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**Illustration 1: Let A, B and C be vectors of length 3, 4, 5 respectively. Let A be perpendicular to B + C, B to C + A and C to A + B. then find the length of the vector A + B + C.****Solution: **Given that |A| = 3, |B| = 4 and |C| = 5

Since, A.(B + C) = B.(C + A) = C. (A + B) ….. (1)

Hence, | A + B + C |^{2} = |A|^{2} + |B|^{2} + |C|^{2} + 2 (A.B + B.C + C.A)

= 9 + 16 + 25 + 0

A.B + B.C + C.A = 0….. (follows from eq (1))

Therefore, | A + B + C |^{2} = 50

Hence, | A + B + C | = 5√2

**Illustration 2: If the vector a = i+j+k, the vector b = 4i + 3j + 4k and c = i + αj + βk are linearly dependent vectors and |c| = √3 then,****1. α = 1, β = -1****2. α = 1, β = ±1****3. α = -1, β = ±1****α = ±1, β = 1 ****Solution: **Since a, b and c are given to be linearly dependent vectors, so this also implies that the vectors are coplanar.

So, a. (b x c) = 0

Solving this with the help of determinants, and applying the column operation

C_{3} – C_{1}, we get

(β-1) (3-4) = 0.

This gives β = 1.

Also, |c| = √3

√ 1+ α^{2} + β^{2} = √3

√ 1+ α^{2} +1 = √3

√α^{2} + 2 = √3

α^{2} + 2 = 3 so α = ±1.

**Illustration 3: If a = (i+j+k), a.b = 1 and a x b = j-k, then b is (2003)**

**1. (i-j+k)****2. 2j-k) ****3. i****4. 2i****Solution: **We know that a x (a x b) = (a.b) a – (a.a) b

Hence, (i+j+k) x (j-k) = (i + j + k) – (√3)^{2} b

This gives -2i + j + k = i + j + k – 3b

Hence, 3b = 3i

And this yields b = i.

*Note : Vectors have been denoted by bold letters.

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