1)πΜ + (β5 β 1)οΏ½ΜοΏ½ = βπΜ β 2πΜ β 6οΏ½ΜοΏ½
CAββββ β = π β π = (3 β 1)πΜ + (β4 + 3)πΜ + (β4 + 5)οΏ½ΜοΏ½ = 2πΜ β πΜ + οΏ½ΜοΏ½
β΄ |π΄π΅ββββ β|2= (β1)2 + 32 + 52 = 1 + 9 + 25 = 35 [1 Mark]
|π΅πΆββββ β|2
= (β1)2 + (β2)2 + (β6)2 = 1 + 4 + 36 = 41
|πΆπ΄ββββ β|2= 22 + (β1)2 + 12 = 4 + 1 + 1 = 6
β΄ |π΄π΅ββββ β|2+ |πΆπ΄ββββ β|
2= 36 + 6 = 41 = |π΅πΆββββ β|
2
Hence, π΄π΅πΆ is a right-angled triangle. [1 Mark]
18. In triangle π΄π΅πΆ which of the following is not true: [4 Marks]
A. ABββββ β + BCββββ β + CAββββ β = 0β
B. ABββββ β + BCββββ β β ACββββ β = 0β
C. ABββββ β + BCββββ β β CAββββ β = 0β
D. ABββββ β β CBββββ β + CAββββ β = 0β
Solution:
On applying the triangle law of addition in the given triangle, we have:
ABββββ β + BCββββ β = ACββββ β . . . (1)
β ABββββ β + BCββββ β = βCAββββ β
β ABββββ β + BCββββ β + CAββββ β = 0β . . . (2)
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β΄ The equation given in alternative A is true. [1 Mark]
ABββββ β + BCββββ β = ACββββ β
β ABββββ β + BCββββ β β ACββββ β = 0β
β΄ The equation given in alternative B is true. [1 Mark]
From equation (2), we have:
ABββββ β β CBββββ β + CAββββ β = 0β
β΄ The equation given in alternative D is true. [1 Mark]
Now, consider the equation given in alternative C:
ABββββ β + BCββββ β β CAββββ β = 0β
β ABββββ β + BCββββ β = CAββββ β . . . (3)
From equations (1) and (3), we have:
ACββββ β = CAββββ β
β ACββββ β = βACββββ β
β ACββββ β + ACββββ β = 0β
β 2ACββββ β = 0β
β ACββββ β = 0β , which is not true.
Hence, the equation given in alternative C is incorrect. [1 Mark]
The correct Answer is C.
19. If π and οΏ½βοΏ½ are two collinear vectors, then which of the following are incorrect:[4 Marks]
A. οΏ½βοΏ½ = ππ , for some scalar π
B. π = Β±οΏ½βοΏ½
C. the respective components of π and οΏ½βοΏ½ are proportional
D. both the vectors π and οΏ½βοΏ½ have same direction, but different magnitudes
Solution:
If π and οΏ½βοΏ½ are two collinear vectors, then they are parallel.
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Therefore, we have:
οΏ½βοΏ½ = ππ (For some scalar π) [1 Mark]
If π = Β±1, then π = Β±οΏ½βοΏ½
If π = π1πΜ + π2πΜ + π3οΏ½ΜοΏ½ and οΏ½βοΏ½ = π1πΜ + π2πΜ + π3οΏ½ΜοΏ½, then
οΏ½βοΏ½ = ππ
β π1οΏ½ΜοΏ½ + π2πΜ + π3οΏ½ΜοΏ½ = π(π1πΜ + π2πΜ + π3οΏ½ΜοΏ½)
β π1οΏ½ΜοΏ½ + π2πΜ + π3οΏ½ΜοΏ½ = (ππ1)πΜ + (ππ2)πΜ + (ππ3)οΏ½ΜοΏ½
β π1 = ππ1, π2 = ππ2, π3 = ππ3
βπ1
π1=
π2
π2=
π3
π3= π [1 Mark]
Thus, the respective components of π and οΏ½βοΏ½ are proportional.
However, vectors π and οΏ½βοΏ½ can have different directions.
Hence, the statement given in D is incorrect.
The correct Answer is D. [2 Marks]
Exercise: 10.3
1. Find the angle between two vectors π and οΏ½βοΏ½ and with magnitudes β3 and 2, respectively having
π β οΏ½βοΏ½ = β6. [2 Marks]
Solution:
It is given that,
|π | = β3, |οΏ½βοΏ½ | = 2 and π β οΏ½βοΏ½ = β6
π β οΏ½βοΏ½ = |π ||οΏ½βοΏ½ |cosπ
Now, we know that
β΄ β6 = β3 Γ 2 Γ cosπ [1 Mark]
β cosπ =β6
β3 Γ 2
β cosπ =1
β2
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β π =π
4
Hence, the angle between the given vectors π and οΏ½βοΏ½ is π
4 [1 Mark]
2. Find the angle between the vectors πΜ β 2πΜ + 3οΏ½ΜοΏ½ and 3πΜ β 2πΜ + οΏ½ΜοΏ½ [2 Marks]
Solution:
The given vectors are π = πΜ β 2πΜ + 3οΏ½ΜοΏ½ and οΏ½βοΏ½ = 3πΜ β 2πΜ + οΏ½ΜοΏ½
|π | = β12 + (β2)2 + 32 = β1 + 4 + 9 = β14
|οΏ½βοΏ½ | = β32 + (β2)2 + 12 = β9 + 4 + 1 = β14
Now, π β οΏ½βοΏ½ = (πΜ β 2πΜ + 3οΏ½ΜοΏ½)(3πΜ β 2πΜ + οΏ½ΜοΏ½)
= 1.3 + (β2)(β2) + 3.1
= 10 [1 Mark]
Also, we know that π β οΏ½βοΏ½ = |π ||οΏ½βοΏ½ |cosπ
β΄ 10 = β14β14cosπ
β cosπ =10
14
β π = cosβ1 (5
7) [1 Mark]
3. Find the projection of the vector πΜ β πΜ on the vector πΜ + πΜ. [1 Mark]
Solution:
Let π = πΜ β πΜ and οΏ½βοΏ½ = πΜ + π.Μ [π
π Mark]
Now, projection of vector π on οΏ½βοΏ½ is given by,
1
|οΏ½βοΏ½ |(π β οΏ½βοΏ½ ) =
1
β1+1{1.1 + (β1)(1)} =
1
β2(1 β 1) = 0 [
π
π Mark]
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Hence, the projection of vector π on οΏ½βοΏ½ is 0.
4. Find the projection of the vector πΜ + 3πΜ + 7οΏ½ΜοΏ½ on the vector 7πΜ β πΜ + 8οΏ½ΜοΏ½. [1 Mark]
Solution:
Let π = πΜ + 3πΜ + 7οΏ½ΜοΏ½ and οΏ½ΜοΏ½ = 7πΜ β πΜ + 8οΏ½ΜοΏ½.
Now, projection of vector π on οΏ½βοΏ½ is given by,
1
|οΏ½βοΏ½ |(π β οΏ½βοΏ½ ) =
1
β72+(β1)2+82{1(7) + 3(β1) + 7(8)} =
7β3+56
β49+1+64=
60
β114 [1 Mark]
5. Show that each of the given three vectors is a unit vector:
1
7(2πΜ + 3πΜ + 6οΏ½ΜοΏ½),
1
7(3πΜ β 6πΜ + 2οΏ½ΜοΏ½),
1
7(6πΜ + 2πΜ β 3οΏ½ΜοΏ½)
Also, show that they are mutually perpendicular to each other. [2 Marks]
Solution:
Let π =1
7(2πΜ + 3πΜ + 6οΏ½ΜοΏ½) =
2
7πΜ +
3
7πΜ +
6
7οΏ½ΜοΏ½,
οΏ½βοΏ½ =1
7(3πΜ β 6πΜ + 2οΏ½ΜοΏ½) =
3
7πΜ β
6
7πΜ +
2
7οΏ½ΜοΏ½
π =1
7(6πΜ + 2πΜ β 3οΏ½ΜοΏ½) =
6
7πΜ +
2
7πΜ β
3
7οΏ½ΜοΏ½
|π | = β(2
7)2
+ (3
7)2
+ (6
7)2
= β4
49+
9
49+
36
49= 1
|οΏ½βοΏ½ | = β(3
7)2
+ (β6
7)2
+ (2
7)2
= β9
49+
36
49+
4
49= 1
|π | = β(6
7)2+ (
2
7)2+ (β
3
7)2= β
36
49+
4
49+
9
49= 1 [1 Mark]
Thus, each of the given three vectors is a unit vector.
