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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) â€¦.. (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 = 0â€¦.. (follows from eq (1))

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

Hence, | A + B + C | = 5âˆš2

**Illustration 2: If the vector a = i+j+k, the vector b = 4i + 3j + 4k and c = i + Î±j + Î²k are linearly dependent vectors and |c| = âˆš3 then,****1. Î± = 1, Î² = -1****2. Î± = 1, Î² = Â±1****3. Î± = -1, Î² = Â±1****Î± = Â±1, Î² = 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} â€“ C_{1}, we get

(Î²-1) (3-4) = 0.

This gives Î² = 1.

Also, |c| = âˆš3

âˆš 1+ Î±^{2} + Î²^{2} = âˆš3

âˆš 1+ Î±^{2} +1 = âˆš3

âˆšÎ±^{2} + 2 = âˆš3

Î±^{2} + 2 = 3 so Î± = Â±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 â€“ (a.a) b

Hence, (i+j+k) x (j-k) = (i + j + k) â€“ (âˆš3)^{2} b

This gives -2i + j + k = i + j + k â€“ 3b

Hence, 3b = 3i

And this yields b = i.

*Note : Vectors have been denoted by bold letters.

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