NCERT Solutions for Class 12 Maths Chapter 10: Vector Algebra

Draw a vector of appropriate length making an angle of 30 degrees with the north direction (y-axis) towards the east.

The vector OA represents the displacement where:

• O is the initial point
• A is the terminal point
• Length of OA represents 40 km (to scale)
• The angle from north (positive y-axis) towards east is 30 degrees

So the vector points in the direction that is 30 degrees clockwise from due north, i.e., 60 degrees from the positive x-axis (east direction).

(i) 10 kg - Scalar (mass has only magnitude)
(ii) 2 metres north-west - Vector (displacement has magnitude and direction)
(iii) 40 degrees - Scalar (angle/temperature has only magnitude)
(iv) 40 watts - Scalar (power has only magnitude)
(v) 10^(-19) coulomb - Scalar (charge has only magnitude)
(vi) 20 m/s^2 - Vector (acceleration has both magnitude and direction)

(i) Time period - Scalar (only magnitude, measured in seconds)
(ii) Distance - Scalar (only magnitude, no direction)
(iii) Force - Vector (has both magnitude and direction)
(iv) Velocity - Vector (has both magnitude and direction; speed + direction)
(v) Work done - Scalar (though defined as dot product of force and displacement, work itself is a scalar quantity)

From the given figure:

(i) Coinitial vectors: Vectors a and d are coinitial because they have the same initial point.

(ii) Equal vectors: Vectors b and d are equal because they have the same magnitude and same direction.

(iii) Collinear but not equal: Vectors a and c are collinear (they lie along the same line or parallel lines) but not equal because they have different magnitudes or opposite directions.

(i) True. a and -a are along the same line (opposite directions), so they are collinear.

(ii) False. Collinear vectors are along the same line but can have different magnitudes. For example, 2a and 3a are collinear but not equal in magnitude.

(iii) False. Two vectors can have the same magnitude but different directions (not along the same line). For example, i and j both have magnitude 1 but are not collinear.

(iv) False. Two collinear vectors with the same magnitude can be in opposite directions. For example, a and -a are collinear with same magnitude but not equal.

|a| = sqrt(1^2 + 1^2 + 1^2) = sqrt(3)

|b| = sqrt(2^2 + (-7)^2 + (-3)^2) = sqrt(4 + 49 + 9) = sqrt(62)

|c| = sqrt((1/sqrt(3))^2 + (1/sqrt(3))^2 + (-1/sqrt(3))^2) = sqrt(1/3 + 1/3 + 1/3) = sqrt(1) = 1

Note: c is a unit vector.

Let a = 2i + j - k and b = i + 2j - k

|a| = sqrt(4 + 1 + 1) = sqrt(6)
|b| = sqrt(1 + 4 + 1) = sqrt(6)

Both vectors have magnitude sqrt(6) but are different vectors (different direction).

a = 2i + j - k and b = i + 2j - k (both have magnitude sqrt(6))

Two vectors with the same direction differ only in magnitude.

Let a = 2i + j + k and b = 4i + 2j + 2k = 2a

Both point in the direction of (2, 1, 1). Since b = 2a (positive scalar), they have the same direction.

a = 2i + j + k and b = 4i + 2j + 2k

Two vectors are equal if and only if their corresponding components are equal.

2i + 3j = xi + yj

Comparing i-components: x = 2
Comparing j-components: y = 3

x = 2, y = 3

Vector = Position vector of terminal point - Position vector of initial point
= (-5 - 2)i + (7 - 1)j
= -7i + 6j

Scalar components: -7 and 6
Vector components: -7i and 6j

a + b + c = (1 - 2 + 1)i + (-2 + 4 - 6)j + (1 + 5 - 7)k
= 0i + (-4)j + (-1)k
= -4j - k

or equivalently: 0i - 4j - k

|a| = sqrt(1^2 + 1^2 + 2^2) = sqrt(1 + 1 + 4) = sqrt(6)

Unit vector = a/|a| = (i + j + 2k)/sqrt(6)

a-hat = (1/sqrt(6))i + (1/sqrt(6))j + (2/sqrt(6))k

PQ = (4 - 1)i + (5 - 2)j + (6 - 3)k = 3i + 3j + 3k

|PQ| = sqrt(9 + 9 + 9) = sqrt(27) = 3sqrt(3)