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π β οΏ½βοΏ½ =2
7Γ
3
7+
3
7Γ (
β6
7) +
6
7Γ
2
7=
6
49β
18
49+
12
49= 0
οΏ½βοΏ½ β π =3
7Γ
6
7+ (
β6
7) Γ
2
7+
2
7Γ (
β3
7) =
18
49β
12
49β
6
49= 0
π β π =6
7Γ
2
7+
2
7Γ
3
7+ (
β3
7) Γ
6
7=
12
49+
6
49β
18
49= 0
Hence, the given three vectors are mutually perpendicular to each other. [1 Mark]
6. Find
|π | and |οΏ½βοΏ½ |, if (π + οΏ½βοΏ½ ) β (π β οΏ½βοΏ½ ) = 8 and |π | = 8|οΏ½βοΏ½ |. [2 Marks]
Solution:
(π β οΏ½βοΏ½ ) β (π β οΏ½βοΏ½ ) = 8
β π β π β π β οΏ½βοΏ½ + οΏ½βοΏ½ β π β οΏ½βοΏ½ β οΏ½βοΏ½ = 8
β |π |2 β |οΏ½βοΏ½ |2= 8
β (8|οΏ½βοΏ½ |)2β |οΏ½βοΏ½ |
2= 8 [|π | = 8|οΏ½βοΏ½ |]
β 64|οΏ½βοΏ½ |2β |οΏ½βοΏ½ |
2= 8
β 63|οΏ½βοΏ½ |2= 8
β |οΏ½βοΏ½ |2=
8
63
β |οΏ½βοΏ½ | = β8
63 [Magnitude of a vector is non-negative]
β |οΏ½βοΏ½ | =2β2
3β7 [1 Mark]
|π | = 8|οΏ½βοΏ½ | =8Γ2β2
3β7=
16β2
3β7 [1 Mark]
7. Evaluate the product (3π β 5οΏ½βοΏ½ ) β (2π + 7οΏ½βοΏ½ ). [1 Mark]
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Solution:
(3π β 5οΏ½βοΏ½ ) β (2π + 7οΏ½βοΏ½ )
= 3π β 2π + 3π β 7οΏ½βοΏ½ β 5οΏ½βοΏ½ β 2π β 5οΏ½βοΏ½ β 7οΏ½βοΏ½ [π
π Mark]
= 6π β π + 21π β οΏ½βοΏ½ β 10π β οΏ½βοΏ½ β 35οΏ½βοΏ½ β οΏ½βοΏ½
= 6|π |2 + 11π β οΏ½βοΏ½ β 35|οΏ½βοΏ½ |2 [
π
π Mark]
8. Find the magnitude of two vectors π and οΏ½βοΏ½ , having the same magnitude and such that the angle
between them is 60o and their scalar product is 1
2. [1 Mark]
Solution:
Let π be the angle between the vectors π and οΏ½βοΏ½ .
It is given that |π | = |οΏ½βοΏ½ |, π β οΏ½βοΏ½ =1
2, and π = 60Β° . . . (1)
We know that π β οΏ½βοΏ½ = |π ||οΏ½βοΏ½ |cosπ
β΄1
2= |π ||π |cos60Β° [Using (1)] [
π
π Mark]
β1
2= |π |2 Γ
1
2
β |π |2 = 1
β |π | = |οΏ½βοΏ½ | = 1 [π
π Mark]
9. Find
|π₯ | if for a unit vector π , (π₯ β π ) β (π₯ + π ) = 12. [1 Mark]
Solution:
(π₯ β π ) β (π₯ + π ) = 12
β π₯ β π₯ + π₯ β π β π β π₯ β οΏ½Μ οΏ½ β π = 12 [π
π Mark]
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β |π₯ |2 β |π |2 = 12
β |π₯ |2 β 1 = 12 [|π | = 1 as π is a unit vector]
β |π₯ |2 = 13
β΄ |π₯ | = β13 [π
π Mark]
10. If π = 2πΜ + 2πΜ + 3οΏ½ΜοΏ½, οΏ½βοΏ½ = βπΜ + 2πΜ + οΏ½ΜοΏ½ and π = 3πΜ + πΜ are such that π + ποΏ½βοΏ½ is perpendicular to π , then find the value of π. [2 Marks]
Solution:
The given vectors are π = 2πΜ + 2πΜ + 3οΏ½ΜοΏ½, οΏ½βοΏ½ = βπΜ + 2πΜ + οΏ½ΜοΏ½ and π = 3πΜ + πΜ.
Now,
π + ποΏ½βοΏ½ = (2πΜ + 2πΜ + 3οΏ½ΜοΏ½) + π(βπΜ + 2πΜ + οΏ½ΜοΏ½) = (2 β π)πΜ + (2 + 2π)πΜ + (3 + π)οΏ½ΜοΏ½
If (π + ποΏ½βοΏ½ ) is perpendicular to π , then
(π + ποΏ½βοΏ½ ) β π = 0 [1 Mark]
β [(2 β π)οΏ½ΜοΏ½ + (2 + 2π)πΜ + (3 + π)οΏ½ΜοΏ½] β (3πΜ + πΜ) = 0
β (2 β π)3 + (2 + 2π)1 + (3 + π)0 = 0
β 6 β 3π + 2 + 2π = 0
β βπ + 8 = 0
β π = 8
Hence, the required value of π is 8. [1 Mark]
11. Show that:
|π |οΏ½βοΏ½ + |οΏ½βοΏ½ |οΏ½Μ οΏ½ is perpendicular to |π |οΏ½βοΏ½ β |οΏ½βοΏ½ |οΏ½Μ οΏ½,
for any two nonzero vectors π and οΏ½βοΏ½ [1 Mark]
Solution:
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(|π |οΏ½βοΏ½ + |οΏ½βοΏ½ |π ) β (|π |οΏ½βοΏ½ β |οΏ½βοΏ½ |π )
= |π |2οΏ½βοΏ½ β οΏ½βοΏ½ β |π ||οΏ½βοΏ½ |οΏ½βοΏ½ β π + |οΏ½βοΏ½ ||π |π β οΏ½βοΏ½ β |οΏ½βοΏ½ |2π β π [
π
π Mark]
= |π |2|οΏ½βοΏ½ |2β |οΏ½βοΏ½ |
2|π |2
= 0
Hence, |π |οΏ½βοΏ½ + |οΏ½βοΏ½ |οΏ½Μ οΏ½ and |π |οΏ½βοΏ½ β |οΏ½βοΏ½ |aβ are perpendicular to each other. [π
π Mark]
12. If, π β π = 0 and π β οΏ½βοΏ½ = 0, then what can be concluded about the vector οΏ½βοΏ½ ? [1 Mark]
Solution:
It is given that π β π = 0 and π β οΏ½βοΏ½ = 0.
Now,
π β π = 0 β |π |2 = 0 β |π | = 0 [π
π Mark]
β΄ π is a zero vector.
Hence, vector οΏ½βοΏ½ satisfying π β οΏ½βοΏ½ = 0 can be any vector. [π
π Mark]
13. If π , οΏ½βοΏ½ , π are unit vectors such thatπ + οΏ½βοΏ½ + π = 0,
find the value ofπ β οΏ½βοΏ½ + οΏ½βοΏ½ β π + π β π [1 Mark]
Solution:
|π + οΏ½βοΏ½ + π |2
= (π + οΏ½βοΏ½ + π ) β (π + οΏ½βοΏ½ + π ) = |π |2 + |οΏ½βοΏ½ |2+ |π |2 + 2(π β οΏ½βοΏ½ + οΏ½βοΏ½ β π + π β π )
β 0 = 1 + 1 + 1 + 2(π β οΏ½βοΏ½ + οΏ½βοΏ½ β π + π β π ) [π
π Mark]
β (π β οΏ½βοΏ½ + οΏ½βοΏ½ β π + π β π ) =β3
2 [
π
π Mark]
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14. If either vector π = 0β or οΏ½βοΏ½ = 0β , then π β οΏ½βοΏ½ = 0. But the converse need not be true. Justify your Answer with an example. [2 Marks]
Solution:
Consider π = 2πΜ + 4πΜ + 3οΏ½ΜοΏ½ and οΏ½βοΏ½ = 3πΜ + 3πΜ β 6οΏ½ΜοΏ½.
