Let $\vec a,\vec b,\vec c$ be unit vectors such that $\vec a\cdot\vec b=0=\vec a\cdot\vec c$ and the angle between $\vec b$ and $\vec c$ is $\dfrac\pi3$. Prove that $\vec a=\pm\dfrac{2}{\sqrt3}(\vec b imes\vec c)$.

Question

Let $\vec a,\vec b,\vec c$ be unit vectors such that $\vec a\cdot\vec b=0=\vec a\cdot\vec c$ and the angle between $\vec b$ and $\vec c$ is $\dfrac\pi3$. Prove that $\vec a=\pm\dfrac{2}{\sqrt3}(\vec b	imes\vec c)$.

Accepted Answer

**Step 1.** $\vec a\cdot\vec b=0$ and $\vec a\cdot\vec c=0$ mean $\vec a$ is perpendicular to both $\vec b$ and $\vec c$.

**Step 2.** The vector $\vec b	imes\vec c$ is also perpendicular to both $\vec b$ and $\vec c$. Since (generically) there is only one line perpendicular to both $\vec b$ and $\vec c$, $\vec a$ must be a scalar multiple of $\vec b	imes\vec c$: $\vec a=\lambda(\vec b	imes\vec c)$.

**Step 3.** $|\vec b	imes\vec c|=|\vec b||\vec c|\sin\dfrac\pi3=(1)(1)\dfrac{\sqrt3}2=\dfrac{\sqrt3}2$ (since $\vec b,\vec c$ are unit vectors). …

Q.Let $\vec a,\vec b,\vec c$ be unit vectors such that $\vec a\cdot\vec b=0=\vec a\cdot\vec c$ and the angle between $\vec b$ and $\vec c$ is $\dfrac\pi3$ . Prove that $\vec a=\pm\dfrac{2}{\sqrt3}(\vec b\times\vec c)$ .

Concept understanding — Vector (Cross) Product

Vector (Cross) Product

🎓Unlock everything free for 14 days

Unlock everything free for 14 days