Vector Algebra
Dot Product of Vectors
Grade 12
Question:
<p>The values of \(a\), for which the points \(A\), \(B\), \(C\) with position vectors \(2\hat{i} - \hat{j} + \hat{k}\), \(\hat{i} - 3\hat{j} - 5\hat{k}\) and \(a\hat{i} - 3\hat{j} + \hat{k}\), respectively, are the vertices of a right-angled triangle with \(C = \pi/2\) are</p>
<p>2 and 1</p>
<p>\(-2\) and \(-1\)</p>
<p>\(-2\) and 1</p>
<p>2 and \(-1\)</p>
Step-by-Step Solution
Key Concept: For a right angle at C, the vectors CA and CB must be perpendicular, so their dot product must equal zero. This gives a linear equation in 'a' that yields the required value.
Step 1: Find vectors CA and CB. CA = A - C = (2-a)î + (-1+3)ĵ + (1-1)k̂ = (2-a)î + 2ĵ CB = B - C = (1-a)î + (-3+3)ĵ + (-5-1)k̂ = (1-a)î - 6k̂ Step 2: For right angle at C, apply CA · CB = 0. CA · CB = (2-a)(1-a) + (2)(0) + (0)(-6) = 0 Step 3: Expand and solve for a. (2-a)(1-a) = 0 2 - 2a - a + a^2 = 0 a^2 - 3a + 2 = 0 (a-1)(a-2) = 0 ∴ a = 1 or a = 2 Answer: D (values are 1 and 2)
Correct Answer: D