(a) The position vectors must be coplanar, so their scalar triple product is zero: OA⋅(OB×OC)=t100t110t=t3+1=0. For real t, this gives t=−1.
(b) Then A=(−1,0,1), B=(1,−1,0) and C=(0,1,−1). The direction vectors AB=(2,−1,−1) and AC=(1,1,−2) have cross product (3,3,3). A normal is therefore (1,1,1). Since the plane passes through the origin, its equation is x+y+z=0.
Uses the coplanarity condition.
[1 mark]
Obtains the determinant t3+1.
[1 mark]
Obtains t=−1.
[1 mark]
Determines two direction vectors.
[1 mark]
Determines a normal.
[1 mark]
States the plane equation.
[1 mark]
Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.