# Collinear and Coplanar Vectors

Collinear and Coplanar Vectors

Collinear Vectors: They can have equal or unequal magnitudes and their directions may be same or opposite. Two vectors are collinear if they have the same direction or are parallel or anti-parallel. They can be expressed in the form a = k b where a and b are vectors and ‘ k ‘ is a scalar quantity. If $$\overrightarrow{a}$$, $$\overrightarrow{b}$$ are any two non – collinear vectors and x, y are scalars, then $$x\overrightarrow{a}\,+\,y\overrightarrow{b}\,=\,\overrightarrow{0}\,\Rightarrow \,x\,=\,y\,=\,0$$.Example: if $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$ are non-collinear vectors, find the value of x for which the vectors $$\overrightarrow{\alpha }\,=\,\left( 2x\,+\,1 \right)\overrightarrow{a}\,-\,\overrightarrow{b}$$ and $$\overrightarrow{\beta }\,=\,\left( x\,-\,2 \right)\overrightarrow{a}\,+\,\overrightarrow{b}$$ are collinear.

Solution: Vectors $$\overrightarrow{\alpha }$$ and $$\overrightarrow{\beta }$$ will be collinear, if $$\overrightarrow{\alpha }\,=\,m\overrightarrow{\beta }$$ for some scalar m.

$$\left( 2x\,+\,1 \right)\overrightarrow{a}\,-\,\overrightarrow{b}\,=\,m\left[ \left( x\,-\,2 \right)\overrightarrow{a}\,+\,\overrightarrow{b} \right]$$,

$$\left[ \left( 2x\,+\,1 \right)\,-\,m\left( x\,-\,2 \right) \right]\overrightarrow{a}\,-\,\left( m\,+\,1 \right)\overrightarrow{b}\,=\,\overrightarrow{0}$$,

$$\left( 2x\,+\,1 \right)\,-\,m\left( x\,-\,2 \right)\,=\,0$$ and $$-\left( m\,+\,1 \right)\,=\,0$$,

$$m\,=\,-\,1$$ and $$x\,=\,\frac{1}{3}$$.

Coplanar Vectors: A system of vectors is said to be coplanar, if their supports are parallel to the same plane.Let $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$ be two given non-zero non-collinear vectors. Then any vectors $$\overrightarrow{r}$$ coplanar with $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$ can be uniquely expressed as $$\overrightarrow{r}\,=\,x\overrightarrow{a}\,+\,y\overrightarrow{b}$$ for some scalars x and y.

Example: Show that the vectors$$\overrightarrow{a}\,-\,2\overrightarrow{b}\,+\,3\overrightarrow{c}$$, $$\overrightarrow{a}\,-\,3\overrightarrow{b}\,+\,5\overrightarrow{c}$$and $$-\,2\overrightarrow{a}\,+\,3\overrightarrow{b}\,-\,4\overrightarrow{c}$$ are coplanar.

Solution:

$$A\,=\,\overrightarrow{a}\,-\,2\overrightarrow{b}\,+\,3\overrightarrow{c}$$,

$$B\,=\,\overrightarrow{a}\,-\,3\overrightarrow{b}\,+\,5\overrightarrow{c}$$,

$$C\,=\,-2\overrightarrow{a}\,+\,3\overrightarrow{b}\,-\,4\overrightarrow{c}$$.

$$\bar{A}.\left[ \overline{B}\times \overline{C} \right]=\left|\begin{matrix}1 & -2 & 3 \\1 & -3 & 5 \\-2 & 3 & -4 \\\end{matrix} \right|$$.

1 (12 – 15) + 2 (- 4 + 10) + 3 (3 – 6)

– 3 + 12 – 9 = 0

Hence, the given vectors are coplanar.