Q.Using vectors, find the value of $k$ such that the points $(k, -10, 3)$, $(1, -1, 3)$ and $(3, 5, 3)$ are collinear.

Concept: Collinear Vectors Property — three points are collinear if the vectors between them are parallel (scalar multiples of each other).

Let $A(k, -10, 3)$, $B(1, -1, 3)$, $C(3, 5, 3)$.

Form $\overrightarrow{AB} = (1-k, 9, 0)$ and $\overrightarrow{AC} = (3-k, 15, 0)$.

For collinearity, $\overrightarrow{AC} = \lambda \overrightarrow{AB}$ for some scalar $\lambda$.

Comparing $z$-components: $0 = \lambda \cdot 0$ gives no restriction. …

Let $P(k,-10,3)$, $Q(1,-1,3)$, $R(3,5,3)$.

Form two vectors from $Q$:

The points are collinear precisely when $\vec{QP}$ is parallel to $\vec{QR}$, i.e. $\vec{QP}=\lambda\,\vec{QR}$. Comparing the $y$-components:

$-9=6\lambda \implies \lambda=-\frac{3}{2}.$ …

Method: Testing three points for collinearity to find an unknown coordinate

Use this when three points must be collinear and one of their coordinates ($k$ here) is unknown.

Steps

Step 1: Form two vectors sharing a common point.

From the three points $A,B,C$ build, say, $\vec{AB}$ and $\vec{AC}$ (or use any common vertex). Each is head minus tail.

Step 2: Impose parallelism. …

Common Mistakes

Mistake 1: Trying to find $\lambda$ from the $z$-components.

Why it's wrong: all three points share $z=3$, so both vectors have $z$-component $0$ and $0=\lambda\cdot0$ gives no information. Correct approach: use a component (like $y$) where both entries are known and non-zero to fix $\lambda$.

Mistake 2: Checking only one component ratio and stopping. …