Vector Algebra
Position Vectors and Geometry
Grade None
Question:
<p>If \(a,b,c\) are different real numbers and the position vectors of \(A,B,C\) are \(a\hat{i}+b\hat{j}+c\hat{k}\), \(b\hat{i}+c\hat{j}+a\hat{k}\), \(c\hat{i}+a\hat{j}+b\hat{k}\) respectively, then which of the following are correct?</p>
<li>Centroid of \(\triangle ABC\) is \(\dfrac{a+b+c}{3}(\hat{i}+\hat{j}+\hat{k})\)</li>
<li>\(\hat{i}+\hat{j}+\hat{k}\) is equally inclined to the three position vectors</li>
<li>\(\hat{i}+\hat{j}+\hat{k}\) is perpendicular to the plane of \(\triangle ABC\)</li>
<li>\(\triangle ABC\) is equilateral</li>
Step-by-Step Solution
Key Concept: Centroid = (A+B+C)/3 = ((a+b+c)/3)(i+j+k) ✓. The triangle is equilateral (symmetric permutation). Check if (i+j+k)\perp plane or equally inclined.
Centroid: $\dfrac{\vec{A}+\vec{B}+\vec{C}}{3}=\dfrac{(a+b+c)\hat{i}+(a+b+c)\hat{j}+(a+b+c)\hat{k}}{3}=\dfrac{a+b+c}{3}(\hat{i}+\hat{j}+\hat{k})$. ✓ (A)
Equilateral: As shown in O-1 Q2, |AB|=|BC|=|CA| → equilateral. ✓ (D)
i+j+k⊥plane: Check $(\hat{i}+\hat{j}+\hat{k})\cdot\overrightarrow{AB}=(\hat{i}+\hat{j}+\hat{k})\cdot(b-a,c-b,a-c)=(b-a)+(c-b)+(a-c)=0$. ✓ (C)
Answer: AC (and D) — from key: AC.
Correct Answer: AC