Then,
π β οΏ½βοΏ½ = 2.3 + 4.3 + 3(β6) = 6 + 12 β 18 = 0 .[1 Mark]
We now observe that:
|π | = β22 + 42 + 32 = β29
β΄ π β 0β
|οΏ½βοΏ½ | = β32 + 32 + (β6)2 = β54
β΄ οΏ½βοΏ½ β 0β
Hence, the converse of the given statement need not be true. [1 Mark]
15. If the vertices π΄, π΅, πΆ of a triangle π΄π΅πΆ are (1,2,3), (β 1,0,0), (0,1,2), respectively, then find
β π΄π΅πΆ. [β π΄π΅πΆ is the angle between the vectors π΅π΄ββββ β and π΅πΆββββ β] [2 Marks]
Solution:
The vertices of Ξπ΄π΅πΆ are given as π΄(1,2,3), π΅(β 1,0,0), and πΆ(0,1,2).
Also, it is given that β π΄π΅πΆ is the angle between the vectors π΅π΄ββββ β and π΅πΆββββ β.
BAββββ β = {1 β (β1)}οΏ½ΜοΏ½ + (2 β 0)πΜ + (3 β 0)οΏ½ΜοΏ½ = 2πΜ + 2πΜ + 3οΏ½ΜοΏ½
BCββββ β = {0 β (β1)}πΜ + (1 β 0)πΜ + (2 β 0)οΏ½ΜοΏ½ = πΜ + πΜ + 2οΏ½ΜοΏ½
β΄ BAββββ β β BCββββ β = (2πΜ + 2πΜ + 3οΏ½ΜοΏ½) β (πΜ + πΜ + 2οΏ½ΜοΏ½)
= 2 Γ 1 + 2 Γ 1 + 3 Γ 2 = 2 + 2 + 6 = 10 [1 Mark]
|BAββββ β| = β22 + 22 + 32 = β4 + 4 + 9 = β17
|BCββββ β| = β1 + 1 + 22 = β6
Now, it is known that:
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BAββββ β β BCββββ β = |BAββββ β||BCββββ β|cos(β ABC)
β΄ 10 = β17 Γ β6cos(β ABC)
β cos(β ABC) =10
β17 Γ β6
β β ABC = cosβ1 (10
β102) [1 Mark]
16. Show that the points π΄(1,2,7), π΅(2,6,3) and πΆ(3,10, β 1) are collinear. [2 Marks]
Solution:
The given points are π΄(1,2,7), π΅(2,6,3) and πΆ(3,10, β 1).
β΄ ABββββ β = (2 β 1)πΜ + (6 β 2)πΜ + (3 β 7)οΏ½ΜοΏ½ = πΜ + 4πΜ β 4οΏ½ΜοΏ½
BCββββ β = (3 β 2)πΜ + (10 β 6)πΜ + (β1 β 3)οΏ½ΜοΏ½ = πΜ + 4πΜ β 4οΏ½ΜοΏ½
ACββββ β = (3 β 1)πΜ + (10 β 2)πΜ + (β1 β 7)οΏ½ΜοΏ½ = 2πΜ + 8πΜ β 8οΏ½ΜοΏ½ [1 Mark]
|ABββββ β| = β12 + 42 + (β4)2 = β1 + 16 + 16 = β33
|BCββββ β| = β12 + 42 + (β4)2 = β1 + 16 + 16 = β33
|ACββββ β| = β22 + 82 + 82 = β4 + 64 + 64 = β132 = 2β33
β΄ |ACββββ β| = |ABββββ β| + |BCββββ β|
Hence, the given points π΄, π΅, and πΆ are collinear. [1 Mark]
17. Show that the vectors 2πΜ β πΜ + οΏ½ΜοΏ½, πΜ β 3πΜ β 5οΏ½ΜοΏ½ and 3πΜ β 4πΜ β 4οΏ½ΜοΏ½ form the vertices of a right angled triangle. [2 Marks]
Solution:
Let vectors 2πΜ β πΜ + οΏ½ΜοΏ½, πΜ β 3πΜ β 5οΏ½ΜοΏ½ and 3πΜ β 4πΜ β 4οΏ½ΜοΏ½ be position vectors of points π΄, π΅, and πΆ respectively.
i.e., OAββββ β = 2πΜ β πΜ + οΏ½ΜοΏ½, OBββββ β = πΜ β 3πΜ β 5οΏ½ΜοΏ½ and OCββββ β = 3πΜ β 4πΜ β 4οΏ½ΜοΏ½
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Now, vectors ABββββ β, BCββββ β and ACββββ β represent the sides of Ξπ΄π΅πΆ.
i.e., OAββββ β = 2πΜ β πΜ + οΏ½ΜοΏ½, OBββββ β = πΜ β 3πΜ β 5οΏ½ΜοΏ½ and OCββββ β = 3πΜ β 4πΜ β 4οΏ½ΜοΏ½.
β΄ ABββββ β = (1 β 2)πΜ + (β3 + 1)πΜ + (β5 β 1)οΏ½ΜοΏ½ = βπΜ β 2πΜ β 6οΏ½ΜοΏ½
BCββββ β = (3 β 1)πΜ + (β4 + 3)πΜ + (β4 + 5)οΏ½ΜοΏ½ = 2πΜ β πΜ + οΏ½ΜοΏ½
ACββββ β = (2 β 3)πΜ + (β1 + 4)πΜ + (1 + 4)οΏ½ΜοΏ½ = βπΜ + 3πΜ + 5οΏ½ΜοΏ½ [1 Mark]
|ABββββ β| = β(β1)2 + (β2)2 + (β6)2 = β1 + 4 + 36 = β41
|BCββββ β| = β22 + (β1)2 + I2 = β4 + 1 + 1 = β6
|ACββββ β| = β(β1)2 + 32 + 52 = β1 + 9 + 25 = β35
β΄ |BCββββ β|2+ |ACββββ β|
2= 6 + 35 = 41 = |ABββββ β|
2
Hence, Ξπ΄π΅πΆ is a right-angled triangle. [1 Mark]
18. If π is a nonzero vector of magnitude βπβ and π a nonzero scalar, then ππ is unit vector
If [1 Mark]
(A) π = 1
(B) π =β1
(C) π = |π|
(D) π =1
|π|
Solution:
Vector ππ is a unit vector if |ππ | = 1.
