Question 1

Represent graphically a displacement of 40 km, 30^° east of north.

Accepted Answer

In the adjoining figure, the vector OP represents the required displacement.![](https://cdn.mathpix.com/cropped/476781f9-a560-4705-81ba-98e12b199800-13.jpg?height=441&width=448&top left y=977&top left x=529). A displacement is represented graphically by a directed line segment: its length, on a stated scale, gives the magnitude, and the direction of the arrow gives the direction of the displacement. Choose a scale first, otherwise "length" is meaningless. Take 1 cm ≡ 10 km, so 40 km ≡ 4 cm. Now fix the direction. The phrase 30^° east of north is measured starting from the north direction and turning through 30^° towards the east — it is not 30^° measured from the east, and not 30^° from the horizontal. This is the step most often got wrong. So, from a starting point O draw the north line ON. Rotate ON through 30^° towards OE (east) and along this ray mark P with OP =4 cm, putting the arrow head at P. The vector OP so drawn represents the required displacement: OP =40 km, directed at 30^° east of north (equivalently, making 60^° with the east direction).

Question 2

Classify the following measures as scalars and vectors. (i) 10 kg (ii) 2 meters north-west (iii) 40° (iv) 40 watt (v) 10^-19 coulomb (vi) 20 m / s^2

Accepted Answer

(i) scalar (ii) vector (iii) scalar (iv) scalar (v) scalar (vi) vector. Test used: a quantity that is completely specified by a magnitude (a number with its unit) alone is a scalar; a quantity that needs a magnitude and a direction, and that combines by the triangle law of addition, is a vector. (i) 10 kg — this is a mass. Only a number and a unit are given, and mass has no direction. Scalar. (ii) 2 metres north-west — this is a displacement (or distance in a stated direction). The words "north-west" are part of the answer, so a direction is essential. Vector. (iii) 40^° — a measure of an angle. An angle is a pure magnitude; the plane angle itself does not point anywhere. Scalar. (iv) 40 watt — this is power. Fully described by its magnitude. Scalar. (v) 10^-19 coulomb — electric charge. Charge has sign but not direction, and a sign is not a direction. Scalar. (vi) 20 m / s^2 — this is an acceleration, the rate of change of velocity; velocity is directed, hence so is its rate of change. Vector. Final answer: scalars — (i), (iii), (iv), (v); vectors — (ii), (vi).

Question 3

Classify the following as scalar and vector quantities. (i) time period (ii) distance (iii) force (iv) velocity (v) work done

Accepted Answer

(i) scalar (ii) scalar (iii) vector (iv) vector (v) scalar. Same test: a scalar is fixed by magnitude alone; a vector needs magnitude together with a direction. (i) Time period — an interval of time, specified by a number of seconds alone. Scalar. (ii) Distance — the length of the path actually covered. It carries no direction (this is exactly what separates distance from displacement, which is a vector). Scalar. (iii) Force — a push or pull, and the line and sense along which it acts must be stated; forces add by the triangle/parallelogram law. Vector. (iv) Velocity — displacement per unit time, so it inherits the direction of the displacement (again, contrast speed, which is a scalar). Vector. (v) Work done — obtained as W= F · d, a scalar product of two vectors, and a scalar product is a number. Scalar. Final answer: scalars — (i) time period, (ii) distance, (v) work done; vectors — (iii) force, (iv) velocity.

Question 4

In Fig 10.6 (a square), identify the following vectors. (i) Coinitial (ii) Equal (iii) Collinear but not equal

Accepted Answer

(i) Vectors a and b are coinitial (ii) Vectors b and d are equal (iii) Vectors a and c are collinear but not equal. Read the arrowheads in Fig 10.6. Writing the square's corners as P (top-left), Q (top-right), R (bottom-right) and S (bottom-left): a = PQ, b = QR, c = RS, d = PS (i) Coinitial. Coinitial vectors share an initial point. The initial points are P, Q, R, P respectively, so the pair that begins at the same corner is a and d, both starting at P. (ii) Equal. Equal vectors have the same magnitude AND the same direction, wherever they are drawn. b runs down the right side and d runs down the left side; both have length equal to the side of the square and both point in the same direction, so b = d. (iii) Collinear but not equal. a and c are both horizontal, so they are parallel and hence collinear; they have the same magnitude but point in OPPOSITE directions, so they are not equal. In fact c =- a.

Question 5

Answer the following as true or false. (i) a and - a 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.

Accepted Answer

(i) True (ii) False (iii) False (iv) False. Definitions being tested: two (non-zero) vectors are collinear (parallel) if each is a scalar multiple of the other, i.e. they are parallel to one and the same line, whatever their magnitudes and whichever way they point; two vectors are equal only if they have the same magnitude and the same direction. (i) a and - a are collinear. TRUE. Since - a =(-1) a, it is a scalar multiple of a, so it is parallel to the same line (pointing the opposite way, which collinearity permits). (ii) Two collinear vectors are always equal in magnitude. FALSE. Collinearity restricts direction only, not length. Counterexample: a and 2 a are collinear, but 2 a =2 a ≠ a for any a ≠ 0. (iii) Two vectors having same magnitude are collinear. FALSE. Equality of magnitudes says nothing about direction. Counterexample: i and j both have magnitude 1, yet they are perpendicular, so not parallel to one line. (iv) Two collinear vectors having the same magnitude are equal. FALSE. Collinear leaves two possible senses along the line — same or opposite. Counterexample: a and - a (with a ≠ 0 ) are collinear and - a = a, but their directions are opposite, so a ≠- a. Final answer: (i) True; (ii) False; (iii) False; (iv) False.

Question 6

Compute the magnitude of the following vectors: a = i + j +k; b =2 i -7 j -3 k; c =1/(√ 3 ) i +1/(√ 3 ) j -1/(√ 3 ) k

Accepted Answer

a =√ 3, b =√ 62, c =1. The magnitude of v =x i +y j +z k is v =√ x^2+y^2+z^2. For a = i + j + k: a =√ 1^2+1^2+1^2 =√ 3 For b =2 i -7 j -3 k: b =√ 2^2+(-7)^2+(-3)^2 =√ 4+49+9 =√ 62 For c =1/(√ 3 ) i +1/(√ 3 ) j -1/(√ 3 ) k the minus sign disappears on squaring: c =√ 1/3+1/3+1/3 =√ 1 =1 a =√ 3, b =√ 62, c =1 (so c is a unit vector).

Question 7

Write two different vectors having same magnitude.

