Vector Algebra
Scalar Triple Product and Orthogonality
Grade 12
Question:
<p>The position vectors of vertices of ∆ABC are \(a\), \(b\), \(c\) and \(a \cdot a = b \cdot b = c \cdot c = 3\). If \([a b c] = 0\), then the position vector of the orthocentre of ∆ABC is</p>
<p>(a) \(a + b + c\)</p>
<p>(b) \(\frac{1}{3}(a + b + c)\)</p>
<p>(c) \(0\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: When [abc] = 0, the triangle is coplanar with origin. With equal circumradii, the orthocentre can be found using the division of the line joining circumcentre and centroid.
Step 1: Since \([a b c] = 0\), the points O, A, B and C are coplanar. Step 2: Also, \(OA = OB = OC = \sqrt{3}\), hence origin O is the circumcentre. Step 3: The position vector of the centroid G is \(\frac{a + b + c}{3}\). Step 4: The orthocentre divides OG in the ratio of 3 : 2 externally. Step 5: Therefore, the position vector of the orthocentre is \(a + b + c\).
Correct Answer: A