Unit vector = PQ/|PQ| = (3i + 3j + 3k)/(3sqrt(3)) = (i + j + k)/sqrt(3)

PQ-hat = (1/sqrt(3))i + (1/sqrt(3))j + (1/sqrt(3))k

a + b = (2 - 1)i + (-1 + 1)j + (2 - 1)k = i + 0j + k = i + k

|a + b| = sqrt(1 + 0 + 1) = sqrt(2)

Unit vector = (a + b)/|a + b| = (i + k)/sqrt(2)

(a+b)-hat = (1/sqrt(2))i + (1/sqrt(2))k

Let a = 5i - j + 2k
|a| = sqrt(25 + 1 + 4) = sqrt(30)

Unit vector in direction of a: a-hat = a/|a| = (5i - j + 2k)/sqrt(30)

Vector of magnitude 8 in this direction = 8 * a-hat = 8(5i - j + 2k)/sqrt(30)

= (40/sqrt(30))i - (8/sqrt(30))j + (16/sqrt(30))k

or equivalently: (8/sqrt(30))(5i - j + 2k)

Let a = 2i - 3j + 4k and b = -4i + 6j - 8k.

b = -4i + 6j - 8k = -2(2i - 3j + 4k) = -2a

Since b = -2a (b is a scalar multiple of a), the vectors a and b are collinear.

(They are parallel but point in opposite directions.)

Let a = i + 2j + 3k
|a| = sqrt(1 + 4 + 9) = sqrt(14)

Direction cosines: l = a_x/|a|, m = a_y/|a|, n = a_z/|a|

l = 1/sqrt(14), m = 2/sqrt(14), n = 3/sqrt(14)

Direction cosines: (1/sqrt(14), 2/sqrt(14), 3/sqrt(14))

Verification: l^2 + m^2 + n^2 = 1/14 + 4/14 + 9/14 = 14/14 = 1 ✓

AB = (-1 - 1)i + (-2 - 2)j + (1 - (-3))k = -2i - 4j + 4k

|AB| = sqrt(4 + 16 + 16) = sqrt(36) = 6

Direction cosines:
l = -2/6 = -1/3
m = -4/6 = -2/3
n = 4/6 = 2/3

Verification: 1/9 + 4/9 + 4/9 = 9/9 = 1 ✓

Let a = i + j + k
|a| = sqrt(1 + 1 + 1) = sqrt(3)

Direction cosines:
l = 1/sqrt(3), m = 1/sqrt(3), n = 1/sqrt(3)

Angles with axes:
cos alpha = l = 1/sqrt(3) → alpha = cos^(-1)(1/sqrt(3))
cos beta = m = 1/sqrt(3) → beta = cos^(-1)(1/sqrt(3))
cos gamma = n = 1/sqrt(3) → gamma = cos^(-1)(1/sqrt(3))

Since alpha = beta = gamma, the vector i + j + k is equally inclined to all three axes.

P = i + 2j - k, Q = -i + j + k, ratio m:n = 2:1

(i) Internally: R = (mQ + nP)/(m + n) = (2(-i + j + k) + 1(i + 2j - k))/(2 + 1)
= (-2i + 2j + 2k + i + 2j - k)/3
= (-i + 4j + k)/3
R = (-1/3)i + (4/3)j + (1/3)k

(ii) Externally: R = (mQ - nP)/(m - n) = (2(-i + j + k) - 1(i + 2j - k))/(2 - 1)
= (-2i + 2j + 2k - i - 2j + k)/1
= -3i + 0j + 3k
R = -3i + 3k

Midpoint M = (P + Q)/2

P = 2i + 3j + 4k, Q = 4i + j - 2k

M = ((2 + 4)/2)i + ((3 + 1)/2)j + ((4 + (-2))/2)k
= 3i + 2j + k

Position vector of midpoint = 3i + 2j + k

AB = b - a = (2 - 3)i + (-1 + 4)j + (1 + 4)k = -i + 3j + 5k
BC = c - b = (1 - 2)i + (-3 + 1)j + (-5 - 1)k = -i - 2j - 6k
CA = a - c = (3 - 1)i + (-4 + 3)j + (-4 + 5)k = 2i - j + k

|AB|^2 = 1 + 9 + 25 = 35
|BC|^2 = 1 + 4 + 36 = 41
|CA|^2 = 4 + 1 + 1 = 6

Checking: |AB|^2 + |CA|^2 = 35 + 6 = 41 = |BC|^2 ✓

Since |BC|^2 = |AB|^2 + |CA|^2, by the Pythagoras theorem, triangle ABC is a right-angled triangle with the right angle at vertex A.