Accepted Answer

An infinite number of possible answers.. Two vectors are equal only when all corresponding components agree, while the magnitude √ x^2+y^2+z^2 is unchanged if the components are permuted. So permuting components produces different vectors of the same length. Take a =2 i +3 j +6 k, b =6 i +3 j +2 k Then a =√ 4+9+36 =√ 49 =7, b =√ 36+9+4 =√ 49 =7 They really are different vectors: the i -components are 2 and 6, and b is not even a scalar multiple of a because 6/2≠3/3. a =2 i +3 j +6 k and b =6 i +3 j +2 k, both of magnitude 7. (Any two unequal vectors of equal magnitude are acceptable.)

Question 8

Write two different vectors having same direction.

Accepted Answer

An infinite number of possible answers.. Two nonzero vectors have the same direction when one is a positive scalar multiple of the other, equivalently when they share the same unit vector a =( a )/( a ). Take a = i + j + k, b =3 i +3 j +3 k =3 a Here a =√ 3 and b =√ 9+9+9 =3√ 3, so a =1/(√ 3 )( i + j + k )=(3 i +3 j +3 k )/(3√ 3 )= b The multiplier 3 is positive, so the direction is the same (a negative multiplier would reverse it), while the magnitudes √ 3 and 3√ 3 differ, so the vectors are different. a = i + j + k and b =3 i +3 j +3 k.

Question 9

Find the values of x and y so that the vectors 2 i +3 j and x i +y j are equal.

Accepted Answer

x=2, y=3. Equality of vectors is componentwise: two vectors are equal only if their corresponding components are equal. 2 i +3 j =x i +y j Comparing the coefficients of i and of j separately, x=2, y=3 x=2 and y=3.

Question 10

Find the scalar and vector components of the vector with initial point (2,1) and terminal point (-5, 7).

Accepted Answer

-7 and 6;-7 i and 6 j. For initial point A (x 1,y 1) and terminal point B (x 2,y 2), AB =(x 2-x 1) i +(y 2-y 1) j terminal minus initial, in that order (reversing the order would flip both signs). With A (2,1) and B (-5,7), AB =(-5-2) i +(7-1) j =-7 i +6 j Scalar components: -7 and 6. Vector components: -7 i and 6 j.

Question 11

Find the sum of the vectors a = i -2 j + k, b =-2 i +4 j +5 k and c = i -6 j -7 k.

Accepted Answer

-4 j - k. Vectors are added componentwise: collect the i, j and k coefficients separately. a + b + c =(1-2+1) i +(-2+4-6) j +(1+5-7) k =0 i -4 j - k a + b + c =-4 j - k.

Question 12

Find the unit vector in the direction of the vector a = i + j +2 k.

Accepted Answer

1/(√ 6 ) i +1/(√ 6 ) j +2/(√ 6 ) k. The unit vector in the direction of a is a =( a )/( a ), defined for a ≠ 0. a =√ 1^2+1^2+2^2 =√ 6 a =1/(√ 6 )( i + j +2 k )=1/(√ 6 ) i +1/(√ 6 ) j +2/(√ 6 ) k Check: 1/6+1/6+4/6=1. a =1/(√ 6 )( i + j +2 k ).

Question 13

Find the unit vector in the direction of vector PQ, where P and Q are the points (1, 2, 3) and (4, 5, 6), respectively.

Accepted Answer

1/(√ 3 ) i +1/(√ 3 ) j +1/(√ 3 ) k. PQ = position vector of Q - position vector of P. PQ =(4-1) i +(5-2) j +(6-3) k =3 i +3 j +3 k PQ =√ 3^2+3^2+3^2 =√ 27 =3√ 3 Divide the vector by its magnitude: PQ =(3 i +3 j +3 k )/(3√ 3 )=1/(√ 3 )( i + j + k ) Unit vector along PQ is 1/(√ 3 ) i +1/(√ 3 ) j +1/(√ 3 ) k.

Question 14

For given vectors, a =2 i - j +2 k and b =- i + j - k, find the unit vector in the direction of the vector a + b.

Accepted Answer

1/(√ 2 ) i +1/(√ 2 ) k. Add first, then normalise — the unit vector is required along a + b, not along a or b. a + b =(2-1) i +(-1+1) j +(2-1) k = i +0 j + k a + b =√ 1^2+0^2+1^2 =√ 2 ( a + b )/( a + b )=1/(√ 2 )( i + k ) Required unit vector =1/(√ 2 ) i +1/(√ 2 ) k.

Question 15

Find a vector in the direction of vector 5 i - j +2 k which has magnitude 8 units.

Accepted Answer

40/(√ 30 ) i -8/(√ 30 ) j +16/(√ 30 ) k. A vector of magnitude m in the direction of a is m a =m( a )/( a ); scaling a unit vector by m is what fixes the length. a =5 i - j +2 k, a =√ 25+1+4 =√ 30 8 a =8/(√ 30 )(5 i - j +2 k )=40/(√ 30 ) i -8/(√ 30 ) j +16/(√ 30 ) k Check: its magnitude is 8/(√ 30 )·√ 30 =8. Required vector =8/(√ 30 )(5 i - j +2 k ).

Question 16

Show that the vectors 2 i -3 j +4 k and -4 i +6 j -8 k are collinear.

Accepted Answer

Two nonzero vectors are collinear (parallel) precisely when one is a scalar multiple of the other, i.e. b =λ a for some scalar λ; equivalently, their corresponding components are proportional. Let a =2 i -3 j +4 k and b =-4 i +6 j -8 k. Compare the components: -4/2=-2, 6/-3=-2, -8/4=-2 All three ratios are the same number, so b =-2(2 i -3 j +4 k )=-2 a Hence b =λ a with λ=-2, so the two vectors are collinear. Since λ<0 they lie along the same line but point in opposite directions, and b =2 a.

Question 17

Find the direction cosines of the vector i +2 j +3 k.

Accepted Answer

1/(√ 14 ), 2/(√ 14 ), 3/(√ 14 ). If r =x i +y j +z k makes angles α,β,γ with the coordinate axes, its direction cosines are l= cos α=x/( r ), m= cos β=y/( r ), n= cos γ=z/( r ) that is, they are the components of the unit vector r — so the vector must be divided by its magnitude, not used as it stands. r =√ 1^2+2^2+3^2 =√ 14 l=1/(√ 14 ), m=2/(√ 14 ), n=3/(√ 14 ) Check: l^2+m^2+n^2=(1+4+9)/14=1. Direction cosines: (1/(√ 14 ), 2/(√ 14 ), 3/(√ 14 )).