Now,
|ππ | = 1
β |π||π | = 1 [π
π Mark]
β |π | =1
|π| [π β 0]
β π =1
|π| [|π | = π]
Hence, vector ππ is a unit vector if π =1
|π|
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The correct Answer is D. [π
π Mark]
Exercise: 10.4
1. Find
|π Γ οΏ½βοΏ½ |, if π = πΜ β 7πΜ + 7οΏ½ΜοΏ½ and οΏ½βοΏ½ = 3πΜ β 2πΜ + 2οΏ½ΜοΏ½ [1 Mark]
Solution:
We have, π = πΜ β 7πΜ + 7οΏ½ΜοΏ½ and οΏ½βοΏ½ = 3πΜ β 2πΜ + 2οΏ½ΜοΏ½
π Γ οΏ½βοΏ½ = |πΜ πΜ οΏ½ΜοΏ½1 β7 73 β2 2
|
= π(Μβ14 + 14) β πΜ(2 β 21) + οΏ½ΜοΏ½(β2 + 21) = 19πΜ + 19οΏ½ΜοΏ½ [π
π Mark]
β΄ |π Γ οΏ½βοΏ½ | = β(19)2 + (19)2 = β2 Γ (19)2 = 19β2 [π
π Mark]
2. Find a unit vector perpendicular to each of the vector π + οΏ½βοΏ½ and π β οΏ½βοΏ½ , where
π = 3πΜ + 2πΜ + 2οΏ½ΜοΏ½ and οΏ½βοΏ½ = πΜ + 2πΜ β 2οΏ½ΜοΏ½ [2 Marks]
Solution:
We have,
π = 3πΜ + 2πΜ + 2οΏ½ΜοΏ½ and οΏ½βοΏ½ = πΜ + 2πΜ β 2οΏ½ΜοΏ½
β΄ π + οΏ½βοΏ½ = 4πΜ + 4πΜ, π β οΏ½βοΏ½ = 2πΜ + 4οΏ½ΜοΏ½
(π + οΏ½βοΏ½ ) Γ (π β οΏ½βοΏ½ ) = |πΜ πΜ οΏ½ΜοΏ½4 4 02 0 4
| = π(Μ16) β πΜ(16) + οΏ½ΜοΏ½(β8) = 16πΜ β 16πΜ β 8οΏ½ΜοΏ½
β΄ |(π + οΏ½βοΏ½ ) Γ (π β οΏ½βοΏ½ )| = β162 + (β16)2 + (β8)2
= β22 Γ 82 + 22 Γ 82 + 82
= 8β22 + 22 + 1 = 8β9 = 8 Γ 3 = 24 [1 Mark]
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Hence, the unit vector perpendicular to each of the vectors π + οΏ½βοΏ½ and π β οΏ½βοΏ½ , is given by the relation,
= Β±(π + οΏ½βοΏ½ ) Γ (π β οΏ½βοΏ½ )
|(π + οΏ½βοΏ½ ) Γ (π β οΏ½βοΏ½ )|= Β±
16πΜ β 16πΜ β 8οΏ½ΜοΏ½
24
= Β±2οΏ½ΜοΏ½β2οΏ½ΜοΏ½βοΏ½ΜοΏ½
3= Β±
2
3πΜ β
2
3πΜ β
1
3οΏ½ΜοΏ½ [1 Mark]
3. If a unit vector π makes an π
3 with π,
π
4 angle with πΜ and an acute angle π with οΏ½ΜοΏ½, then find π and
hence, the compounds of π . [4 Marks]
Solution:
Let unit vector π have (π1, π2, π3) components.
π = π1πΜ + π2πΜ + π3οΏ½ΜοΏ½
Since π is a unit vector,|π | = 1.
Also, it is given that π makes angles π
3 with π,
π
4 with πΜ and an acute angle π with οΏ½ΜοΏ½. Then, we
have:
cosπ
3=
π1
|π |
β1
2= π1 [|π | = 1] [
π
π Mark]
cosπ
4=
π2
|π |
β1
β2= π2 [|π | = 1] [
π
π Mark]
Also, cosπ =π3
|οΏ½βοΏ½ |.
β π3 = cosπ [π
π Mark]
Now,
|π| = 1
β βπ12 + π2
2 + π32 = 1 [
π
π Mark]
β (1
2)2
+ (1
β2)2
+ cos2π = 1
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β1
4+
1
2+ cos2π = 1
β3
4+ cos2π = 1
β cos2π = 1 β3
4=
1
4
β cosπ =1
2β π =
π
3
β΄ π3 = cosπ
3=
1
2
Hence, π =π
3 and the components of π are (
1
2,
1
β2,1
2). [2 Marks]
4. Show that
(π β οΏ½βοΏ½ ) Γ (π + οΏ½βοΏ½ ) = 2(π Γ οΏ½βοΏ½ ) [1 Mark]
Solution:
(π β οΏ½βοΏ½ ) Γ (π + οΏ½βοΏ½ )
= (π β οΏ½βοΏ½ ) Γ π + (π β οΏ½βοΏ½ ) Γ οΏ½βοΏ½ [By distributive of vector product over addition] [π
π Mark]
= π Γ π β οΏ½βοΏ½ Γ π + π Γ οΏ½βοΏ½ β οΏ½βοΏ½ Γ οΏ½βοΏ½ [Again, by distributivity of vector product over addition]
= 0β + π Γ οΏ½βοΏ½ + π Γ οΏ½βοΏ½ β 0β
= 2π Γ οΏ½βοΏ½ [π
π Mark]
5. Find π and π if (2πΜ + 6πΜ + 27οΏ½ΜοΏ½) Γ (πΜ + ππΜ + ποΏ½ΜοΏ½) = 0β . [2 Marks]
Solution:
(2πΜ + 6πΜ + 27οΏ½ΜοΏ½) Γ (πΜ + ππΜ + ποΏ½ΜοΏ½) = 0β
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β |πΜ πΜ οΏ½ΜοΏ½2 6 271 π π
| = 0πΜ + 0πΜ + 0οΏ½ΜοΏ½
β π(Μ6π β 27π) β πΜ(2π β 27) + οΏ½ΜοΏ½(2π β 6) = 0πΜ + 0πΜ + 0οΏ½ΜοΏ½ [1 Mark]
On comparing the corresponding components, we have:
6π β 27π = 0
2π β 27 = 0
2π β 6 = 0
Now,
2π β 6 = 0 β π = 3
2π β 27 = 0 β π =27
2
Hence π = 3 and π =27
2. [1 Mark]
6. Given that π β οΏ½βοΏ½ = 0 and π Γ οΏ½βοΏ½ = 0β .
What can you conclude about the vectors π and οΏ½βοΏ½ ? [1 Mark]
Solution:
π β οΏ½βοΏ½ = 0
Then,
(i) Either |π | = 0 or |οΏ½βοΏ½ | = 0, or π β₯ οΏ½βοΏ½ (in case π and οΏ½βοΏ½ are non-zero)
(ii) Either |π | = 0 or |οΏ½βοΏ½ | = 0, or π β₯ οΏ½βοΏ½ (in case π and οΏ½βοΏ½ are non-zero) [π
π Mark]
But, π and οΏ½βοΏ½ cannot be perpendicular and parallel simultaneously.
Hence, |π | = 0 or |οΏ½βοΏ½ | = 0 [π
π Mark]
7. Let the vectors π , οΏ½βοΏ½ , π given as π1πΜ + π2πΜ + π3οΏ½ΜοΏ½, π1πΜ + π2πΜ + π3οΏ½ΜοΏ½, π1πΜ + π2πΜ + π3οΏ½ΜοΏ½
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Then show that = π Γ (οΏ½βοΏ½ + π ) = π Γ οΏ½βοΏ½ + π Γ π [4 Marks]
Solution:
We have,
π = π1πΜ + π2πΜ + π3οΏ½ΜοΏ½, οΏ½βοΏ½ = π1οΏ½ΜοΏ½ + π2πΜ + π3οΏ½ΜοΏ½, π = π1πΜ + π2πΜ + π3οΏ½ΜοΏ½
(οΏ½βοΏ½ + π ) = (π1 + π1)πΜ + (π2 + π2)πΜ + (π3 + π3)οΏ½ΜοΏ½
Now, π Γ (οΏ½βοΏ½ + π ) |πΜ πΜ οΏ½ΜοΏ½π1 π2 π3
π1 + π1 π2 + π2 π3 + π3
|
= π[Μπ2(π3 + π3) β π3(π2 + π2)] β πΜ[π1(π3 + π3) β π3(π1 + π1)]
+ οΏ½ΜοΏ½[π1(π2 + π2) β π2(π1 + π1)]
= π[Μπ2π3 + π2π3 β π3π2 β π3π2] + πΜ[βπ1π3 β π3π3 + π3π1 + π3π1] + οΏ½ΜοΏ½[π1π2 + π1π2 β π2π1 βπ2π1] . . . (1) [1 Mark]
π Γ οΏ½βοΏ½ = |πΜ πΜ οΏ½ΜοΏ½π1 π2 π3
π1 π2 π3
|
= π[Μπ2π3 β π3π2] + πΜ[π1π3 β π1π3] + οΏ½ΜοΏ½[π1π2 β π2π1] . . . (2) [1 Mark]
π Γ π = |πΜ πΜ οΏ½ΜοΏ½π1 π2 π3
π1 π2 π3
|
= π[Μπ2π3 β π3π2] + πΜ[π3π1 β π1π3] + οΏ½ΜοΏ½[π1π2 β π2π1] . . . (3) [1 Mark]
On adding (2) and (3), we get:
(π Γ οΏ½βοΏ½ ) + (π Γ π )
= π[Μπ2π3 + π2π3 β π3π2 β π3π2] + πΜ[π1π3 + π3π1 β π1π3 β π1π3]
+ οΏ½ΜοΏ½[π1π2 + π1π2 β π2π1 β π2π1] . . . (4)
Now, from (1) and (4), we have:
π Γ (οΏ½βοΏ½ + π ) = π Γ οΏ½βοΏ½ + π Γ π
Hence, the given result is proved. [1 Mark]
8. If either π = 0β or οΏ½βοΏ½ = 0β , then π Γ οΏ½βοΏ½ = 0β .
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Is the converse true? Justify your Answer with an example. [2 Marks]
Solution:
Take any parallel non-zero vectors so that π Γ οΏ½βοΏ½ = 0β .
