It is given that aa, bb and cc are non-zero vectors. If ∣a+b∣=∣a−b∣|a+b| = |a-b|, show that the two vectors aa and bb are perpendicular to each other. (i) Explain why the result of a×(b+c)+b×(c+a)+c×(a+b)a\times(b+c)+b\times(c+a)+c\times(a+b) is a vector. (ii) Simplify a×(b+c)+b×(c+a)+c×(a+b)a\times(b+c)+b\times(c+a)+c\times(a+b). Show your workings clearly. Explain why the result of a×(b+c)+b×(c+a)+c×(a+b)a\times(b+c)+b\times(c+a)+c\times(a+b) is a vector. Simplify a×(b+c)+b×(c+a)+c×(a+b)a\times(b+c)+b\times(c+a)+c\times(a+b). Show your workings clearly.

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Worked solution

5(a)

1. Square both sides of the given equation to relate magnitude to scalar product.

∣a+b∣=∣a−b∣|a+b| = |a-b|

∣a+b∣2=∣a−b∣2|a+b|^2 = |a-b|^2

2. Expand both sides using the scalar product property $|v|^2 = v \cdot v$.

(a+b)⋅(a+b)=(a−b)⋅(a−b)(a+b)\cdot(a+b) = (a-b)\cdot(a-b)

a⋅a+a⋅b+b⋅a+b⋅b=a⋅a−a⋅b−b⋅a+b⋅ba\cdot a + a\cdot b + b\cdot a + b\cdot b = a\cdot a - a\cdot b - b\cdot a + b\cdot b

∣a∣2+2(a⋅b)+∣b∣2=∣a∣2−2(a⋅b)+∣b∣2|a|^2 + 2(a\cdot b) + |b|^2 = |a|^2 - 2(a\cdot b) + |b|^2

3. Simplify the equation to solve for $a \cdot b$.

2(a⋅b)=−2(a⋅b)2(a\cdot b) = -2(a\cdot b)

4(a⋅b)=04(a\cdot b) = 0

a⋅b=0a\cdot b = 0

4. Conclude the geometric relationship between the vectors.

Since a⋅b=0a \cdot b = 0 and it is given that a≠0a \neq 0 and b≠0b \neq 0, the angle θ\theta between them satisfies cos⁡θ=0\cos \theta = 0.

Thus, θ=90∘\theta = 90^\circ, which means aa and bb are perpendicular to each other. (Shown)

Answer

Shown

5(b)(i)

1. Explain how the vector operations in the expression guarantee a vector result

The addition of two vectors (e.g., b+cb+c) results in a vector. The vector product (cross product) of two vectors (e.g., a×(b+c)a \times (b+c)) also yields a vector. Since the expression is a sum of these individual vector products, the final result is a vector.

Answer

The addition of vectors and the cross product of vectors both result in a vector. Therefore, the entire expression evaluates to a vector.

5(b)(ii)

1. Expand the expression using the distributive property of the vector product over vector addition.

a×(b+c)+b×(c+a)+c×(a+b)a\times(b+c)+b\times(c+a)+c\times(a+b)
=a×b+a×c+b×c+b×a+c×a+c×b= a\times b + a\times c + b\times c + b\times a + c\times a + c\times b

2. Apply the anti-commutative property of the vector product to match like terms.

Since u×v=−v×uu\times v = -v\times u for any vectors uu and vv:
b×a=−a×bb\times a = -a\times b
c×a=−a×cc\times a = -a\times c
c×b=−b×cc\times b = -b\times c
Substitute these into the expression:
=a×b+a×c+b×c−a×b−a×c−b×c= a\times b + a\times c + b\times c - a\times b - a\times c - b\times c

3. Simplify the expression to find the final resulting vector.

=(a×b−a×b)+(a×c−a×c)+(b×c−b×c)= (a\times b - a\times b) + (a\times c - a\times c) + (b\times c - b\times c)
=0= \mathbf{0}

Answer

0\mathbf{0}

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