Question 18

Find the direction cosines of the vector joining the points A (1,2,-3) and B (-1, -2, 1), directed from A to B.

Accepted Answer

-1/3,-2/3, 2/3. "Directed from A to B" fixes the order: use AB = B - A. Taking B to A instead would reverse the sign of all three direction cosines. AB =(-1-1) i +(-2-2) j + (1-(-3) ) k =-2 i -4 j +4 k AB =√ (-2)^2+(-4)^2+4^2 =√ 4+16+16 =√ 36 =6 The direction cosines are the components divided by this magnitude: l=-2/6=-1/3, m=-4/6=-2/3, n=4/6=2/3 Check: 1/9+4/9+4/9=1. Direction cosines: (-1/3, -2/3, 2/3).

Question 19

Show that the vector i + j + k is equally inclined to the axes OX, OY and OZ.

Accepted Answer

Let r = i + j + k make angles α,β,γ with OX, OY and OZ. By the definition of direction cosines, cos α=x/( r ), cos β=y/( r ), cos γ=z/( r ). r =√ 1^2+1^2+1^2 =√ 3 cos α=1/(√ 3 ), cos β=1/(√ 3 ), cos γ=1/(√ 3 ) The three cosines are equal, and the angles a vector makes with the axes satisfy 0≤α,β,γ≤π, an interval on which cosine is one-to-one. Hence equal cosines force equal angles: α=β=γ= cos^-1 (1/(√ 3 ))≈ 54.74^° Therefore i + j + k is equally inclined to the axes OX, OY and OZ.

Question 20

Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are i +2 j - k and - i + j + k respectively, in the ratio 2: 1 (i) internally (ii) externally

Accepted Answer

(i) -1/3 i +4/3 j +1/3 k (ii) -3 i +3 k. Section formula. If R divides the line joining P (position vector p ) and Q (position vector q ) in the ratio m:n, then internally: r =(m q +n p )/(m+n), externally: r =(m q -n p )/m-n Here p = i +2 j - k, q =- i + j + k, m=2, n=1. Note it is q (the far point) that carries the weight m=2. (i) Internally: r =(2(- i + j + k )+1( i +2 j - k ))/(2+1)=((-2+1) i +(2+2) j +(2-1) k )/3=(- i +4 j + k )/3 (ii) Externally (the same formula with n replaced by -n ): r =(2(- i + j + k )-1( i +2 j - k ))/2-1=(-2-1) i +(2-2) j +(2+1) k =-3 i +3 k (i) r =1/3(- i +4 j + k ) (ii) r =-3 i +3 k.

Question 21

Find the position vector of the mid point of the vector joining the points P (2,3,4) and Q(4, 1, -2).

Accepted Answer

3 i +2 j + k. The mid point is the section formula with ratio 1:1: r =( p + q )/2 With P (2,3,4) and Q (4,1,-2), i.e. p =2 i +3 j +4 k and q =4 i + j -2 k, r =((2+4) i +(3+1) j +(4-2) k )/2=(6 i +4 j +2 k )/2=3 i +2 j + k Position vector of the mid point: 3 i +2 j + k, i.e. the point (3,2,1).

Question 22

Show that the points A, B and C with position vectors, a =3 i -4 j -4 k, b =2 i - j + k and c = i -3 j -5 k, respectively form the vertices of a right angled triangle.

Accepted Answer

Form the side vectors as (terminal position vector - initial position vector), then apply the converse of Pythagoras' theorem to their squared lengths. AB = b - a =(2-3) i +(-1+4) j +(1+4) k =- i +3 j +5 k BC = c - b =(1-2) i +(-3+1) j +(-5-1) k =- i -2 j -6 k CA = a - c =(3-1) i +(-4+3) j +(-4+5) k =2 i - j + k Squared lengths (keep them squared — no surds are needed): AB ^2=1+9+25=35, BC ^2=1+4+36=41, CA ^2=4+1+1=6 First check that a triangle exists: AB =- i +3 j +5 k is not a scalar multiple of CA =2 i - j + k (since -1/2≠3/-1 ), so A, B, C are not collinear. Now AB ^2+ CA ^2=35+6=41= BC ^2 By the converse of Pythagoras' theorem, the angle opposite the longest side BC — the angle at vertex A — is a right angle. Hence A, B and C are the vertices of a right angled triangle, right angled at A.

Question 23

If a and b are two collinear vectors, then which of the following are incorrect: (A) b =λ a, for some scalar λ (B) a = ± b (C) the respective components of a and b are not proportional (D) both the vectors a and b have same direction, but different magnitudes.

Accepted Answer

(B), (C), (D). Take a and b to be nonzero collinear (parallel) vectors. The defining property is b =λ a for some scalar λ≠ 0, equivalently that their corresponding components are proportional. Test each option against this definition, using the concrete collinear pairs a = i, b =2 i and a = i, b =- i. (A) b =λ a for some scalar λ: this is the definition itself, so it is a correct statement. (B) a =± b: this would force a = b, which collinearity does not require. For a = i, b =2 i the vectors are collinear yet a ≠± b. Incorrect in general (it holds only in the special case λ=± 1 ). (C) the respective components are not proportional: from b =λ a we get b 1=λ a 1, b 2=λ a 2, b 3=λ a 3, so the components ARE proportional, with constant of proportionality λ. Incorrect. (D) same direction but different magnitudes: λ may be negative — a = i and b =- i are collinear with opposite directions — and λ=±1 gives equal magnitudes. Neither half of the claim is forced. Incorrect. Only (A) is always true. The incorrect statements are (C) and (D) — and (B) as well, since it holds only when a = b.

Question 24

Find the angle between two vectors a and b with magnitudes √ 3 and 2, respectively having a · b =√ 6.

Accepted Answer

(π)/4. By the definition of the scalar (dot) product, a · b = a b cos θ, where θ is the angle between the two vectors and 0≤θ≤π. Substituting a =√ 3, b =2 and a · b =√ 6, √ 6 =√ 3 × 2× cos θ cos θ=(√ 6 )/(2√ 3 )=(√ 2 )/2=1/(√ 2 ) Since θ is restricted to [0,π], the only angle whose cosine is 1/(√ 2 ) is (π)/4. The angle between a and b is (π)/4 (that is, 45^° ).