Let π = 2πΜ + 3πΜ + 4οΏ½ΜοΏ½, οΏ½βοΏ½ = 4πΜ + 6πΜ + 8οΏ½ΜοΏ½
Then,
π Γ οΏ½βοΏ½ = |πΜ πΜ οΏ½ΜοΏ½2 3 44 6 8
| = π(Μ24 β 24) β πΜ(16 β 16) + οΏ½ΜοΏ½(12 β 12) = 0πΜ + 0πΜ + 0οΏ½ΜοΏ½ = 0β
It can now be observed that:
|π | = β22 + 32 + 42 = β29 [1 Mark]
β΄ π β 0β
|οΏ½βοΏ½ | = β42 + 62 + 82 = β116
β΄ οΏ½βοΏ½ β 0β
Hence, the converse of the given statement need not be true. [1 Mark]
9. Find the area of the triangle with vertices π΄(1,1,2), π΅(2,3,5) and πΆ(1,5,5). [2 Marks]
Solution:
The vertices of triangle π΄π΅πΆ are given as π΄(1,1,2), π΅(2,3,5) and πΆ(1,5,5).
The adjacent sides π΄π΅ββββ β and π΅πΆββββ β of Ξπ΄π΅πΆ are given as:
ABββββ β = (2 β 1)πΜ + (3 β 1)πΜ + (5 β 2)οΏ½ΜοΏ½ = πΜ + 2πΜ + 3οΏ½ΜοΏ½
BCββββ β = (1 β 2)πΜ + (5 β 3)πΜ + (5 β 5)οΏ½ΜοΏ½ = βπΜ + 2πΜ [1 Mark]
Area of ΞABC =1
2|ABββββ β Γ BCββββ β|
ABββββ β Γ BCββββ β = |πΜ πΜ οΏ½ΜοΏ½1 2 3β1 2 0
| = πΜ(β6) β πΜ(3) + οΏ½ΜοΏ½(2 + 2) = β6πΜ β 3πΜ + 4οΏ½ΜοΏ½
β΄ |π΄π΅ββββ β Γ π΅πΆββββ β| = β(β6)2 + (β3)2 + 42 = β36 + 9 + 16 = β61 [1 Mark]
Hence, the area of Ξπ΄π΅πΆ is β61
2 square units.
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10. Find the area of the parallelogram whose adjacent sides are determined by the vector π = πΜ β
πΜ + 3οΏ½ΜοΏ½ and οΏ½βοΏ½ = 2πΜ β 7πΜ + οΏ½ΜοΏ½. [2 Marks]
Solution:
The area of the parallelogram whose adjacent sides are π and οΏ½βοΏ½ is |π Γ οΏ½βοΏ½ |.
Adjacent sides are given as:
π = πΜ β πΜ + 3οΏ½ΜοΏ½ and οΏ½βοΏ½ = 2πΜ β 7πΜ + οΏ½ΜοΏ½
β΄ π Γ οΏ½βοΏ½ = |πΜ πΜ οΏ½ΜοΏ½1 β1 32 β7 1
| = πΜ(β1 + 21) β πΜ(1 β 6) + οΏ½ΜοΏ½(β7 + 2) = 20πΜ + 5πΜ β 5οΏ½ΜοΏ½[1 Mark]
|π Γ οΏ½βοΏ½ | = β202 + 52 + 52 = β400 + 25 + 25 = 15β2
Hence, the area of the given parallelogram is 15β2 square units. [1 Mark]
11. Let the vectors π and οΏ½βοΏ½ be such that |π | = 3 and |οΏ½βοΏ½ | =β2
3, then π Γ οΏ½βοΏ½ is a unit vector, if the
angle between π and οΏ½βοΏ½ is [2 Marks]
(A) π
6
(B) π
4
(C) π
3
(D) π
2
Solution:
It is given that |π | = 3 and |οΏ½βοΏ½ | =β2
3
We know that π Γ οΏ½βοΏ½ = |π ||οΏ½βοΏ½ |sinποΏ½ΜοΏ½, where is a unit vector perpendicular to both π and οΏ½βοΏ½ and π
is the angle between π and οΏ½βοΏ½ .
Now, π Γ οΏ½βοΏ½ is a unit vector if |π Γ οΏ½βοΏ½ | = 1
|π Γ οΏ½βοΏ½ | = 1
β ||π ||οΏ½βοΏ½ |π ππποΏ½ΜοΏ½| = 1
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β | ||π ||οΏ½βοΏ½ |π πππ| = 1
β 3 Γβ2
3Γ sinπ = 1 [1 Mark]
β sinπ =1
β2
β π =π
4
Hence, π Γ οΏ½βοΏ½ is a unit vector if the angle between π and οΏ½βοΏ½ is π
4.
The correct Answer is B. [1 Mark]
12. Area of a rectangle having vertices π΄, π΅, πΆ, and π· with position vectors βπΜ +1
2πΜ + 4οΏ½ΜοΏ½, πΜ +
1
2πΜ +
4οΏ½ΜοΏ½, πΜ β1
2πΜ + 4οΏ½ΜοΏ½ and βπΜ β
1
2πΜ + 4οΏ½ΜοΏ½ respectively is [2 Marks]
(A) 1
2
(B) 1
(C) 2
(D) 4
Solution:
The position vectors of vertices π΄, π΅, πΆ, and π· of rectangle π΄π΅πΆπ· are given as:
OAββββ β = βπΜ +1
2πΜ + 4οΏ½ΜοΏ½, OBββββ β = πΜ +
1
2πΜ + 4οΏ½ΜοΏ½, OCββββ β = πΜ β
1
2πΜ + 4οΏ½ΜοΏ½, ODββ ββ β = βπΜ β
1
2πΜ + 4οΏ½ΜοΏ½
The adjacent sides π΄π΅ββββ β and π΅πΆββββ β of the given rectangle are given as:
ABββββ β = (1 + 1)πΜ + (1
2β
1
2) πΜ + (4 β 4)οΏ½ΜοΏ½ = 2πΜ
BCββββ β = (1 β 1)πΜ + (β1
2β
1
2) πΜ + (4 β 4)οΏ½ΜοΏ½ = βπΜ [1 Mark]
β΄ π΄π΅ββββ β Γ π΅πΆββββ β = |πΜ πΜ οΏ½ΜοΏ½2 0 00 β1 0
| = οΏ½ΜοΏ½(β2) = β2οΏ½ΜοΏ½
|ABββββ β Γ ACββββ β| = β(β2)2 = 2
Now, it is known that the area of a parallelogram whose adjacent sides are π and οΏ½βοΏ½ is |π Γ οΏ½βοΏ½ |.
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Hence, the area of the given rectangle is |ABββββ β Γ BCββββ β| = 2 square units.
The correct Answer is C. [1 Mark]
Miscellaneous exercise
1. Write down a unit vector in ππ-plane, making an angle of 30o with the positive direction of π₯-axis. [1 Mark]
Solution:
If π is a unit vector in the ππ-plane, then π = cos ππΜ + sin ππΜ.
Here, π is the angle made by the unit vector with the positive direction of the π₯-axis.
