Strange Cross Products

Geometry Level 1

Let $\vec{a},\vec{b}$ and $\vec{c}$ be three non-zero vectors such that no two of them are collinear and $(\vec{a} \times \vec{b}) \times \vec{c}=\frac{1}{3} |\vec{b}||\vec{c}| \vec{a}$ . If $\theta$ is the angle between vectors $\vec{b}$ and $\vec{c}$ , then the value of $\sin\theta$ is :

\frac{-\sqrt{2}}{3}

\frac{2}{3}

\frac{2\sqrt{2}}{3}

\frac{-2\sqrt{2}}{3}

1 solution

Sandeep Bhardwaj
Oct 2, 2015

We have $(\vec{a} \times \vec{b}) \times \vec{c}=\frac{1}{3} |\vec{b}||\vec{c}| \vec{a}$ .

Using the notion of vector triple product, we can say that

$\left( \vec{a} \cdot \vec{c} \right) \vec{b} - \left( \vec{b} \cdot \vec{c} \right) \vec{a} = \frac{1}{3} |\vec{b}||\vec{c}| \vec{a}$

$\Rightarrow \left( \vec{a} \cdot \vec{c} \right) \vec{b} - \left( \vec{b} \cdot \vec{c} + \frac{1}{3} |\vec{b}||\vec{c}| \right) \vec{a} = \vec{0}$

$\Rightarrow \left( \vec{a} \cdot \vec{c} \right)=0$ and $\left( \vec{b} \cdot \vec{c} + \frac{1}{3} |\vec{b}||\vec{c}| \right)=0$ [Since $\vec{a}$ and $\vec{b}$ are non-collinear.]

Now using the definition of Dot Product , we can say that

$|\vec{b}| \cdot |\vec{c}| \cos \theta + \frac{1}{3} |\vec{b}||\vec{c}| =0$

$\Rightarrow \cos \theta=-\frac 13$

$\therefore \sin \theta=\sqrt{1-\cos^2 \theta}=\sqrt{\frac 89}=\frac{2 \sqrt{2}}{3}$ . $\square$

Moderator note:

Great approach using up the facts stated in the question.

Great usage of vector properties sir

Ashrith Appani - 5 years, 1 month ago

Strange Cross Products

1 solution

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