Question

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^Answer both the Questions (15,16) from the Attachment^

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Answer 1

hope it helps u ....

Answer 2

The answer is given below :

[My keyboard doesn't have vector sign. So I am using A' to denote A vector, B' to denote B vector and C' to denote C vector.]

15.

Given,

A' + B' = C' .....(i)

Now, taking dot product of A' to both sides of (i), we get

A' . (A' + B') = A' . C'

⇒ A' . A' + A' . B' = A' . C'

⇒ A² + A'. B' = A' . C', since A' . A' = A²
.....(ii)

Again, taking dot product of B' to both sides of (i), we get

B' . (A' + B') = B' . C'

⇒ B' . A' + B' . B' = B' . C'

⇒ A' . B' + B² = B' . C', since A' . B' = B' . A' and B' . B' = B²
.....(iii)

Now, adding (ii) and (iii) no equations, we get

A² + A' . B' + A' . B' + B² = A' . C' + B' + C'

⇒ 2 A' . B' + A² + B² = (A' + B') . C'

⇒ 2 A' . B' + A² + B² = C' . C', by (i)

⇒ 2 A' . B' + A² + B² = C², since C' . C' = C²

⇒ 2 A' . B' + (A + B)² - 2AB = C²

⇒ 2 A' . B' + C² - 2AB = C², since A + B = C

⇒ 2AB cosθ = 2AB, where θ is the angle between A' and B', A' . B' = AB cosθ

⇒ cosθ = 1

⇒ cosθ = cos 0

⇒ θ = 0

Since, the angle between the vectors A' and B' is 0°, the two vectors A' and B' are parallel to each other.

16.

Now,

L.H.S.

= (A' + 2B') . (2A' - 3B')

= A' . 2A' - A' . 3B' + 2B' . 2A' - 2B' . 3B'

= 2 A' . A' - 3 A' . B' + 4 A' . B' - 6 B' . B',
since A' . B' = B' . A'

= 2 A² - 3 AB cosθ + 4 AB cosθ - 6 B²,
since A' . A' = A² and θ being the angle between A' and B', A' . B' = AB cosθ

= 2 A² + AB cosθ - 6 B²

= R.H.S. [Proved]

(These solutions are similar to the previous answerer. Sorry for that.)

Thank you for your question.