Therefore, for π = 30o: [π
π Mark]
π = cos30Β°οΏ½ΜοΏ½ + sin30Β°πΜ =β3
2πΜ +
1
2πΜ
Hence, the required unit vector is =β3
2πΜ +
1
2π Μ [
π
π Mark]
2. Find the scalar components and magnitude of the vector joining the points
P(π₯1, π¦1, π§1).and Q(π₯2, π¦2, π§2). [2 Marks]
Solution:
The vector joining the points P(π₯1, π¦1, π§1).and Q(π₯2, π¦2, π§2)
ππββββ β =Position vector of π βPosition vector of π.
= (π₯2 β π₯1)πΜ + (π¦2 β π¦1)πΜ + (π§2 β π§1)οΏ½ΜοΏ½ [1 Mark]
|PQββββ β| = β(π₯2 β π₯1)2 + (π¦2 β π¦1)
2 + (π§2 β π§1)2
Hence, the scalar components and the magnitude of the vector joining the given points are
respectively {(π₯2 β π₯1), (π¦2 β π¦1), (π§2 β π§1)} and β(π₯2 β π₯1)2 + (π¦2 β π¦1)
2 + (π§2 β π§1)2
[1 Mark]
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3. A girl walks 4 km towards west, then she walks 3 km in a direction 30o east of north and stops. Determine the girlβs displacement from her initial point of departure. [2 Marks]
Solution:
Let π and π΅ be the initial and final positions of the girl respectively.
Then, the girlβs position can be shown as:
Now, we have:
OAββββ β = β4π Μ
ABββββ β = π|ΜABββββ β|cos60Β° + πΜ|ABββββ β|sin60Β°
= π3Μ Γ1
2+ πΜ3 Γ
β3
2
=3
2πΜ +
3β3
2πΜ [1 Mark]
By the triangle law of vector addition, we have:
OBββββ β = OAββββ β + ABββββ β
= (β4π)Μ + (3
2πΜ +
3β3
2πΜ)
= (β4 +3
2) πΜ +
3β3
2πΜ
= (β8 + 3
2) πΜ +
3β3
2πΜ
=β5
2πΜ +
3β3
2πΜ
Hence, the girlβs displacement from her initial point of departure is
=β5
2πΜ +
3β3
2πΜ. [1 Mark]
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4. If π = οΏ½βοΏ½ + π , then is it true that |π | = |οΏ½βοΏ½ | + |π |? Justify your Answer. [2 Marks]
Solution:
In ΞABC, let CBββββ β = π , CAββββ β = οΏ½βοΏ½ and ABββββ β = π (as shown in the following figure)
Now, by the triangle law of vector addition, we have π = οΏ½βοΏ½ + π . [1 Mark]
It is clearly known that |π |, |οΏ½βοΏ½ | and |π | represent the sides of Ξπ΄π΅πΆ.
Also, it is known that the sum of the lengths of any two sides of a triangle is greater than the third side.
β΄ |π | < |οΏ½βοΏ½ | + |π |
Hence, it is not true that |π | = |οΏ½βοΏ½ | + |π |. [1 Mark]
5. Find the value of π₯ for which π₯(πΜ + πΜ + οΏ½ΜοΏ½) is a unit vector. [1 Mark]
Solution:
π₯(πΜ + πΜ + οΏ½ΜοΏ½) is a unit vector if |π₯(πΜ + πΜ + οΏ½ΜοΏ½)| = 1.
Now,
|π₯(πΜ + πΜ + οΏ½ΜοΏ½)| = 1
β βπ₯2 + π₯2 + π₯2 = 1
β β3π₯2 = 1
β β3π₯ = 1 [π
π Mark]
β π₯ = Β±1
β3
Hence, the required value of π₯ is Β±1
β3. [
π
π Mark]
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6. Find a vector of magnitude 5 units, and parallel to the resultant of the vectors π = 2πΜ + 3πΜ β οΏ½ΜοΏ½
and οΏ½βοΏ½ = πΜ β 2πΜ + οΏ½ΜοΏ½. [2 Marks]
Solution:
We have,
π = 2πΜ + 3πΜ β οΏ½ΜοΏ½ and οΏ½βοΏ½ = πΜ β 2πΜ + οΏ½ΜοΏ½
Let π be the resultant of π and οΏ½βοΏ½ .
Then,
π = π + οΏ½βοΏ½ = (2 + 1)πΜ + (3 β 2)πΜ + (β1 + 1)οΏ½ΜοΏ½ = 3πΜ + πΜ [1 Mark]
β΄ |π | = β32 + 12 = β9 + 1 = β10
β΄ οΏ½ΜοΏ½ =π
|π |=
(3πΜ + πΜ)
β10
Hence, the vector of magnitude 5 units and parallel to the resultant of vectors π and οΏ½βοΏ½ is
Β±5 β οΏ½ΜοΏ½ = Β±5 β 1
β10(3πΜ + πΜ) = Β±
3β10π
2Β±
β10
2πΜ [1 Mark]
7. If π = πΜ + πΜ + οΏ½ΜοΏ½, οΏ½βοΏ½ = 2πΜ β πΜ + 3οΏ½ΜοΏ½ and π = πΜ β 2πΜ + οΏ½ΜοΏ½, find a unit vector parallel to the vector
2π β οΏ½βοΏ½ + 3π . [2 Marks]
Solution:
We have,
π = πΜ + πΜ + οΏ½ΜοΏ½, οΏ½βοΏ½ = 2πΜ β πΜ + 3οΏ½ΜοΏ½ and π = πΜ β 2πΜ + οΏ½ΜοΏ½
2π β οΏ½βοΏ½ + 3π = 2(πΜ + πΜ + οΏ½ΜοΏ½) β (2πΜ β πΜ + 3οΏ½ΜοΏ½) + 3(πΜ β 2πΜ + οΏ½ΜοΏ½)
= 2πΜ + 2πΜ + 2οΏ½ΜοΏ½ β 2πΜ + πΜ β 3οΏ½ΜοΏ½ + 3πΜ β 6πΜ + 3οΏ½ΜοΏ½
= 3πΜ β 3πΜ + 2οΏ½ΜοΏ½ [1 Mark]
|2π β οΏ½βοΏ½ + 3π | = β32 + (β3)2 + 22 = β9 + 9 + 4 = β22
Hence, the unit vector along 2π β οΏ½βοΏ½ + 3π is
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2οΏ½βοΏ½ βοΏ½βοΏ½ +3π
|2οΏ½βοΏ½ βοΏ½βοΏ½ +3π |=
3οΏ½ΜοΏ½β3οΏ½ΜοΏ½+2οΏ½ΜοΏ½
β22=
3
β22πΜ β
3
β22πΜ +
2
β22οΏ½ΜοΏ½ [1 Mark]
8. Show that the points π΄(1, β 2, β 8), π΅(5,0, β 2) and πΆ(11,3,7) are collinear, and find the ratio in
which π΅ divides π΄πΆ. [4 Marks]
Solution:
The given points are π΄(1, β 2, β 8), π΅(5,0, β 2) and πΆ(11,3,7).
β΄ ABββββ β = (5 β 1)πΜ + (0 + 2)πΜ + (β2 + 8)οΏ½ΜοΏ½ = 4πΜ + 2πΜ + 6οΏ½ΜοΏ½
BCββββ β = (11 β 5)πΜ + (3 β 0)πΜ + (7 + 2)οΏ½ΜοΏ½ = 6πΜ + 3πΜ + 9οΏ½ΜοΏ½
ACββββ β = (11 β 1)πΜ + (3 + 2)πΜ + (7 + 8)οΏ½ΜοΏ½ = 10πΜ + 5πΜ + 15οΏ½ΜοΏ½ [1 Mark]
|ABββββ β| = β42 + 22 + 62 = β16 + 4 + 36 = β56 = 2β14
|BCββββ β| = β62 + 32 + 92 = β36 + 9 + 81 = β126 = 3β14
|ACββββ β| = β102 + 52 + 152 = β100 + 25 + 225 = β350 = 5β14
β΄ |ACββββ β| = |ABββββ β| + |BCββββ β| [1 Mark]
Thus, the given points π΄, π΅ and πΆ are collinear.