Question 25

Find the angle between the vectors i -2 j +3 k and 3 i -2 j + k

Accepted Answer

cos^-1(5/7). Let a = i -2 j +3 k and b =3 i -2 j + k. Use cos θ=( a · b )/( a b ), the dot product being computed componentwise. a · b =(1)(3)+(-2)(-2)+(3)(1)=3+4+3=10 a =√ 1^2+(-2)^2+3^2 =√ 14, b =√ 3^2+(-2)^2+1^2 =√ 14 Therefore cos θ=10/(√ 14 ·√ 14 )=10/14=5/7 The required angle is θ= cos^-1(5/7).

Question 26

Find the projection of the vector i - j on the vector i + j.

Accepted Answer

0. The projection of a on b is ( a · b )/( b ). The step most often got wrong is which magnitude divides: it is the magnitude of the vector being projected onto, i.e. b. Here a = i - j and b = i + j, so a · b =(1)(1)+(-1)(1)=1-1=0, b =√ 1^2+1^2 =√ 2 Projection =0/(√ 2 )=0 The projection is 0. This is consistent: the dot product vanishing means the two vectors are perpendicular, so i - j has no component along i + j.

Question 27

Find the projection of the vector i +3 j +7 k on the vector 7 i - j +8 k.

Accepted Answer

60/(√ 114 ). Projection of a on b is ( a · b )/( b ), with a = i +3 j +7 k and b =7 i - j +8 k. a · b =(1)(7)+(3)(-1)+(7)(8)=7-3+56=60 b =√ 7^2+(-1)^2+8^2 =√ 49+1+64 =√ 114 Hence Projection =60/(√ 114 )=(60√ 114 )/114=(10√ 114 )/19 The projection is 60/(√ 114 ) (=(10√ 114 )/19).

Question 28

Show that each of the given three vectors is a unit vector: 1/7(2 i +3 j +6 k ), 1/7(3 i -6 j +2 k ), 1/7(6 i +2 j -3 k ) Also, show that they are mutually perpendicular to each other.

Accepted Answer

A vector is a unit vector when its magnitude equals 1; two nonzero vectors are perpendicular when their dot product is 0. Write a =1/7(2 i +3 j +6 k ), b =1/7(3 i -6 j +2 k ), c =1/7(6 i +2 j -3 k ). A scalar multiple pulls out of the magnitude as k v = k v. a =1/7√ 2^2+3^2+6^2 =1/7√ 4+9+36 =(√ 49 )/7=7/7=1 b =1/7√ 3^2+(-6)^2+2^2 =1/7√ 9+36+4 =7/7=1 c =1/7√ 6^2+2^2+(-3)^2 =1/7√ 36+4+9 =7/7=1 So each of the three is a unit vector. For the dot products, the two factors of 1/7 give a common 1/49: a · b =1/49 [(2)(3)+(3)(-6)+(6)(2) ]=1/49(6-18+12)=0 b · c =1/49 [(3)(6)+(-6)(2)+(2)(-3) ]=1/49(18-12-6)=0 c · a =1/49 [(6)(2)+(2)(3)+(-3)(6) ]=1/49(12+6-18)=0 Every pair has zero dot product and none of the vectors is 0; hence the three given vectors are mutually perpendicular unit vectors.

Question 29

Find a and b, if ( a + b ) ·( a - b )=8 and a =8 b.

Accepted Answer

(16 √ 2 )/(3 √ 7 ), (2 √ 2 )/(3 √ 7 ). Expand the product using distributivity of the dot product together with its commutativity a · b = b · a and v · v = v ^2: ( a + b )·( a - b )= a · a - a · b + b · a - b · b = a ^2- b ^2 So the first condition is a ^2- b ^2=8 Now substitute a =8 b: (8 b )^2- b ^2=8 ⇒ 64 b ^2- b ^2=8 ⇒ 63 b ^2=8 ⇒ b ^2=8/63 A magnitude is non-negative, so only the positive square root is admissible: b =√ 8/63 =(2√ 2 )/(3√ 7 )=(2√ 14 )/21 a =8 b =(16√ 14 )/21 Hence a =(16√ 14 )/21 and b =(2√ 14 )/21.

Question 30

Evaluate the product (3 a -5 b ) ·(2 a +7 b ).

Accepted Answer

6 a ^2+11 a · b -35 b ^2. Expand using distributivity of the dot product over addition, then use a · a = a ^2, b · b = b ^2 and commutativity b · a = a · b: (3 a -5 b )·(2 a +7 b )=3 a · 2 a +3 a · 7 b -5 b · 2 a -5 b · 7 b =6( a · a )+21( a · b )-10( b · a )-35( b · b ) The middle two terms may be combined only because b · a = a · b: =6 a ^2+(21-10) a · b -35 b ^2 =6 a ^2+11 a · b -35 b ^2 The value of the product is 6 a ^2+11 a · b -35 b ^2.

Question 31

Find the magnitude of two vectors a and b, having the same magnitude and such that the angle between them is 60^ o and their scalar product is 1/2.

Accepted Answer

a =1, b =1. Let the common magnitude be a = b =k, where k 0. By the definition a · b = a b cos θ with θ=60^° and a · b =1/2, 1/2=k· k· cos 60^°=k^2·1/2 k^2=1 ⇒ k=± 1 A magnitude cannot be negative, so k=-1 is rejected and k=1. Hence a = b =1.

Question 32

Find x, if for a unit vector a,( x - a ) ·( x + a )=12.

Accepted Answer

√ 13. Expand the product, using distributivity and x · a = a · x: ( x - a )·( x + a )= x · x + x · a - a · x - a · a = x ^2- a ^2 Since a is a unit vector, a =1, so a ^2=1. Therefore x ^2-1=12 ⇒ x ^2=13 Taking the positive root (a magnitude is non-negative), x =√ 13

Question 33

If a =2 i +2 j +3 k, b =- i +2 j + k and c =3 i + j are such that a +λ b is perpendicular to c, then find the value of λ.

Accepted Answer

8. Two vectors are perpendicular exactly when their dot product is zero. First form a +λ b with a =2 i +2 j +3 k and b =- i +2 j + k: a +λ b =(2-λ) i +(2+2λ) j +(3+λ) k Note that c =3 i + j has k -component 0, so that term drops out of the dot product. Setting ( a +λ b )· c =0: 3(2-λ)+1(2+2λ)+0(3+λ)=0 6-3λ+2+2λ=0 8-λ=0 λ=8

Question 34

Show that a b + b a is perpendicular to a b - b a, for any two nonzero vectors a and b.