By the triangle law of vector addition:
AB + BC = AC → AB + BC - AC = 0 → Option (B) is true ✓

Also: AB + BC + CA = AB + BC - AC = 0 (since CA = -AC) → Option (A) is true ✓

Option (D): AB - CB + CA = AB + BC + CA = 0 (since -CB = BC) → Option (D) is true ✓

Option (C): AB + BC - CA = AB + BC + AC. This equals AC + AC = 2AC (not zero in general).

Answer: (C) AB + BC - CA = 0 is NOT true.

If a and b are collinear, then b = lambda*a for some scalar lambda.

(A) b = lambda*a - Correct. This is the definition of collinear vectors.

(B) a = +-b - Incorrect. Collinear vectors need not have the same magnitude. For example, a = i and b = 3i are collinear but a ≠ +-b.

(C) Components are proportional - Correct. If b = lambda*a, then each component of b is lambda times the corresponding component of a.

(D) Same direction - Incorrect. Collinear vectors can have opposite directions (if lambda < 0). For example, a = i and b = -2i are collinear but have opposite directions.

Incorrect options: (B) and (D)

a.b = |a||b|cos(theta)
sqrt(6) = sqrt(3) * 2 * cos(theta)
sqrt(6) = 2sqrt(3) cos(theta)
cos(theta) = sqrt(6)/(2sqrt(3)) = sqrt(2)/2 = 1/sqrt(2)

theta = pi/4 (or 45 degrees)

a = i - 2j + 3k, b = 3i - 2j + k

a.b = (1)(3) + (-2)(-2) + (3)(1) = 3 + 4 + 3 = 10

|a| = sqrt(1 + 4 + 9) = sqrt(14)
|b| = sqrt(9 + 4 + 1) = sqrt(14)

cos(theta) = a.b/(|a||b|) = 10/(sqrt(14) * sqrt(14)) = 10/14 = 5/7

theta = cos^(-1)(5/7)

Projection of a on b = (a.b)/|b|

a = i - j, b = i + j

a.b = (1)(1) + (-1)(1) = 1 - 1 = 0

Projection = 0/|b| = 0

The vectors are perpendicular, so the projection is zero.

a = i + 3j + 7k, b = 7i - j + 8k

a.b = (1)(7) + (3)(-1) + (7)(8) = 7 - 3 + 56 = 60

|b| = sqrt(49 + 1 + 64) = sqrt(114)

Projection = a.b/|b| = 60/sqrt(114)

Let a = (1/7)(2i + 3j + 6k), b = (1/7)(3i - 6j + 2k), c = (1/7)(6i + 2j - 3k)

Unit vectors:
|a| = (1/7)sqrt(4 + 9 + 36) = (1/7)sqrt(49) = (1/7)(7) = 1 ✓
|b| = (1/7)sqrt(9 + 36 + 4) = (1/7)(7) = 1 ✓
|c| = (1/7)sqrt(36 + 4 + 9) = (1/7)(7) = 1 ✓

Mutually perpendicular:
a.b = (1/49)(6 - 18 + 12) = (1/49)(0) = 0 ✓
b.c = (1/49)(18 - 12 - 6) = (1/49)(0) = 0 ✓
a.c = (1/49)(12 + 6 - 18) = (1/49)(0) = 0 ✓

Since all dot products are zero, the vectors are mutually perpendicular.

