Vectors & 3D Geometry
Areas of triangles from position vectors
MJAT_TS6_P1
Grade 12
Question:
Let $\vec{a},\vec{b},\vec{c},\vec{d}$ be position vectors of $A,B,C,D$ with $|\vec{a}|=1$, $|\vec{b}|=2$, $|\vec{c}|=3$, $|\vec{d}|=4$, $(\vec{a}\times\vec{b})\cdot(\vec{c}\times\vec{d})=24$, $\vec{a}\cdot\vec{c}=\frac{3}{2}$, $\vec{b}\cdot\vec{c}>0$. Area of $\triangle ABC=\frac{1}{4}(2\alpha+\beta\sqrt{3})$ and area of $\triangle BCD=\frac{1}{2}(m-n\sqrt{3})$ where $\alpha,\beta,m,n\in\mathbb{N}$. Match P)$\alpha$, Q)$\beta$, R)$m$, S)$n$ with 1)1, 2)3, 3)4, 4)9.
A) P-1; Q-2; R-3; S-4
B) P-2; Q-3; R-4; S-3
C) P-4; Q-3; R-1; S-2
D) P-1; Q-2; R-4; S-3
Step-by-Step Solution
Key Concept: Use $(\vec{a}\times\vec{b})\cdot(\vec{c}\times\vec{d})=\det\begin{pmatrix}\vec{a}\cdot\vec{c}&\vec{a}\cdot\vec{d}\\\vec{b}\cdot\vec{c}&\vec{b}\cdot\vec{d}\end{pmatrix}=24$. With $\vec{a}\cdot\vec{c}=3/2$ and other dot products from $|\vec{a}|\cdot|\vec{c}|\cdot\cos\theta=3/2\Rightarrow\cos\theta=1/2$. Find areas using the cross product.
$\alpha=1,\beta=2,m=4,n=3$. Answer: **D**.
Correct Answer: D