Accepted Answer

Two nonzero vectors are perpendicular if and only if their dot product is zero, so compute the dot product of the two given vectors. Using distributivity, and writing a, b as scalars that may be moved out: ( a b + b a )· ( a b - b a ) = a ^2( b · b )- a b ( b · a )+ b a ( a · b )- b ^2( a · a ) The two middle terms cancel because b · a = a · b, and b · b = b ^2, a · a = a ^2: = a ^2 b ^2- b ^2 a ^2=0 Since a and b are nonzero, the dot product being 0 shows that a b + b a is perpendicular to a b - b a.

Question 35

If a · a =0 and a · b =0, then what can be concluded about the vector b?

Accepted Answer

Vector b can be any vector. Use a · a = a ^2. The first condition gives a ^2=0 ⇒ a =0 ⇒ a = 0 So a is forced to be the zero vector. But the zero vector has zero dot product with every vector: 0 · b =0 for all b. Hence the second condition a · b =0 is automatically satisfied and imposes no restriction whatever on b. Conclusion: nothing can be concluded about b — it may be any vector at all.

Question 36

If a, b, c are unit vectors such that a + b + c = 0, find the value of a · b + b · c + c · a.

Accepted Answer

-3/2. The standard move is to take the square of the modulus of the given vector equation, using v ^2= v · v. From a + b + c = 0, a + b + c ^2=0 ( a + b + c )·( a + b + c )=0 Expanding, and pairing terms with commutativity of the dot product, a ^2+ b ^2+ c ^2+2 ( a · b + b · c + c · a )=0 Since a, b, c are unit vectors, a ^2= b ^2= c ^2=1: 3+2 ( a · b + b · c + c · a )=0 a · b + b · c + c · a =-3/2

Question 37

If either vector a = 0 or b = 0, then a · b =0. But the converse need not be true. Justify your answer with an example.

Accepted Answer

Take any two non-zero perpendicular vectors a and b. The converse would say: if a · b =0 then a = 0 or b = 0. This is false, because a · b = a b cos θ can vanish through the third factor cos θ — that is, when the vectors are perpendicular (θ=90^°) — without either magnitude being zero. Counter-example. Take a = i + j, b = i - j Then a =√ 1^2+1^2 =√ 2 ≠ 0, b =√ 1^2+(-1)^2 =√ 2 ≠ 0 so neither vector is the zero vector, yet a · b =(1)(1)+(1)(-1)=1-1=0 Hence a · b =0 does not force a = 0 or b = 0; the converse need not be true.

Question 38

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 and BC ].

Accepted Answer

∠ ABC is the angle at the vertex B, so both vectors must start at B: use BA and BC (not AB, which would give the supplement of the answer). With A (1,2,3), B (-1,0,0), C (0,1,2), BA = (1-(-1) ) i +(2-0) j +(3-0) k =2 i +2 j +3 k BC = (0-(-1) ) i +(1-0) j +(2-0) k = i + j +2 k Then BA · BC =(2)(1)+(2)(1)+(3)(2)=2+2+6=10 BA =√ 4+4+9 =√ 17, BC =√ 1+1+4 =√ 6 Using cos θ= BA · BC BA BC, cos (∠ ABC )=10/(√ 17 ·√ 6 )=10/(√ 102 ) ∠ ABC = cos^-1 (10/(√ 102 ))

Question 39

Show that the points A (1,2,7), B (2,6,3) and C (3,10,-1) are collinear.

Accepted Answer

Three points are collinear when the sum of two of the distances between them equals the third, i.e. AB + BC = AC (equivalently, when AB and BC are parallel with a common point). From A (1,2,7), B (2,6,3), C (3,10,-1), AB =(2-1) i +(6-2) j +(3-7) k = i +4 j -4 k BC =(3-2) i +(10-6) j +(-1-3) k = i +4 j -4 k AC =(3-1) i +(10-2) j +(-1-7) k =2 i +8 j -8 k Their magnitudes are AB =√ 1+16+16 =√ 33, BC =√ 1+16+16 =√ 33 AC =√ 4+64+64 =√ 132 =2√ 33 Since AB + BC =√ 33 +√ 33 =2√ 33 = AC the points A, B, C are collinear. (Indeed AC =2 AB, so B is the midpoint of AC.)

Question 40

Show that the vectors 2 i - j + k, i -3 j -5 k and 3 i -4 j -4 k form the vertices of a right angled triangle.

Accepted Answer

Let the three points A, B, C have position vectors A =2 i - j + k, B = i -3 j -5 k, C =3 i -4 j -4 k A triangle is right angled when one pair of sides meeting at a vertex has zero dot product; equivalently when Pythagoras' relation holds among the squared side lengths. Side vectors: AB = B - A =- i -2 j -6 k, AB ^2=1+4+36=41 BC = C - B =2 i - j + k, BC ^2=4+1+1=6 CA = A - C =- i +3 j +5 k, CA ^2=1+9+25=35 All three lengths are nonzero and BC ^2+ CA ^2=6+35=41= AB ^2 So the longest side AB is the hypotenuse and the right angle is at the opposite vertex C. Confirm it directly with the dot product of the two sides meeting at C: CB = B - C =-2 i + j - k, CA =- i +3 j +5 k CB · CA =(-2)(-1)+(1)(3)+(-1)(5)=2+3-5=0 Hence the three points form the vertices of a triangle that is right angled at C.

Question 41

If a is a nonzero vector of magnitude ' a ' and λ a nonzero scalar, then λ a is unit vector if (A) λ=1 (B) λ=-1 (C) a= λ (D) a=1 / λ

Accepted Answer

For a scalar λ, λ a = λ a. The vector λ a is a unit vector precisely when this magnitude equals 1. With a =a, λ a =1 ⇒ λ a=1 ⇒ a=1/( λ ) (The division is legitimate since λ≠ 0, so λ ≠ 0.) Note that λ must be used, not λ: a negative λ reverses direction but cannot make a length negative. The correct option is (D) a=1/ λ.

Question 42

Find a × b, if a = i -7 j +7 k and b =3 i -2 j +2 k.

Accepted Answer

19 √ 2. By definition, the cross product of a =a 1 i+a 2 j+a 3 k and b =b 1 i+b 2 j+b 3 k is the determinant a × b =. Here a = i-7 j+7 k and b =3 i-2 j+2 k, so a × b =. Expanding along the first row (note the minus sign attached to the j term): a × b = i [(-7)(2)-(7)(-2) ]- j [(1)(2)-(7)(3) ]+ k [(1)(-2)-(-7)(3) ] = i(-14+14)- j(2-21)+ k(-2+21)=0 i+19 j+19 k. Hence a × b =√ 0^2+19^2+19^2 =√ 2× 361 =19√ 2. Final answer: a × b =19√ 2.