(a + b).(a - b) = a.a - a.b + b.a - b.b = |a|^2 - |b|^2

Given: |a|^2 - |b|^2 = 8 ... (i)
Given: |a| = 8|b| → |a|^2 = 64|b|^2 ... (ii)

Substituting (ii) in (i):
64|b|^2 - |b|^2 = 8
63|b|^2 = 8
|b|^2 = 8/63
|b| = sqrt(8/63) = 2sqrt(2)/(3sqrt(7)) = 2sqrt(2/63)

|a| = 8|b| = 8 * 2sqrt(2/63) = 16sqrt(2/63)

Simplified: |b| = 2sqrt(2)/sqrt(63) = 2sqrt(2)/(3sqrt(7)), |a| = 16sqrt(2)/(3sqrt(7))

(3a - 5b).(2a + 7b)
= 3a.2a + 3a.7b - 5b.2a - 5b.7b
= 6(a.a) + 21(a.b) - 10(b.a) - 35(b.b)
= 6|a|^2 + 21(a.b) - 10(a.b) - 35|b|^2
= 6|a|^2 + 11(a.b) - 35|b|^2

Given: |a| = |b| = x (say), theta = 60 degrees, a.b = 1/2

a.b = |a||b|cos(theta)
1/2 = x * x * cos 60
1/2 = x^2 * (1/2)
x^2 = 1
x = 1 (magnitude is positive)

|a| = |b| = 1

(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.

|x|^2 - 1 = 12
|x|^2 = 13

|x| = sqrt(13)

a + lambda*b = (2 - lambda)i + (2 + 2lambda)j + (3 + lambda)k

For (a + lambdab) perpendicular to c: (a + lambdab).c = 0

(2 - lambda)(3) + (2 + 2lambda)(1) + (3 + lambda)(0) = 0
6 - 3lambda + 2 + 2lambda + 0 = 0
8 - lambda = 0

lambda = 8

Let p = |a|b + |b|a and q = |a|b - |b|a

p.q = (|a|b + |b|a).(|a|b - |b|a)
= |a|^2(b.b) - |a||b|(b.a) + |b||a|(a.b) - |b|^2(a.a)
= |a|^2|b|^2 - |a||b|(a.b) + |a||b|(a.b) - |b|^2|a|^2
= |a|^2|b|^2 - |b|^2|a|^2
= 0

Since p.q = 0, the vectors p and q are perpendicular.

a.a = 0 means |a|^2 = 0, which implies a = 0 (zero vector).

If a is the zero vector, then a.b = 0.b = 0 for ANY vector b.

Therefore, b can be any vector (no restriction on b).

Given: a + b + c = 0 and |a| = |b| = |c| = 1

Squaring both sides (taking dot product with itself):
(a + b + c).(a + b + c) = 0
|a|^2 + |b|^2 + |c|^2 + 2(a.b + b.c + c.a) = 0
1 + 1 + 1 + 2(a.b + b.c + c.a) = 0
3 + 2(a.b + b.c + c.a) = 0

a.b + b.c + c.a = -3/2

Converse: If a.b = 0, then a = 0 or b = 0. This is NOT necessarily true.

Counterexample: Let a = i and b = j
a.b = (1)(0) + (0)(1) = 0

But a ≠ 0 (|a| = 1) and b ≠ 0 (|b| = 1).

Here a.b = 0 because a and b are perpendicular (not because either is zero).

So the converse does not hold - a.b = 0 can also mean the vectors are perpendicular.

BA = A - B = (1-(-1))i + (2-0)j + (3-0)k = 2i + 2j + 3k
BC = C - B = (0-(-1))i + (1-0)j + (2-0)k = i + j + 2k

BA.BC = (2)(1) + (2)(1) + (3)(2) = 2 + 2 + 6 = 10

|BA| = sqrt(4 + 4 + 9) = sqrt(17)
|BC| = sqrt(1 + 1 + 4) = sqrt(6)

cos(angle ABC) = BA.BC/(|BA||BC|) = 10/(sqrt(17) * sqrt(6)) = 10/sqrt(102)

Angle ABC = cos^(-1)(10/sqrt(102))

AB = B - A = (2-1)i + (6-2)j + (3-7)k = i + 4j - 4k
AC = C - A = (3-1)i + (10-2)j + (-1-7)k = 2i + 8j - 8k

AC = 2(i + 4j - 4k) = 2 AB

Since AC = 2 AB (AC is a scalar multiple of AB), the vectors AB and AC are parallel and share the common point A.

Therefore, the points A, B, C are collinear.

