Vector Algebra
Distance and Triangle Properties
Grade 12
Question:
<p>The point having position vectors 2<strong>i</strong> + 3<strong>j</strong> + 4<strong>k</strong>, 3<strong>i</strong> + 4<strong>j</strong> + 2<strong>k</strong> and 4<strong>i</strong> + 2<strong>j</strong> + 3<strong>k</strong> are the vertices of</p>
<p>(a) right angled triangle</p>
<p>(b) isosceles triangle</p>
<p>(c) equilateral triangle</p>
<p>(d) None of the above</p>
Step-by-Step Solution
Key Concept: Calculate distances between vertices using the distance formula; compare to classify the triangle.
Solution: Let A = 2 i + 3 j + 4 k , B = 3 i + 4 j + 2 k , C = 4 i + 2 j + 3 k AB = B - A = (3-2) i + (4-3) j + (2-4) k = i + j - 2 k |AB| = \(\sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{6}\) BC = C - B = (4-3) i + (2-4) j + (3-2) k = i - 2 j + k |BC| = \(\sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6}\) CA = A - C = (2-4) i + (3-2) j + (4-3) k = -2 i + j + k |CA| = \(\sqrt{(-2)^2 + 1^2 + 1^2} = \sqrt{6}\) Since |AB| = |BC| = |CA| = \(\sqrt{6}\), the triangle is equilateral.
Correct Answer: B