Question 43

Find a unit vector perpendicular to each of the vector a + b and a - b, where a =3 i +2 j +2 k and b = i +2 j -2 k.

Accepted Answer

The cross product of two vectors is perpendicular to both of them, so ( a + b )×( a - b ) is perpendicular to a + b and to a - b; dividing it by its own magnitude makes it a unit vector. With a =3 i+2 j+2 k and b = i+2 j-2 k, a + b =4 i+4 j+0 k, a - b =2 i+0 j+4 k. Then ( a + b )×( a - b )= = i(4· 4-0· 0)- j(4· 4-0· 2)+ k(4· 0-4· 2) =16 i-16 j-8 k. Its magnitude is √ 16^2+(-16)^2+(-8)^2 =√ 256+256+64 =√ 576 =24. Therefore the required unit vector is (16 i-16 j-8 k)/24=2/3 i-2/3 j-1/3 k. Final answer: ±1/3(2 i-2 j- k) — the negative of this vector is equally perpendicular to both, so either sign is acceptable.

Question 44

If a unit vector a makes angles (π)/3 with i, (π)/4 with j and an acute angle θ with k, then find θ and hence, the components of a.

Accepted Answer

(π)/3; 1/2, 1/(√ 2 ), 1/2. If a unit vector makes angles α,β,γ with i, j, k, then cos α, cos β, cos γ are its direction cosines and they satisfy cos^2α+ cos^2β+ cos^2γ=1. Here α=(π)/3, β=(π)/4, γ=θ, so cos^2(π)/3+ cos^2(π)/4+ cos^2θ=1 ⇒ (1/2)^2+(1/(√ 2 ))^2+ cos^2θ=1. That is 1/4+1/2+ cos^2θ=1 ⇒ cos^2θ=1/4 ⇒ cos θ=±1/2. The sign is decided by the condition that θ is acute: an acute angle has a positive cosine, so cos θ= 12 and θ=(π)/3. The components of a unit vector are exactly its direction cosines, so a = cos (π)/3 i+ cos (π)/4 j+ cos (π)/3 k=1/2 i+1/(√ 2 ) j+1/2 k. Final answer: θ=(π)/3 and a = 12 i+1/(√2) j+ 12 k, i.e. the components are 12,1/(√2), 12.

Question 45

Show that ( a - b ) ×( a + b )=2( a × b )

Accepted Answer

Use the distributive property of the cross product over addition, together with the two facts u × u = 0 (the cross product of a vector with itself vanishes) and v × u =-( u × v ) (the cross product is anti-commutative). Expanding the left side, being careful to keep the order of the factors in each term: ( a - b )×( a + b )= a × a + a × b - b × a - b × b. Now a × a = 0 and b × b = 0, and by anti-commutativity - b × a =+ a × b. Hence ( a - b )×( a + b )= 0 + a × b + a × b - 0 =2( a × b ). Hence ( a - b )×( a + b )=2( a × b ), as required. (Geometrically: the diagonals of the parallelogram on a, b span a parallelogram of twice its area.)

Question 46

Find λ and μ if (2 i +6 j +27 k ) ×( i +λ j +μ k )= 0.

Accepted Answer

3, 27/2. Compute the cross product as a determinant and set every component equal to zero. (2 i+6 j+27 k)×( i+λ j+μ k)= = i (6μ-27λ)- j (2μ-27)+ k (2λ-6). A vector is 0 only when each of its components is 0, so 6μ-27λ=0, 2μ-27=0, 2λ-6=0. The last two give directly μ=27/2, λ=3. Check these in the first equation (this is the step worth not skipping — the three equations must be consistent): 6(27/2)-27(3)=81-81=0. Final answer: λ=3, μ=27/2. (Equivalently, a vanishing cross product of two non-zero vectors means they are parallel, so 2/1=6/(λ)=27/(μ), giving the same values.)

Question 47

Given that a · b =0 and a × b = 0. What can you conclude about the vectors a and b?

Accepted Answer

Either a =0 or b =0. Suppose, if possible, that both a≠ 0 and b≠ 0, and let θ∈[0,π] be the angle between them. From a· b= a b cos θ=0 with a ≠ 0, b ≠ 0, we get cos θ=0 ⇒ θ=(π)/2. From a× b = a b sin θ=0 with the same non-zero magnitudes, sin θ=0 ⇒ θ=0 or θ=π. These two conclusions contradict each other: θ cannot be π/2 and simultaneously 0 or π. So the supposition is false. Conclusion: at least one of a and b must be the zero vector, i.e. a= 0 or b= 0 (possibly both).

Question 48

Let the vectors a, b, c be given as a 1 i +a 2 j +a 3 k, b 1 i +b 2 j +b 3 k, c 1 i +c 2 j +c 3 k. Then show that a ×( b + c )= a × b + a × c.

Accepted Answer

Write b+ c=(b 1+c 1) i+(b 2+c 2) j+(b 3+c 3) k and expand the cross product as a determinant. a×( b+ c)= = i [a 2(b 3+c 3)-a 3(b 2+c 2) ]- j [a 1(b 3+c 3)-a 3(b 1+c 1) ]+ k [a 1(b 2+c 2)-a 2(b 1+c 1) ]. Now split each bracket into a b -part and a c -part and regroup: = i(a 2b 3-a 3b 2)- j(a 1b 3-a 3b 1)+ k(a 1b 2-a 2b 1) + i(a 2c 3-a 3c 2)- j(a 1c 3-a 3c 1)+ k(a 1c 2-a 2c 1). The first brace is precisely = a× b, and the second brace is precisely = a× c. Hence a×( b+ c)= a× b+ a× c, which is the required distributive law.

Question 49

If either a = 0 or b = 0, then a × b = 0. Is the converse true? Justify your answer with an example.

Accepted Answer

No; take any two nonzero collinear vectors. The converse would read: "if a× b= 0, then a= 0 or b= 0." This is not true. Reason: a× b = a b sin θ, and this vanishes not only when a magnitude is zero but also when sin θ=0, i.e. when the two non-zero vectors are parallel or anti-parallel. Counter-example. Take a=2 i+3 j+4 k, b=4 i+6 j+8 k (=2 a). Both are clearly non-zero, yet a× b= = i(3· 8-4· 6)- j(2· 8-4· 4)+ k(2· 6-3· 4) = i(24-24)- j(16-16)+ k(12-12)= 0. So a× b= 0 with neither vector zero. Conclusion: the converse is false. The correct statement is that a× b= 0 if and only if a= 0, or b= 0, or a and b are parallel.

Question 50

Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).