Let a = 2i - j + k, b = i - 3j - 5k, c = 3i - 4j - 4k

AB = b - a = -i - 2j - 6k, |AB|^2 = 1 + 4 + 36 = 41
BC = c - b = 2i - j + k, |BC|^2 = 4 + 1 + 1 = 6
CA = a - c = -i + 3j + 5k, |CA|^2 = 1 + 9 + 25 = 35

|BC|^2 + |CA|^2 = 6 + 35 = 41 = |AB|^2 ✓

By the Pythagoras theorem, triangle ABC is right-angled at C.

Verification using dot product: At vertex C, the two sides are CB and CA.
CB = b - c = -2i + j - k
CA = a - c = -i + 3j + 5k
CB.CA = (-2)(-1) + (1)(3) + (-1)(5) = 2 + 3 - 5 = 0 ✓

Since CB.CA = 0, CB is perpendicular to CA, confirming right angle at C.

|lambda*a| = |lambda| * |a| = |lambda| * a

For lambda*a to be a unit vector: |lambda| * a = 1
a = 1/|lambda|

Answer: (D) a = 1/|lambda|

a x b = |i j k|
|1 -7 7|
|3 -2 2|

= i((-7)(2) - (7)(-2)) - j((1)(2) - (7)(3)) + k((1)(-2) - (-7)(3))
= i(-14 + 14) - j(2 - 21) + k(-2 + 21)
= i(0) - j(-19) + k(19)
= 0i + 19j + 19k
= 19j + 19k

|a x b| = sqrt(0 + 361 + 361) = sqrt(722) = 19sqrt(2)

a + b = 4i + 4j + 0k
a - b = 2i + 0j + 4k

(a+b) x (a-b) = |i j k|
|4 4 0|
|2 0 4|

= i(44 - 00) - j(44 - 02) + k(40 - 42)
= i(16) - j(16) + k(-8)
= 16i - 16j - 8k
= 8(2i - 2j - k)

|8(2i - 2j - k)| = 8sqrt(4 + 4 + 1) = 83 = 24

Unit vector = (16i - 16j - 8k)/24 = (2/3)i - (2/3)j - (1/3)k

or equivalently: (1/3)(2i - 2j - k)

Since a is a unit vector, if it makes angles alpha, beta, gamma with i, j, k:
cos^2(alpha) + cos^2(beta) + cos^2(gamma) = 1

cos^2(pi/3) + cos^2(pi/4) + cos^2(theta) = 1
(1/2)^2 + (1/sqrt(2))^2 + cos^2(theta) = 1
1/4 + 1/2 + cos^2(theta) = 1
cos^2(theta) = 1 - 3/4 = 1/4
cos(theta) = 1/2 (since theta is acute)

theta = pi/3

Components of a:
a = cos(pi/3)i + cos(pi/4)j + cos(pi/3)k
a = (1/2)i + (1/sqrt(2))j + (1/2)k

(a - b) x (a + b)
= a x a + a x b - b x a - b x b
= 0 + a x b - (-a x b) - 0 [since a x a = 0, b x b = 0, and b x a = -(a x b)]
= a x b + a x b
= 2(a x b) ✓

Hence proved.

If a x b = 0, then a and b are parallel (collinear).

So (2i + 6j + 27k) and (i + lambdaj + muk) must be parallel.

For parallel vectors, corresponding components are proportional:
2/1 = 6/lambda = 27/mu

From 2/1 = 6/lambda: lambda = 6/2 = 3
From 2/1 = 27/mu: mu = 27/2 = 27/2

a.b = 0 means either:

• a = 0 or b = 0, OR
• a is perpendicular to b

a x b = 0 means either:

• a = 0 or b = 0, OR
• a is parallel to b

Two nonzero vectors cannot be both perpendicular AND parallel simultaneously.