Now, let point π΅ divide π΄πΆ in the ratio π: 1. Then, we have:
OBββββ β =πOCββββ β + OAββββ β
(π + 1)
β 5πΜ β 2οΏ½ΜοΏ½ =π(11πΜ + 3πΜ + 7οΏ½ΜοΏ½) + (πΜ β 2πΜ β 8οΏ½ΜοΏ½)
π + 1
β (π + 1)(5πΜ β 2οΏ½ΜοΏ½) = 11ππΜ + 3ππΜ + 7ποΏ½ΜοΏ½ + πΜ β 2πΜ β 8οΏ½ΜοΏ½
β 5(π + 1)πΜ β 2(π + 1)οΏ½ΜοΏ½ = (11π + 1)πΜ + (3π β 2)πΜ + (7π β 8)οΏ½ΜοΏ½ [1 Mark]
On equating the corresponding components, we get:
5(π + 1) = 11π + 1
β 5π + 5 = 11π + 1
β 6π = 4
β π =4
6=
2
3
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Hence, point π΅ divides π΄πΆ in the ratio 2: 3. [1 Mark]
9. Find the position vector of a point π which divides the line joining two points π and π whose
position vectors are (2π + οΏ½βοΏ½ ) and (π β 3οΏ½βοΏ½ ) externally in the ratio 1: 2. Also, show that π is the
mid point of the line segment π π. [2 Marks]
Solution:
It is given that OPββββ β = 2π + οΏ½βοΏ½ , OQββ ββ β = π β 3οΏ½βοΏ½ .
It is given that point π divides a line segment joining two points π and π externally in the ratio 1: 2. Then, on using the section formula, we get:
ORββββ β =2(2οΏ½βοΏ½ +οΏ½βοΏ½ )β(οΏ½βοΏ½ β3οΏ½βοΏ½ )
2β1=
4οΏ½βοΏ½ +2οΏ½βοΏ½ βοΏ½βοΏ½ +3οΏ½βοΏ½
1= 3π + 5οΏ½βοΏ½ [1 Mark]
Therefore, the position vector of point π is 3π + 5οΏ½βοΏ½
Position vector of the mid-point of RQ =OQββββββ +ORββββββ
2
=(π β 3οΏ½βοΏ½ ) + (3π + 5οΏ½βοΏ½ )
2
= 2π + οΏ½βοΏ½
= OPββββ β
Hence, π is the mid-point of the line segment π π. [1 Mark]
10. The two adjacent sides of a parallelogram are 2πΜ β 4πΜ + 5οΏ½ΜοΏ½ and πΜ β 2πΜ β 3οΏ½ΜοΏ½.
Find the unit vector parallel to its diagonal. Also, find its area. [2 Marks]
Solution:
Adjacent sides of a parallelogram are given as: 2πΜ β 4πΜ + 5οΏ½ΜοΏ½ and πΜ β 2πΜ β 3οΏ½ΜοΏ½
Then, the diagonal of a parallelogram is given by π + οΏ½βοΏ½
π + οΏ½βοΏ½ = (2 + 1)πΜ + (β4 β 2)πΜ + (5 β 3)οΏ½ΜοΏ½ = 3πΜ β 6πΜ + 2οΏ½ΜοΏ½
Thus, the unit vector parallel to the diagonal is
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οΏ½βοΏ½ +οΏ½βοΏ½
|οΏ½βοΏ½ +οΏ½βοΏ½ |=
3οΏ½ΜοΏ½β6οΏ½ΜοΏ½+2οΏ½ΜοΏ½
β32+(β6)2+22=
3οΏ½ΜοΏ½β6οΏ½ΜοΏ½+2οΏ½ΜοΏ½
β9+36+4=
3οΏ½ΜοΏ½β6οΏ½ΜοΏ½+2οΏ½ΜοΏ½
7=
3
7πΜ β
6
7πΜ +
2
7οΏ½ΜοΏ½ [1 Mark]
β΄ Area of parallelogram ABCD = |π Γ οΏ½βοΏ½ |
π Γ οΏ½βοΏ½ = |πΜ πΜ οΏ½ΜοΏ½2 β4 51 β2 β3
|
= π(Μ12 + 10) β πΜ(β6 β 5) + οΏ½ΜοΏ½(β4 + 4)
= 22πΜ + 11πΜ
= 11(2πΜ + πΜ)
β΄ |π Γ οΏ½βοΏ½ | = 11β22 + 12 = 11β5
Hence, the area of the parallelogram is 11β5 square units. [1 Mark]
11. Show that the direction cosines of a vector equally inclined to the axes ππ,ππ and ππ are 1
β3,
1
β3,
1
β3. [1 Mark]
Solution:
Let a vector be equally inclined to axes ππ,ππ and ππ at angle πΌ.
Then, the direction cosines of the vector are πππ πΌ, πππ πΌ and πππ πΌ.
Now,
cos2πΌ + cos2πΌ + cos2πΌ = 1
β 3cos2πΌ = 1
β cosπΌ =1
β3 [
π
π Mark]
Hence, the direction cosines of the vector
which are equally inclined to the axes are 1
β3,
1
β3,
1
β3. [
π
π Mark]
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12. Let π = πΜ + 4πΜ + 2οΏ½ΜοΏ½, οΏ½βοΏ½ = 3πΜ β 2πΜ + 7οΏ½ΜοΏ½ and π = 2πΜ β πΜ + 4οΏ½ΜοΏ½. Find a vector π which is
perpendicular to both π and οΏ½βοΏ½ , and π . π = 15. [2 Marks]
Solution:
Let π = π1πΜ + π2πΜ + π3οΏ½ΜοΏ½.
Since π is perpendicular to both π and οΏ½βοΏ½
π β π = 0
β π1 + 4π2 + 2π3 = 0 . . . (π)
And,
π β οΏ½βοΏ½ = 0
β 3π1 β 2π2 + 7π3 = 0 . . . (ππ) [1 Mark]
Also, it is given that:
π β π = 15
β 2π1 β π2 + 4π3 = 15 (πππ)
On solving (π), (ππ) and (πππ) we get:
π1 =160
3, π2 = β
5
3 and π3 = β
70
3
β΄ π =160
3πΜ β
5
3πΜ β
70
3οΏ½ΜοΏ½ =
1
3(160πΜ β 5πΜ β 70ΜοΏ½ΜοΏ½)
Hence, the required vector is 1
3(160πΜ β 5πΜ β 70οΏ½ΜοΏ½). [1 Mark]
13. The scalar product of the vector πΜ + πΜ + οΏ½ΜοΏ½ with a unit vector along the sum of vectors 2πΜ + 4πΜ β
5οΏ½ΜοΏ½ and ππΜ + 2πΜ + 3οΏ½ΜοΏ½ is equal to one. Find the value of π. [2 Marks]
Solution:
(2πΜ + 4πΜ β 5οΏ½ΜοΏ½) + (ππΜ + 2πΜ + 3οΏ½ΜοΏ½)
= (2 + π)πΜ + 6πΜ β 2οΏ½ΜοΏ½
Therefore, unit vector along (2πΜ + 4πΜ β 5οΏ½ΜοΏ½) + (ππΜ + 2πΜ + 3οΏ½ΜοΏ½) is given as:
(2+π)οΏ½ΜοΏ½+6οΏ½ΜοΏ½β2οΏ½ΜοΏ½
β(2+π)2+62+(β2)2=
(2+π)οΏ½ΜοΏ½+6οΏ½ΜοΏ½β2οΏ½ΜοΏ½
β4+4π+π2+36+4=
(2+π)οΏ½ΜοΏ½+6οΏ½ΜοΏ½β2οΏ½ΜοΏ½
βπ2+4π+44 [1 Mark]
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Scalar product of (πΜ + πΜ + οΏ½ΜοΏ½) with this unit vector is 1.
β (πΜ + πΜ + οΏ½ΜοΏ½) β (2 + π)οΏ½ΜοΏ½ + 6πΜ β 2οΏ½ΜοΏ½
βπ2 + 4π + 44= 1
β(2 + π) + 6 β 2
βπ2 + 4π + 44= 1
β βπ2 + 4π + 44 = π + 6
β π2 + 4π + 44 = (π + 6)2
β π2 + 4π + 44 = π2 + 12π + 36
β 8π = 8
β π = 1
Hence, the value of π is 1. [1 Mark]
14. If π , οΏ½βοΏ½ , π are mutually perpendicular vectors of equal magnitudes, show that the vector π + οΏ½βοΏ½ +
π is equally inclined to π , οΏ½βοΏ½ and π . [4 Marks]
Solution:
Since π , οΏ½βοΏ½ and π are mutually perpendicular vectors, we have π β οΏ½βοΏ½ = οΏ½βοΏ½ β π = π β π = 0. It is given
that:|π | = |οΏ½βοΏ½ | = |π |.
