Let $\vec a,\vec b,\vec c$ be three non-zero vectors such that $\vec c$ is a unit vector perpendicular to both $\vec a$ and $\vec b$. If the angle between $\vec a$ and $\vec b$ is $\dfrac\pi6$, show that $[\vec a,\vec b,\vec c]^2=\dfrac14|\vec a|^2|\vec b|^2$.

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Accepted Answer

> [!TLDR]
>
> Since $\vec c$ is a unit normal to both $\vec a$ and $\vec b$, it must point along $\pm\vec a	imes\vec b$, so the scalar triple product collapses to $\pm|\vec a	imes\vec b|$, whose square is exactly $|\vec a|^2|\vec b|^2\sin^2(	ext{angle})$.

**Step 1. Identify the direction of $\vec c$.** $\vec c$ is a unit vector perpendicular to BOTH $\vec a$ and $\vec b$. In 3-dimensional space there is (up to sign) only ONE such direction: $\dfrac{\vec a	imes\vec b}{|\vec a	imes\vec b|}$. So $\vec c=\pm\dfrac{\vec a	imes\vec b}{|\vec a	imes\vec b|}$.

**Step 2. Compute $[\vec a,\vec b,\vec c]$.**
$$[\vec a,\vec b,\vec c]=(\vec a	imes\vec b)\cdot\vec c=(\vec a	imes\vec b)\cdot\left(\pm\frac{\vec a	imes\vec b}{|\vec a	imes\vec b|}ight)=\pm\frac{|\vec a	imes\vec b|^2}{|\vec a	imes\vec b|}=\pm|\vec a	imes\vec b|.$$

**Step 3. Square both sides.**
$$[\vec a,\vec b,\vec c]^2=|\vec a	imes\vec b|^2.$$ …

Q.Let $\vec a,\vec b,\vec c$ be three non-zero vectors such that $\vec c$ is a unit vector perpendicular to both $\vec a$ and $\vec b$ . If the angle between $\vec a$ and $\vec b$ is $\dfrac\pi6$ , show that $[\vec a,\vec b,\vec c]^2=\dfrac14|\vec a|^2|\vec b|^2$ .

Concept understanding — Scalar Triple Product and Coplanarity

Scalar Triple Product and Coplanarity

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