Therefore, if a.b = 0 and a x b = 0, then at least one of a or b must be the zero vector.

b + c = (b1 + c1)i + (b2 + c2)j + (b3 + c3)k

a x (b + c) = |i j k|
|a1 a2 a3|
|b1+c1 b2+c2 b3+c3|

= i(a2(b3+c3) - a3(b2+c2)) - j(a1(b3+c3) - a3(b1+c1)) + k(a1(b2+c2) - a2(b1+c1))

Expanding:
= i(a2b3 + a2c3 - a3b2 - a3c2) - j(a1b3 + a1c3 - a3b1 - a3c1) + k(a1b2 + a1c2 - a2b1 - a2c1)

= [i(a2b3-a3b2) - j(a1b3-a3b1) + k(a1b2-a2b1)] + [i(a2c3-a3c2) - j(a1c3-a3c1) + k(a1c2-a2c1)]

= (a x b) + (a x c) ✓

Hence proved: Cross product distributes over addition.

Converse: If a x b = 0, then a = 0 or b = 0. This is NOT necessarily true.

Counterexample: Let a = i and b = 2i
a x b = i x 2i = 2(i x i) = 2(0) = 0

But a ≠ 0 and b ≠ 0.

Here a x b = 0 because a and b are parallel (not because either is zero).

So the converse does not hold - a x b = 0 can also mean the vectors are parallel.

AB = B - A = i + 2j + 3k
AC = C - A = 0i + 4j + 3k

AB x AC = |i j k|
|1 2 3|
|0 4 3|

= i(23 - 34) - j(13 - 30) + k(14 - 20)
= i(6 - 12) - j(3 - 0) + k(4 - 0)
= -6i - 3j + 4k

|AB x AC| = sqrt(36 + 9 + 16) = sqrt(61)

Area of triangle = (1/2)|AB x AC| = sqrt(61)/2 square units

Area = |a x b|

a x b = |i j k|
|1 -1 3|
|2 -7 1|

= i((-1)(1) - (3)(-7)) - j((1)(1) - (3)(2)) + k((1)(-7) - (-1)(2))
= i(-1 + 21) - j(1 - 6) + k(-7 + 2)
= 20i + 5j - 5k

|a x b| = sqrt(400 + 25 + 25) = sqrt(450) = 15sqrt(2)

Area = 15sqrt(2) square units

|a x b| = |a||b|sin(theta)

For a x b to be a unit vector: |a x b| = 1

3 * (sqrt(2)/3) * sin(theta) = 1
sqrt(2) * sin(theta) = 1
sin(theta) = 1/sqrt(2)

theta = pi/4

Answer: (B) pi/4

AB = B - A = (1-(-1))i + (1/2-1/2)j + (4-4)k = 2i
AD = D - A = (-1-(-1))i + (-1/2-1/2)j + (4-4)k = -j

AB x AD = |i j k|
|2 0 0|
|0 -1 0|

= i(0-0) - j(0-0) + k(2*(-1) - 0*0)
= -2k

Area = |AB x AD| = |-2k| = 2

Answer: (C) 2

A unit vector in the XY-plane making angle theta with the positive x-axis:
a-hat = cos(theta)i + sin(theta)j

For theta = 30 degrees:
a-hat = cos 30 i + sin 30 j = (sqrt(3)/2)i + (1/2)j

Unit vector = (sqrt(3)/2)i + (1/2)j

PQ = (x2 - x1)i + (y2 - y1)j + (z2 - z1)k

Scalar components: (x2 - x1), (y2 - y1), (z2 - z1)

Magnitude: |PQ| = sqrt((x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2)

Taking east as positive x-axis and north as positive y-axis:

Step 1: 4 km west = -4i
Step 2: 3 km at 30 degrees east of north = 3 sin 30 i + 3 cos 30 j = (3/2)i + (3sqrt(3)/2)j

Total displacement = -4i + (3/2)i + (3sqrt(3)/2)j
= (-4 + 3/2)i + (3sqrt(3)/2)j
= (-5/2)i + (3sqrt(3)/2)j

Magnitude = sqrt(25/4 + 27/4) = sqrt(52/4) = sqrt(13)

Displacement = (-5/2)i + (3sqrt(3)/2)j, magnitude = sqrt(13) km

No, this is NOT true in general.

By the triangle inequality: |a| = |b + c| <= |b| + |c|

Equality holds only when b and c have the same direction.

