Let $\vec {a}$ and $\vec {b}$ be unit vectors. If $\vec{c}$ is a vector such that $\vec{c}+(\vec{c} \times \vec{a})=\vec {b}$ then maximum value of $|(\vec {a} \times\vec {b} ).\vec{c}|$ is $\frac{A}{10}$, the find $A$.

Note: Here |.| represents absolute value

Could someone give me as how to solve this problem. I am not getting how to use given data

  • $\begingroup$ @achillehui No there was a mistake. I have made the correction. Thank you for pointing it out. $\endgroup$ – Akira Apr 29 '16 at 0:59
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    $\begingroup$ Hint: 1) Express the condition $|\vec{b}|^2 = 1$ and the expression $|( \vec{a} \times \vec{b} ) \cdot \vec{c}|$ in terms of $\vec{a}$ and $\vec{c}$ alone. 2) Use $|\vec{a}\times\vec{c}|^2 \le |\vec{a}|^2 |\vec{c}|^2 = |\vec{c}|^2$ to bound $A$ from above. 3) Find a $c$ that achieve that bound. $\endgroup$ – achille hui Apr 29 '16 at 1:11

$$|(\vec {a} \times(\vec{c}+(\vec{c} \times \vec{a}) ))\cdot\vec{c}|=|(\vec {a} \times\vec{c}+\vec {a} \times(\vec{c} \times \vec{a} ))\cdot\vec{c}|=|(\vec {a} \times\vec{c})\cdot\vec{c}+(\vec {a} \times(\vec{c} \times \vec{a} ))\cdot\vec{c}|$$

The first quantity in the last equality vanishes by orthogonality. Then we can use the BAC-CAB identity to evaluate the triple vector product. $$=|(\vec {a} \times(\vec{c} \times \vec{a} ))\cdot\vec{c}|=|(\vec {c}(\vec{a}\cdot\vec{a}) -\vec{a}(\vec{a}\cdot\vec{c}))\cdot\vec{c}|=\left||c|^2 -(\vec{a}\cdot\vec{c})^2\right|\le A/10$$

Since $|a|$=1, the maximum value is where $\vec{a}\cdot\vec{c}=0$. Then $|c|^2=A/10$, or $A=10|c|^2$.

The condition $|b|=1$ gives

$$|(\vec{c}+(\vec{c} \times \vec{a})| = 1$$ $$|c|^2+(\vec{c} \times \vec{a})\cdot(\vec{c} \times \vec{a}) = 1$$

Using another identity, $(A\times B)\cdot(C\times D)=(A\cdot C)(B\cdot D)-(B\cdot C)(A\cdot D)$,$^1$ we get $$|c|^2+|c|^2|a|^2-(\vec{a}\cdot\vec{c})^2 = 1$$ $$|c|^2+|c|^2-(\vec{a}\cdot\vec{c})^2 = 1$$

Again, at the maximum $\vec{a}\cdot\vec{c}=0$, so $|c|^2 = 1/2$.

Thus $A=5$.

$^1$ I just recognized that this identity is essentially the "epsilon killer" of index notation.

  • $\begingroup$ @achillehui . Thanks. I had to look up my identities for the rest. I feel there should be a more efficient way. $\endgroup$ – zahbaz Apr 29 '16 at 1:34
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    $\begingroup$ You can shorten your proof a little bit by keeping $|\vec{a}\times\vec{c}|^2$ as is without splitting it out as $|\vec{c}|^2 - |\vec{a}\cdot\vec{c}|^2$ $\endgroup$ – achille hui Apr 29 '16 at 1:40

Here is another solution. Choosing an appropriate orthogonal coordinate frame, we may assume that $\vec{a}$ and $\vec{b}$ are of the form $\vec{a} = (1, 0, 0)$ and $\vec{b} = (\cos\theta, \sin\theta, 0)$. Letting $\vec{c} = (x, y, z)$, the given condition satisfies

$$ (x, y, z) + (0, -z, y) = (\cos\theta, \sin\theta, 0). $$

Solving this relation gives $x = \cos\theta$ and $y = -z = \frac{1}{2}\sin\theta$. Now since

$$ |(\vec{a} \times \vec{b})\cdot\vec{c}| = |(0, 0, \sin\theta)\cdot \vec{c}| = \tfrac{1}{2}\sin^2\theta, $$

it follows that the maximum possible value of $|(\vec{a} \times \vec{b})\cdot\vec{c}|$ is $\frac{1}{2}$. Equating this with $\frac{1}{10}A$ gives $A = 5$.


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