Vector Algebra
Cross and scalar triple product; angle between vectors
nta_pyq_2023_jan
Grade 12
Question:
Let $\vec{a}$, $\vec{b}$, $\vec{c}$ be three vectors such that $|\vec{a}|=\sqrt{31}$, $4|\vec{b}|=|\vec{c}|=2$ and $2(\vec{a}\times\vec{b}) = 3(\vec{c}\times\vec{a})$. If the angle between $\vec{b}$ and $\vec{c}$ is $\frac{2\pi}{3}$, then $\left(\frac{\vec{a}\times\vec{c}}{\vec{a}\cdot\vec{b}}\right)^2$ is equal to _____.
Step-by-Step Solution
Key Concept: From $2(\vec{a}\times\vec{b}) = 3(\vec{c}\times\vec{a})$: $(2\vec{b}+3\vec{c})\times\vec{a}=0 \Rightarrow \vec{a}\parallel (2\vec{b}+3\vec{c})$. Compute magnitudes.
$|\vec{a}|^2 = k^2(4\cdot1/4 + 12(-1/2) + 9\cdot4) = k^2(1-6+36) = 31k^2 = 31 \Rightarrow k=1$. $\vec{a}\times\vec{c} = (2\vec{b}+3\vec{c})\times\vec{c} = 2(\vec{b}\times\vec{c})$. $|\vec{a}\times\vec{c}|^2 = 4|\vec{b}|^2|\vec{c}|^2\sin^2(2\pi/3) = 4\cdot(1/4)(4)(3/4) = 3$. $\vec{a}\cdot\vec{b} = (2\vec{b}+3\vec{c})\cdot\vec{b} = 2|\vec{b}|^2+3\vec{b}\cdot\vec{c} = 1/2 - 3/2 = -1$. Result $= 3/1 = 3$. Answer: 3 (per answer key says Q19=(3))
Correct Answer: 3