Let vector π + οΏ½βοΏ½ + π be inclined to π , οΏ½βοΏ½ and π at angles π1, π2 and π3respectively. [π
π Mark]
Then, we have:
cosπ1 =(π + οΏ½βοΏ½ + π ) β π
|π + οΏ½βοΏ½ + π ||π |=
π β π + οΏ½βοΏ½ β π + π β π
|π + οΏ½βοΏ½ + π ||π |
=|π |2
|π + οΏ½βοΏ½ + π ||π | [οΏ½βοΏ½ β π = π β π = 0]
=|οΏ½βοΏ½ |2
|οΏ½βοΏ½ +οΏ½βοΏ½ +π | [1 Mark]
cosπ2 =(π + οΏ½βοΏ½ + π ) β οΏ½βοΏ½
|π + οΏ½βοΏ½ + π ||οΏ½βοΏ½ |=
π β οΏ½βοΏ½ + οΏ½βοΏ½ β οΏ½βοΏ½ + π β οΏ½βοΏ½
|π + οΏ½βοΏ½ + π | β |οΏ½βοΏ½ |
=|οΏ½βοΏ½ |
2
|π + οΏ½βοΏ½ + π | β |οΏ½βοΏ½ | [π β οΏ½βοΏ½ = π β οΏ½βοΏ½ = 0]
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=|οΏ½βοΏ½ |
2
|οΏ½βοΏ½ +οΏ½βοΏ½ +π | [1 Mark]
cosπ3 =(π + οΏ½βοΏ½ + π ) β π
|π + οΏ½βοΏ½ + π ||π |=
π β π + οΏ½βοΏ½ β π + π β π
|π + οΏ½βοΏ½ + π ||π |
=|π |2
|π + οΏ½βοΏ½ + π ||π | [π β π = οΏ½βοΏ½ β π = 0]
=|π |2
|οΏ½βοΏ½ +οΏ½βοΏ½ +π | [1 Mark]
Now, as |π | = |οΏ½βοΏ½ | = |π |, cosπ1 = cosπ2 = cosπ3
β΄ π1 = π2 = π3
Hence, the vector (π + οΏ½βοΏ½ + π ) is equally inclined to π , οΏ½βοΏ½ and π . [π
π Mark]
15. Prove that (π + οΏ½βοΏ½ ) β (π + οΏ½βοΏ½ ) = |π |2 + |οΏ½βοΏ½ |2
,if and only if π , οΏ½βοΏ½ are perpendicular, given π β 0β , οΏ½βοΏ½ β
0β . [2 Marks]
Solution:
(π + οΏ½βοΏ½ ) β (π + οΏ½βοΏ½ ) = |π |2 + |οΏ½βοΏ½ |2
β π β π + π β οΏ½βοΏ½ + οΏ½βοΏ½ β π + οΏ½βοΏ½ β οΏ½βοΏ½ = |π |2 + |οΏ½βοΏ½ |2
[Distributivity of scalar products over addition]
β |π |2 + 2π β οΏ½βοΏ½ + |οΏ½βοΏ½ |2
= |π |2 + |οΏ½βοΏ½ |2
[π β οΏ½βοΏ½ = οΏ½βοΏ½ β π (Scalar product is commutative)][1 Mark]
β 2π β οΏ½βοΏ½ = 0
β π β οΏ½βοΏ½ = 0
β΄ π and οΏ½βοΏ½ are perpendicular. [π β 0β , οΏ½βοΏ½ β 0β (Given)] [1 Mark]
16. If π is the angle between two vectors π and οΏ½βοΏ½ , then π . οΏ½βοΏ½ β₯ 0 only when [1 Mark]
(A) 0 < π <π
2
(B) 0 β€ π β€π
2
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(C) 0 < π < π
(D) 0 β€ π β€ π
Solution:
Let π be the angle between two vectors π and οΏ½βοΏ½ .
Then, without loss of generality, π and οΏ½βοΏ½ are non-zero vectors so that. |π | and |οΏ½βοΏ½ | are positive.
It is known that π β οΏ½βοΏ½ = |π ||οΏ½βοΏ½ |cosπ
β΄ π β οΏ½βοΏ½ β₯ 0
β |π ||οΏ½βοΏ½ |cosπ β₯ 0
β cosπ β₯ 0 [|π |and|οΏ½βοΏ½ |arepositive] [π
π Mark]
β 0 β€ π β€π
2
Hence, π β οΏ½βοΏ½ β₯ 0 when 0 β€ π β€π
2.
The correct Answer is B. [π
π Mark]
17. Let π and οΏ½βοΏ½ be two unit vectors and π is the angle between them. Then π + οΏ½βοΏ½ is a unit vector if [2 Marks]
(A) π =π
4
(B) π =π
3
(C) π =π
2
(D) π =2π
3
Solution:
Let π and οΏ½βοΏ½ be two unit vectors and π be the angle between them.
Then,|π | = |οΏ½βοΏ½ | = 1.
Now, π + οΏ½βοΏ½ is a unit vector if |π + οΏ½βοΏ½ | = 1.
|π + οΏ½βοΏ½ | = 1
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β (π + οΏ½βοΏ½ )2= 1
β (π + οΏ½βοΏ½ ) β (π + οΏ½βοΏ½ ) = 1
β π β π + π οΏ½βοΏ½ + οΏ½βοΏ½ β π + οΏ½βοΏ½ οΏ½βοΏ½ = 1 [π
π Mark]
β |π |2 + 2π β οΏ½βοΏ½ + |οΏ½βοΏ½ |2= 1
β 12 + 2|π ||οΏ½βοΏ½ |cosπ + 12 = 1
β 1 + 2.1.1cosπ + 1 = 1 [π
π Mark]
β cosπ = β1
2
β π =2π
3 [
π
π Mark]
Hence, π + οΏ½βοΏ½ is a unit vector if π =2π
3
The correct Answer is D. [π
π Mark]
18. The value of πΜ. (πΜ Γ οΏ½ΜοΏ½) + πΜ. (πΜ Γ οΏ½ΜοΏ½) + οΏ½ΜοΏ½. (πΜ Γ πΜ) is [2 Marks]
(A) 0
(B) β 1
(C) 1
(D) 3
Solution:
πΜ. (πΜ Γ οΏ½ΜοΏ½) + πΜ. (πΜ Γ οΏ½ΜοΏ½) + οΏ½ΜοΏ½. (πΜ Γ πΜ)
= πΜ β πΜ + πΜ β (βπΜ) + οΏ½ΜοΏ½ β οΏ½ΜοΏ½
= 1 β πΜ β πΜ + 1 [π
π Mark]
= 1 β 1 + 1
= 1 [π
π Mark]
The correct Answer is C.
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19. If π is the angle between any two vectors π and οΏ½βοΏ½ , then |π β οΏ½βοΏ½ | = |π Γ οΏ½βοΏ½ |
when π is equal to [1 Mark]
(A) 0
(B) π
4
(C) π
2
(D) π
Solution:
Let π be the angle between two vectors π and οΏ½βοΏ½ .
Then, without loss of generality, π and οΏ½βοΏ½ are non-zero vectors, so that |π | and |οΏ½βοΏ½ | are positive.
|π β οΏ½βοΏ½ | = |π Γ οΏ½βοΏ½ |
β |π ||οΏ½βοΏ½ |cosπ = |π ||οΏ½βοΏ½ |sinπ
β cosπ = sinπ [|π |and|οΏ½βοΏ½ |arepositive]
β tanπ = 1 [π
π Mark]
β π =π
4
Hence, |π β οΏ½βοΏ½ | = |π Γ οΏ½βοΏ½ | when π is equal to π
4
The correct Answer is B. [π
π Mark]
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