Let {u1,u2,u3} be an orthonormal basis for an inner product space V. If

v=au1+bu2+cu3
is so that ∥v∥=42, v is orthogonal to u3, and ⟨v,u2⟩=−42, find the possible values for a, b, and c.

• ||v|| = 42, which is to say

||v||² = 〈v, v 〉

…  = 〈a u₁ + b u₂ + c u₃, a u₁ + b u₂ + c u₃〉

… = a ² 〈u₁, u₁〉 + b ² 〈u₂, u₂〉 + c ² 〈u₃, u₃〉 + 2(ab 〈u₁, u₂〉 + ac 〈u₁, u₃〉 + bc 〈u₂, u₃〉)

… = a ² ||u₁||² + b ² ||u₂||² + c ² ||u₃||²

[since each vector in the basis for V is orthogonal to any other vector in the basis, and 〈x, x〉 = ||x||² for any vector x ]

42² = a ² + b ² + c ²

[since each vector in the basis has unit length]

42 = √(a ² + b ² + c ²)

• v is orthogonal to u₃, so 〈v, u₃〉 = 0. Expanding v gives the relation

〈v, u₃〉 = 〈a u₁ + b u₂ + c u₃, u₃〉

… = a 〈u₁, u₃〉 + b 〈u₂, u₃〉 + c 〈u₃, u₃〉

… = c ||u₃||²

… = c

which gives c = 0, and so

42 = √(a ² + b ²)

• Lastly, 〈v, u₂〉 = -42, which means

〈v, u₂〉 = 〈a u₁ + b u₂ + c u₃, u₂〉

… = a 〈u₁, u₂〉 + b 〈u₂, u₂〉 + c 〈u₃, u₂〉

… = b ||u₂||²

… = b

so that b = -42. Then

42 = √(a ² + (-42)²)   →   a = 0

So we have a = 0, b = -42, and c = 0.

The required values are, [tex]a=0,b=-42 ,c=0[/tex]

Given,

[tex]v=au_1+bu_2+cu_3[/tex]

[tex]\left\| V\right\|=42[/tex]

Computation:

Since, [tex]v[/tex] is orthogonal to [tex]u_3[/tex] then we have,

[tex]\left<v,u_3 \right> =0\\\left< v,u_2\right> =-42[/tex]

Then,

[tex]\left\| V\right\|^2=\left<v,v \right>\\=\left<au_1+bu_2+cu_3,au_1+bu_2+cu_3 \right>\\=a_2\left\|u_1 \right\|^2+b_2\left\|u_2 \right\|^2+c_2\left\|u_3 \right\|^2\\=a^2+b^2+c^2\\=a^2+b^2+c^2=42^2[/tex]

As we know,

[tex]a=\left<v_1u_1 \right>\\b=\left< v_1u_2\right>= -42\\c=\left<v_1u_3 \right> =0[/tex]

[tex]a_2+b_2+c_2=42\\a=0[/tex]

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