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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
C3 – C1, we get
(β-1) (3-4) = 0.
This gives β = 1.
Also, |c| = √3
√ 1+ α2 + β2 = √3
√ 1+ α2 +1 = √3
√α2 + 2 = √3
α2 + 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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