Accepted Answer

(√ 61 )/2. If two sides of a triangle are the vectors AB and AC, its area is half the area of the parallelogram on them: Area =1/2 AB × AC. With A(1,1,2), B(2,3,5), C(1,5,5), AB =(2-1) i+(3-1) j+(5-2) k= i+2 j+3 k, AC =(1-1) i+(5-1) j+(5-2) k=0 i+4 j+3 k. Then AB × AC = = i(2· 3-3· 4)- j(1· 3-3· 0)+ k(1· 4-2· 0) =-6 i-3 j+4 k. Its magnitude is AB × AC =√ (-6)^2+(-3)^2+4^2 =√ 36+9+16 =√ 61. Final answer: area of ABC=(√ 61 )/2 square units.

Question 51

Find the area of the parallelogram whose adjacent sides are determined by the vectors a = i - j +3 k and b =2 i -7 j + k.

Accepted Answer

15 √ 2. For a parallelogram whose adjacent sides are the vectors a and b, Area = a× b. With a= i- j+3 k and b=2 i-7 j+ k, a× b= = i [(-1)(1)-(3)(-7) ]- j [(1)(1)-(3)(2) ]+ k [(1)(-7)-(-1)(2) ] = i(-1+21)- j(1-6)+ k(-7+2)=20 i+5 j-5 k. Hence a× b =√ 20^2+5^2+(-5)^2 =√ 400+25+25 =√ 450 =15√ 2. Final answer: the area of the parallelogram is 15√ 2 square units.

Question 52

Let the vectors a and b be such that a =3 and b =(√ 2 )/3, then a × b is a unit vector, if the angle between a and b is (A) π / 6 (B) π / 4 (C) π / 3 (D) π / 2

Accepted Answer

(B). Use a× b = a b sin θ, where θ∈[0,π] is the angle between the vectors. " a× b is a unit vector" means a× b =1. With a =3 and b =(√2)/3, 3·(√2)/3· sin θ=1 ⇒ √ 2 sin θ=1 ⇒ sin θ=1/(√2). For 0≤θ≤π this gives θ=(π)/4 or θ=(3π)/4; only (π)/4 appears among the options. Final answer: option (B) (π)/4.

Question 53

Area of a rectangle having vertices A, B, C and D with position vectors - i +1/2 j +4 k, i +1/2 j +4 k, i -1/2 j +4 k and - i -1/2 j +4 k, respectively is (A) 1/2 (B) 1 (C) 2 (D) 4

Accepted Answer

(C). Take the position vectors as points: A(-1, 12,4), B(1, 12,4), C(1,- 12,4), D(-1,- 12,4). The adjacent sides are AB = (1-(-1) ) i+( 12- 12) j+(4-4) k=2 i, BC =(1-1) i+(- 12- 12) j+(4-4) k=- j. (Indeed AB · BC =0, confirming the right angle at B, so the figure really is a rectangle.) The area of the rectangle with adjacent sides AB and BC is AB × BC = = i(0-0)- j(0-0)+ k(-2-0) = -2 k =2. Final answer: option (C) 2.

Question 54

Write down a unit vector in XY-plane, making an angle of 30° with the positive direction of x -axis.

Accepted Answer

(√ 3 )/2 i +1/2 j. A unit vector lying in the XY-plane has no k component, and if it makes an angle θ with the positive x -axis its direction cosines along x and y are cos θ and sin θ. So r= cos θ i+ sin θ j. Putting θ=30^°, r= cos 30^° i+ sin 30^° j=(√ 3 )/2 i+1/2 j. Check: ((√ 3 )/2)^2+(1/2)^2=3/4+1/4=1, so r =1. Required unit vector: (√ 3 )/2 i+1/2 j.

Question 55

Find the scalar components and magnitude of the vector joining the points P (x 1, y 1, z 1) and Q (x 2, y 2, z 2).

Accepted Answer

x 2-x 1, y 2-y 1, z 2-z 1; √ (x 2-x 1)^2+(y 2-y 1)^2+(z 2-z 1)^2. For any two points, the vector joining them is (position vector of the terminal point) - (position vector of the initial point): PQ = OQ - OP. Here OP =x 1 i+y 1 j+z 1 k and OQ =x 2 i+y 2 j+z 2 k, so PQ =(x 2-x 1) i+(y 2-y 1) j+(z 2-z 1) k. The scalar components are the coefficients of i, j, k: x 2-x 1, y 2-y 1, z 2-z 1. The magnitude is the square root of the sum of their squares: PQ =√ (x 2-x 1)^2+(y 2-y 1)^2+(z 2-z 1)^2.

Question 56

A girl walks 4 km towards west, then she walks 3 km in a direction 30° east of north and stops. Determine the girl's displacement from her initial point of departure.

Accepted Answer

-5/2 i +(3 √ 3 )/2 j. Take i pointing due east and j due north, with O her starting point. First leg, 4 km west (the direction - i ): OA =-4 i. Second leg, 3 km at 30^° east of north. "East of north" means the angle is measured from the north direction turning towards the east, so this direction makes 30^° with j — the north component carries the cosine and the east component the sine: AB =3 sin 30^° i+3 cos 30^° j=3/2 i+(3√ 3 )/2 j. By the triangle law of addition, the displacement from the start is OB = OA + AB =(-4+3/2) i+(3√ 3 )/2 j=-5/2 i+(3√ 3 )/2 j. Its magnitude: OB =√ (-5/2)^2+((3√ 3 )/2)^2 =√ 25/4+27/4 =√ 52/4 =√ 13. Displacement =-5/2 i+(3√ 3 )/2 j, i.e. √ 13 ≈ 3.6 km in a direction north-west of O (the i component is negative, the j component positive).

Question 57

If a = b + c, then is it true that a = b + c? Justify your answer.

Accepted Answer

No; take a, b and c to represent the sides of a triangle.. No, it is not true in general. With θ the angle between b and c, a ^2= b+ c ^2=( b+ c)·( b+ c)= b ^2+ c ^2+2 b c cos θ, whereas ( b + c )^2= b ^2+ c ^2+2 b c. Since cos θ≤ 1, we always have a ≤ b + c (the triangle inequality), and the two sides are equal only when cos θ=1, i.e. only when b and c are like vectors (collinear and in the same sense), or one of them is 0. Counterexample: take b= i and c=- i. Then a= b+ c= 0, so a =0 but b + c =2. Hence a = b + c is not true in general; it holds exactly when b and c have the same direction, and otherwise a < b + c.

Question 58

Find the value of x for which x( i + j + k ) is a unit vector.

