Vector Algebra
Position Vectors and Geometry
Grade None
Question:
<p>Let \(\alpha,\beta,\gamma\) be distinct real numbers. The points whose position vectors are \(\vec{a}=\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k}\), \(\vec{b}=\beta\hat{i}+\gamma\hat{j}+\alpha\hat{k}\), \(\vec{c}=\gamma\hat{i}+\alpha\hat{j}+\beta\hat{k}\)</p>
<li>form an equilateral triangle</li>
<li>form a scalene triangle</li>
<li>form a right-angled triangle</li>
<li>form a right-angled isosceles triangle</li>
Step-by-Step Solution
Key Concept: Compute |a-b|^2, |b-c|^2, |c-a|^2. Show all three equal 2(\alpha^2+\beta^2+\gamma^2-\alpha\beta-\beta\gamma-\gamma\alpha).
$|\vec{a}-\vec{b}|^2=(\alpha-\beta)^2+(\beta-\gamma)^2+(\gamma-\alpha)^2$
By symmetry, $|\vec{b}-\vec{c}|^2=(\beta-\gamma)^2+(\gamma-\alpha)^2+(\alpha-\beta)^2$ and $|\vec{c}-\vec{a}|^2=(\gamma-\alpha)^2+(\alpha-\beta)^2+(\beta-\gamma)^2$.
All three expressions are identical: $|\vec{a}-\vec{b}|=|\vec{b}-\vec{c}|=|\vec{c}-\vec{a}|$.
Since $\alpha,\beta,\gamma$ are distinct, these distances are non-zero. The triangle is equilateral. Answer: (A)
Correct Answer: A