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If $\vec{a}$ and $\vec{b}$ are two unit vectors and $\vec{c}$ be a vector such that $2(\vec{a}\times \vec{b})+\vec{c}=\vec{b}\times \vec{c}$. Then maximum value of $|(\vec{a}\times \vec{c})\cdot \vec{b}|$

Try: From $2(\vec{a}\times \vec{b})+ \vec{c}=\vec{b}\times \vec{c}$.

Taking dot product of $\vec{c}$ on both side, we have

$2[\vec{a}\vec{b}\vec{c}]=|\vec{c}|^2/2$

So $$|(\vec{a}\times \vec{c})\cdot \vec{b}|=|[\vec{a}\vec{b}\vec{c}]|=\frac{|\vec{c}|^2}{2}$$

Could some help me to solve it, thanks

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    $\begingroup$ The statement $2(\vec a \times \vec b) \cdot \vec c = \vec b \times \vec c$ doesn't make sense. Do you know about the scalar triple product? $\endgroup$ – Umberto P. Apr 2 '18 at 16:34
  • $\begingroup$ Yes Umberto i have a knowledge of triple product $\endgroup$ – DXT Apr 2 '18 at 16:35
  • $\begingroup$ In that case write $(\vec a \times \vec c) \cdot \vec b = (\vec c \times \vec b) \cdot \vec a$ and work from there. $\endgroup$ – Umberto P. Apr 2 '18 at 16:38
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    $\begingroup$ Where did $2(\vec{a}\times \vec{b})\cdot \vec{c}=\vec{b}\times \vec{c}$ come from? In the highlighted question, you have a plus instead of a dot product. $\endgroup$ – amd Apr 2 '18 at 16:51
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Let $\vec{c}=\alpha\vec{a}+\beta\vec{b}+\gamma\vec{a}\times\vec{b}$.

\begin{align*} \vec{b}\times\vec{c}&=\alpha\vec{b}\times\vec{a}+\beta\vec{b}\times\vec{b}+\gamma\vec{b}\times(\vec{a}\times\vec{b})\\ &=-\alpha\vec{a}\times\vec{b}+\gamma[(\vec{b}\cdot\vec{b})\vec{a}-(\vec{b}\cdot\vec{a})\vec{b}]\\ &=\gamma\vec{a}-\gamma(\vec{a}\cdot\vec{b})\vec{b}-\alpha\vec{a}\times\vec{b} \end{align*}

$$2(\vec{a}\times\vec{b})+\vec{c}=\alpha\vec{a}+\beta\vec{b}+(\gamma+2)\vec{a}\times\vec{b}$$

Therefore, $\gamma=\alpha$, $-\gamma(\vec{a}\cdot\vec{b})=\beta$ and $-\alpha=\gamma+2$.

Solving, $\alpha=\gamma=-1$ and $\beta=\vec{a}\cdot\vec{b}$.

So, $\vec{c}=-\vec{a}+(\vec{a}\cdot\vec{b})\vec{b}-\vec{a}\times\vec{b}$.

\begin{align*} \vec{c}\cdot\vec{c}&=\vec{a}\cdot\vec{a}+(\vec{a}\cdot\vec{b})^2(\vec{b}\cdot\vec{b})+(\vec{a}\times\vec{b})\cdot(\vec{a}\times\vec{b})-2(\vec{a}\cdot\vec{b})(\vec{a}\cdot\vec{b})\\ |\vec{c}|^2&=1-(\vec{a}\cdot\vec{b})^2+|\vec{a}\times\vec{b}|^2\\ &=2|\vec{a}\times\vec{b}|^2 \end{align*}

and its greatest value is $2$.

The greatest value of $|(\vec{a}\times\vec{c})\cdot\vec{b}|$ is $1$.

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The maximum value is $1$, and this occurs when $\underline{a}\cdot\underline{b}=0$

You might like to try this yourself: Starting in each case with $$2(\underline{a}\times\underline{b})+\underline{c}=\underline{b}\times\underline{c}$$

Step 1: do the dot product with $\underline{a}$

Step 2: do the dot product with $\underline{b}$

Step 3: do the cross product with $\underline{a}$

Step 4: do the dot product of the result of step 3 with $\underline{b}$

You will end up with $$|(\underline{a}\times\underline{c})\cdot\underline{b}|=1-(\underline{a}\cdot\underline{b})^2$$

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