Solution:

Since a + b + c = 0

On squaring both sides, we get

|a|² + |b|² + |c|² + 2 (a . b + b . c + c . a) = 0

⇒ 2 (a . b + b . c + c . a) = − (9 + 16 + 25)

⇒ a . b + b . c + c . a = −25

Solution:

If x, y are the original components; X, Y the new components and α is the angle of rotation, then x = X cosα − Y sinα and y = X sinα + Y cosα

Therefore, 2p = (p + 1) cosα − sinα and 1 = (p + 1) sinα + cosα

Squaring and adding, we get 4p² + 1 = (p + 1)² + 1

⇒ p + 1 = ± 2p

⇒ p = 1 or −1 / 3

Solution:

Let a = li + mj + nk, where l² + m² + n² = 1. a makes an angle π / 4 with z−axis.

Hence, n = 1 / √2, l² + m² = 1 / 2 …..(i)

Therefore, a = li + mj + k / √2

a + i + j = (l + 1) i + (m + 1) j + k / √2

Its magnitude is 1, hence (l + 1)² + (m + 1)² = 1 / 2 …..(ii)

From (i) and (ii),

2lm = 1 / 2

⇒ l = m = −1 / 2

Hence, a = [−i / 2] − [j / 2] + [k / √2].

Solution:

|p| = |q| = |r| = c, (say) and

p . q = 0 = p . r = q . r

p × |( x − q) × p |+ q × |(x − r) × q| + r × |( x − p) × r| = 0

⇒ (p . p) (x − q) − {p . (x − q)} p + . . . . . . . . . = 0

⇒ c² (x − q + x − r + x − p) − (p . x) p − (q . x) q − (r . x) r = 0

⇒ c² {3x − (p + q + r)} − [(p . x) p + (q . x) q + (r . x) r] = 0

which is satisfied by x = [1 / 2] (p + q+ r).

Solution:

|a − b|² + |b − c|² + |c − a|² = 2 (a² + b² + c²) − 2 (a * b + b * c + c * a)

= 2 * 3 − 2 (a * b + b * c + c * a)

= 6 − {(a + b + c)² − a²− b² − c²}

= 9 − |a + b + c| 2 ≤ 9

Solution:

a. [(b + c) × (a + b + c)]

= a . (b × a + b × b + b × c) + a . (c × a + c × b + c × c)

= [aba] + [abb] + [a b c] + [aca] + [a c b] + [a c c]

= 0 + 0 + [abc] + 0 − [abc] + 0

= 0

Solution:

Let r = λb + μc and c = ± (xi + yj).

Since c and b are perpendicular, we have 4x + 3y = 0

⇒ c = ±x (i − 43j), {Because, y = [−4 / 3]x}

Now, projection of r on b = [r . b] / [|b|] = 1

⇒ [(λb + μc) . b] / [|b|]

= [λb . B] / [|b|] = 1

⇒ λ = 1 / 5

Again, projection of r on c = [r . c] / [|c|] = 2

This gives μx = [6 / 5]

⇒ r = [1 / 5] (4i + 3j) + [6 / 5] (i − [4 / 3]j)

= 2i−j or

r = [1 / 5] (4i + 3j) − [6 / 5] (i − [4 / 3]j)

= [−2 / 5] i + [11 / 5] j

Solution:

R = √[2² + 10² + 11²].

= √[4 + 100 + 121]

= 15

Solution:

Let i be a unit vector in the direction of b, j in the direction of c.

Note that b = i and c = j

We have b × c = |b| |c| sinαk = sinαk, where k is a unit vector perpendicular to b and c. ⇒ |b × c| = sinα

⇒ k = [b × c] / [|b × c|]

Any vector a can be written as a linear combination of i, j and k.

Let a = a₁i + a₂j + a₃k

Now a . b = a . i = a₁, a . c = a . j = a₂ and {[a] . [b × c] / [|b × c|]} = a . k = a₃

Thus, (a . b) b + (a . c) c + {[a] . [(b × c) / |b × c|] * [(b × c)|]}

= a₁b + a₂c + a₃ [b × c] / [|b × c|]

= a₁i + a₂j + a₃k

= a

Solution:

a × b = \begin{vmatrix} i & j & k \\ 2 &1 &-2 \\ 1&1 &0 \end{vmatrix}∣∣∣∣∣∣∣​i21​j11​k−20​∣∣∣∣∣∣∣​ = 2i − 2j + k

Therefore, |a × b| = √[4 + 4 + 1] = 3

|c − a| = 2√2

⇒ |c − a|² = (c − a)² = 8

⇒ |c|² − 2 a . c + |a|² = 8

⇒|c|² − 2|c| + 9 = 8

⇒|c|² − 2|c| + 1 = 0

⇒|c| = 1

Hence, |(a × b) × c| = |a × b| |c|sin 30^∘ = 3 / 2.