Vectors
Vectors
Allen Star Batch
Grade 12
Question:
Three points having position vectors $\vec{a}, \vec{b}$ and $\vec{c}$ will be collinear if:
$\lambda \vec{a} + \mu\vec{b} = (\lambda + \mu)\vec{c}$
$[\vec{a} \vec{b} \vec{c}] = 0$
$\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} = 0$
$\vec{a} \times \vec{c} = \vec{b}$
Step-by-Step Solution
Key Concept: Three points with position vectors $\vec{a}, \vec{b}, \vec{c}$ are collinear if and only if vectors $\vec{AB} = \vec{b} - \vec{a}$ and $\vec{AC} = \vec{c} - \vec{a}$ are parallel, which means either their cross product is zero or one is a scalar multiple of the other. Option 1 represents the weighted average form where $\vec{c}$ lies on the line through points A and B, and Option 3 uses the scalar triple product property that three vectors are coplanar when positioned from origin.
Using the section formula, a point dividing the line segment in ratio $m:n$ is given by $\vec{r} = \frac{n\vec{r}_1 + m\vec{r}_2}{m+n}$. The cross product of two collinear vectors is zero, confirming internal division. When applying to plane intersections, verify that the point satisfies both plane equations simultaneously.
Correct Answer: 1,3