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
(β1) (34) = 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 = jk, then b is (2003)
1. (ij+k)
2. 2jk)
3. i
4. 2i
Solution: We know that a x (a x b) = (a.b) a – (a.a) b
Hence, (i+j+k) x (jk) = (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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