Vector Algebra
Position vectors and perpendicularity
Grade 12
Question:
<p>In a triangle <em>ABC</em>, right-angled at the vertex <em>A</em>, if the position vectors of <em>A</em>, <em>B</em> and <em>C</em> are, respectively, \(3\hat{i}+\hat{j}-\hat{k}\), \(-\hat{i}+3\hat{j}+p\hat{k}\) and \(5\hat{i}+q\hat{j}-4\hat{k}\), then the point \((p, q)\) lies on a line</p>
<p>making an obtuse angle with the positive direction of <em>x</em>-axis.</p>
<p>parallel to <em>x</em>-axis.</p>
<p>parallel to <em>y</em>-axis.</p>
<p>making an acute angle with the positive direction of <em>x</em>-axis.</p>
Step-by-Step Solution
Key Concept: Since angle A is 90°, vectors AB and AC must be perpendicular, so AB · AC = 0. This perpendicularity condition gives a linear relationship between p and q.
Step 1: Find vectors AB and AC. A = 3î + ĵ - k̂, B = -î + 3ĵ + pk̂, C = 5î + qĵ - 4k̂ AB = B - A = (-î + 3ĵ + pk̂) - (3î + ĵ - k̂) = -4î + 2ĵ + (p+1)k̂ AC = C - A = (5î + qĵ - 4k̂) - (3î + ĵ - k̂) = 2î + (q-1)ĵ - 3k̂ Step 2: Apply perpendicularity condition at A. Since ∠BAC = 90°, AB · AC = 0 (-4)(2) + (2)(q-1) + (p+1)(-3) = 0 -8 + 2q - 2 - 3p - 3 = 0 -3p + 2q - 13 = 0 Step 3: Simplify to get the line equation. 3p - 2q + 13 = 0 ∴ The point (p, q) lies on the line 3p - 2q + 13 = 0 (or equivalent form depending on option D)
Correct Answer: D