Counterexample: Let b = i and c = j
a = b + c = i + j
|a| = sqrt(2) ≈ 1.414
|b| + |c| = 1 + 1 = 2
sqrt(2) ≠ 2

So |a| ≠ |b| + |c| in general. The correct relation is the triangle inequality: |a| <= |b| + |c|.

|x(i + j + k)| = |x| * |i + j + k| = |x| * sqrt(3)

For a unit vector: |x| * sqrt(3) = 1
|x| = 1/sqrt(3)

x = +- 1/sqrt(3)

Resultant r = a + b = 3i + j + 0k = 3i + j

|r| = sqrt(9 + 1) = sqrt(10)

Unit vector in direction of r: r-hat = (3i + j)/sqrt(10)

Vector of magnitude 5: 5 * r-hat = 5(3i + j)/sqrt(10) = (15/sqrt(10))i + (5/sqrt(10))j

= (15i + 5j)/sqrt(10) = (5/sqrt(10))(3i + j)

Rationalizing: = (3sqrt(10)/2)i + (sqrt(10)/2)j

2a = 2i + 2j + 2k
-b = -2i + j - 3k
3c = 3i - 6j + 3k

2a - b + 3c = (2-2+3)i + (2+1-6)j + (2-3+3)k = 3i - 3j + 2k

|2a - b + 3c| = sqrt(9 + 9 + 4) = sqrt(22)

Unit vector = (3i - 3j + 2k)/sqrt(22)
= (3/sqrt(22))i - (3/sqrt(22))j + (2/sqrt(22))k

AB = B - A = 4i + 2j + 6k
AC = C - A = 10i + 5j + 15k = (5/2)(4i + 2j + 6k) = (5/2)AB

Since AC = (5/2)AB, vectors AB and AC are parallel with common point A.
Therefore A, B, C are collinear.

Now AB = (2/5)AC, so B divides AC in the ratio AB:BC.
Since AC = AB + BC: BC = AC - AB = (5/2)AB - AB = (3/2)AB
AB:BC = AB : (3/2)AB = 1 : 3/2 = 2:3

B divides AC internally in the ratio 2:3.

For external division in ratio m:n = 1:2:

R = (mQ - nP)/(m - n) = (1(a - 3b) - 2(2a + b))/(1 - 2)
= (a - 3b - 4a - 2b)/(-1)
= (-3a - 5b)/(-1)
= 3a + 5b

Let a = 2i - 4j + 5k and b = i - 2j - 3k

Diagonal: d = a + b = 3i - 6j + 2k
|d| = sqrt(9 + 36 + 4) = sqrt(49) = 7

Unit vector parallel to diagonal = d/|d| = (3i - 6j + 2k)/7
= (3/7)i - (6/7)j + (2/7)k

Area: a x b = |i j k|
|2 -4 5|
|1 -2 -3|

= i((-4)(-3) - (5)(-2)) - j((2)(-3) - (5)(1)) + k((2)(-2) - (-4)(1))
= i(12 + 10) - j(-6 - 5) + k(-4 + 4)
= 22i + 11j + 0k

|a x b| = sqrt(484 + 121) = sqrt(605) = 11sqrt(5)

Area = 11sqrt(5) square units

Let the vector make equal angles alpha with each axis.
Then: l = cos alpha, m = cos alpha, n = cos alpha

Using l^2 + m^2 + n^2 = 1:
3cos^2(alpha) = 1
cos^2(alpha) = 1/3
cos(alpha) = +- 1/sqrt(3)

Taking positive values (assuming the vector is in the first octant):

Direction cosines = 1/sqrt(3), 1/sqrt(3), 1/sqrt(3) ✓

d is perpendicular to both a and b, so d is parallel to a x b.

a x b = |i j k|
|1 4 2|
|3 -2 7|

= i(47 - 2(-2)) - j(17 - 23) + k(1*(-2) - 4*3)
= i(28 + 4) - j(7 - 6) + k(-2 - 12)
= 32i - j - 14k

So d = lambda(32i - j - 14k) for some scalar lambda.