Accepted Answer

± 1/(√ 3 ). x( i+ j+ k)=x i+x j+x k, so by the formula for the magnitude of a vector, x( i+ j+ k) =√ x^2+x^2+x^2 =√ 3x^2 =√ 3 x. Note √ x^2 = x, not x — that is what keeps the negative root in play. For the vector to be a unit vector, √ 3 x =1 ⇒ x =1/(√ 3 ) ⇒ x=±1/(√ 3 ). Hence x=±1/(√ 3 ).

Question 59

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

Accepted Answer

3/2 √ 10 i +(√ 10 )/2 j. The resultant of the two vectors is their sum: a+ b=(2+1) i+(3-2) j+(-1+1) k=3 i+ j. a+ b =√ 3^2+1^2+0^2 =√ 10. A unit vector along the resultant is c=( a+ b)/( a+ b )=1/(√ 10 )(3 i+ j). A vector of magnitude 5 parallel to it is ± 5 c (both signs qualify as "parallel", one along the resultant and one opposite to it): ±5/(√ 10 )(3 i+ j)=±(15/(√ 10 ) i+5/(√ 10 ) j)=±((3√ 10 )/2 i+(√ 10 )/2 j). Required vector: ±((3√ 10 )/2 i+(√ 10 )/2 j).

Question 60

If a = i + j + k, b =2 i - j +3 k and c = i -2 j + k, find a unit vector parallel to the vector 2 a - b +3 c.

Accepted Answer

3/(√ 22 ) i -3/(√ 22 ) j +2/(√ 22 ) k. First form the combination term by term: 2 a=2 i+2 j+2 k, - b=-2 i+ j-3 k, 3 c=3 i-6 j+3 k. (The sign trap is - b: every component of b flips, so -(- j)=+ j.) Adding, 2 a- b+3 c=(2-2+3) i+(2+1-6) j+(2-3+3) k=3 i-3 j+2 k. Its magnitude: 2 a- b+3 c =√ 3^2+(-3)^2+2^2 =√ 9+9+4 =√ 22. A unit vector parallel to it is the vector divided by its magnitude: ±1/(√ 22 )(3 i-3 j+2 k)=±(3/(√ 22 ) i-3/(√ 22 ) j+2/(√ 22 ) k). The required unit vector is 3/(√ 22 ) i-3/(√ 22 ) j+2/(√ 22 ) k (its negative is also parallel).

Question 61

Show that the points A(1, -2, -8), B(5, 0, -2) and C(11, 3, 7) are collinear, and find the ratio in which B divides AC.

Accepted Answer

2: 3. Three points are collinear if two of the vectors joining them are parallel, i.e. one is a scalar multiple of the other, since they share a common point. AB =(5-1) i+(0-(-2)) j+(-2-(-8)) k=4 i+2 j+6 k, BC =(11-5) i+(3-0) j+(7-(-2)) k=6 i+3 j+9 k, AC =(11-1) i+(3-(-2)) j+(7-(-8)) k=10 i+5 j+15 k. Observe AB =2/3(6 i+3 j+9 k)=2/3 BC, so AB ∥ BC and they have the common point B. Hence A, B, C are collinear. (Equivalently, AB =√ 56 =2√ 14, BC =√ 126 =3√ 14, AC =√ 350 =5√ 14, and AB + BC = AC.) Since the scalar 2/3 is positive, AB and BC point the same way, so B lies between A and C and divides AC internally in the ratio AB: BC =2√ 14:3√ 14 =2:3. B divides AC internally in the ratio 2:3.

Question 62

Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are (2 a + b ) and ( a -3 b ) externally in the ratio 1: 2. Also, show that P is the mid point of the line segment RQ.

Accepted Answer

3 a +5 b. Section formula (external division): if R divides the line joining P and Q externally in the ratio m:n, then r=(m q-n p)/m-n. Here p=2 a+ b, q= a-3 b, m=1, n=2. The denominator m-n=-1 is negative — that sign is where the working usually goes wrong: r=(1·( a-3 b)-2(2 a+ b))/1-2=( a-3 b-4 a-2 b)/-1=(-3 a-5 b)/-1=3 a+5 b. So the position vector of R is 3 a+5 b. Now the midpoint of RQ has position vector ( r+ q)/2=((3 a+5 b)+( a-3 b))/2=(4 a+2 b)/2=2 a+ b= p. This is exactly the position vector of P, so P is the mid point of the line segment RQ.

Question 63

The two adjacent sides of a parallelogram are 2 i -4 j +5 k and i -2 j -3 k. Find the unit vector parallel to its diagonal. Also, find its area.

Accepted Answer

1/7(3 i -6 j +2 k ); 11 √ 5. Let the adjacent sides be a=2 i-4 j+5 k and b= i-2 j-3 k, drawn from the same vertex. Then the diagonals of the parallelogram are a+ b and a- b. Diagonal through the common vertex: a+ b=3 i-6 j+2 k, a+ b =√ 9+36+4 =√ 49 =7, so a unit vector parallel to it is 1/7(3 i-6 j+2 k). The other diagonal: a- b= i-2 j+8 k, a- b =√ 1+4+64 =√ 69, giving the unit vector 1/(√ 69 )( i-2 j+8 k). Area of a parallelogram with adjacent sides a, b is a× b: a× b= = i(12+10)- j(-6-5)+ k(-4+4)=22 i+11 j. a× b =√ 22^2+11^2 =√ 484+121 =√ 605 =11√ 5. Unit vector parallel to the diagonal a+ b: 1/7(3 i-6 j+2 k) (the other diagonal gives 1/(√ 69 )( i-2 j+8 k) ); area =11√ 5 square units.

Question 64

Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are ±(1/(√ 3 ), 1/(√ 3 ), 1/(√ 3 )).

Accepted Answer

Let r be equally inclined to OX, OY and OZ, making the same angle α with each. By definition its direction cosines are l= cos α, m= cos α, n= cos α, so l=m=n. The direction cosines of any vector satisfy l^2+m^2+n^2=1. Hence 3 cos^2α=1 ⇒ cos^2α=1/3 ⇒ cos α=±1/(√ 3 ). Both signs must be kept: taking the square root of cos^2α admits an acute angle α (for a vector in the direction of i+ j+ k ) and an obtuse one (for the oppositely directed vector -( i+ j+ k) ); in each case the three cosines are equal to one another. Therefore the direction cosines are (l,m,n)=±(1/(√ 3 ),1/(√ 3 ),1/(√ 3 )).

EXERCISE 10.1 1–5 (part 1 of 8)

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5