Using c.d = 15:
(2i - j + 4k).(lambda(32i - j - 14k)) = 15
lambda(64 + 1 - 56) = 15
lambda(9) = 15
lambda = 5/3

d = (5/3)(32i - j - 14k) = (160/3)i - (5/3)j - (70/3)k

Sum = (2 + lambda)i + (4 + 2)j + (-5 + 3)k = (2 + lambda)i + 6j - 2k

Magnitude of sum = sqrt((2+lambda)^2 + 36 + 4) = sqrt((2+lambda)^2 + 40)

Unit vector along sum = [(2+lambda)i + 6j - 2k] / sqrt((2+lambda)^2 + 40)

Scalar product with (i + j + k):
[(2+lambda) + 6 - 2] / sqrt((2+lambda)^2 + 40) = 1
(lambda + 6) / sqrt((2+lambda)^2 + 40) = 1

Squaring: (lambda + 6)^2 = (2 + lambda)^2 + 40
lambda^2 + 12lambda + 36 = lambda^2 + 4lambda + 4 + 40
12lambda + 36 = 4lambda + 44
8lambda = 8
lambda = 1

Let |a| = |b| = |c| = m (say). Since they are mutually perpendicular:
a.b = b.c = c.a = 0

Let r = a + b + c
|r|^2 = |a + b + c|^2 = |a|^2 + |b|^2 + |c|^2 + 2(a.b + b.c + c.a)
= m^2 + m^2 + m^2 + 0 = 3m^2
|r| = m*sqrt(3)

Angle with a:
cos(alpha) = r.a / (|r||a|) = (a.a + b.a + c.a)/(msqrt(3) * m)
= (m^2 + 0 + 0)/(m^2sqrt(3)) = 1/sqrt(3)

Angle with b:
cos(beta) = r.b / (|r||b|) = (a.b + b.b + c.b)/(m^2sqrt(3))
= (0 + m^2 + 0)/(m^2sqrt(3)) = 1/sqrt(3)

Angle with c:
cos(gamma) = r.c / (|r||c|) = 1/sqrt(3) (similarly)

Since alpha = beta = gamma = cos^(-1)(1/sqrt(3)), the vector a + b + c is equally inclined to a, b, and c.

Angle = cos^(-1)(1/sqrt(3))

(a + b).(a + b)
= a.a + a.b + b.a + b.b
= |a|^2 + a.b + a.b + |b|^2 [since b.a = a.b]
= |a|^2 + 2a.b + |b|^2 ✓

Hence proved.

a + b + c = 0 → a + b = -c

Taking dot product of both sides with themselves:
(a + b).(a + b) = (-c).(-c)
|a|^2 + 2a.b + |b|^2 = |c|^2
9 + 2a.b + 25 = 49
2a.b = 49 - 34 = 15
a.b = 15/2

cos(theta) = a.b/(|a||b|) = (15/2)/(3*5) = (15/2)/15 = 1/2

theta = pi/3 (or 60 degrees)

Verifying unit vectors:
|a-hat| = (1/sqrt(26))*sqrt(1 + 9 + 16) = (1/sqrt(26))*sqrt(26) = 1 ✓
|b-hat| = (1/7)*sqrt(4 + 9 + 36) = (1/7)*sqrt(49) = 1 ✓

Dot product:
a-hat.b-hat = (1/(7sqrt(26)))[(-1)(2) + (3)(3) + (4)(-6)]
= (1/(7sqrt(26)))[-2 + 9 - 24]
= -17/(7sqrt(26))

Angle between them:
cos(theta) = a-hat.b-hat = -17/(7sqrt(26))

theta = cos^(-1)(-17/(7sqrt(26)))

|a.b| = |a||b||cos(theta)|
|a x b| = |a||b|sin(theta)

Setting them equal:
|a||b||cos(theta)| = |a||b|sin(theta)
|cos(theta)| = sin(theta)

For 0 <= theta <= pi:
sin(theta) >= 0, so |cos(theta)| = sin(theta)

Case 1: cos(theta) >= 0 → cos(theta) = sin(theta) → tan(theta) = 1 → theta = pi/4
Case 2: cos(theta) < 0 → -cos(theta) = sin(theta) → tan(theta) = -1 → theta = 3pi/4

Among the options: theta = pi/4

Answer: (B) pi/4

a.b = |a||b|cos(theta)

Since |a| >= 0 and |b| >= 0:
a.b >= 0 when cos(theta) >= 0
cos(theta) >= 0 when 0 <= theta <= pi/2

Answer: (B) 0 <= theta <= pi/2