# Examples On Basics Of Vectors Set-5

Go back to  'Vectors and 3-D Geometry'

Example – 15

Justify the following tests for collinearity and coplanarity

(a) Three points with position vectors $$\vec a,\vec b,\vec c,$$ are collinear iff there exist scalars p, q, r not all zero such that

\begin{align}&\quad\quad p\vec a + \,q\vec b + \,r\vec c = \vec 0 \hfill \\&\;\; where\,\,p + \,q + \,r = 0 \hfill \\ \end{align}

(b) Four points with position vectors $$\vec a,\,\,\vec b,\,\,\vec c,\,\,\vec d$$ are coplanar iff there exist scalars p, q, r, s not all zero such that

\begin{align}&\;\;\;p\vec a + q\vec b + r\vec c + s\vec d = \vec 0 \hfill \\&where\,\,p + q + r + s = 0 \hfill \\ \end{align}

Solution: There is nothing new in these tests; we’ve already seen their justification in the discussion preceeding the examples. These tests are just the same facts put into slightly different language.

(a) Since p + q + r = 0, we have r = – ( p + q )

Assume $$r \ne 0$$

Now,$$p\vec a + q\vec b + r\vec c = \vec 0$$

\begin{align}& \Rightarrow \quad p\vec a + q\vec b - \left( {p + q} \right)\vec c = \vec 0 \hfill \\\\ &\Rightarrow \quad \vec c = \frac{{p\vec a + q\vec b}}{{p + q}} \hfill \\ \end{align}

This implies that  $$\vec c$$ is the position vector of a point which divides the points $$\vec a\,\,{\text{and}}\,\,\vec b$$ in the ratio p : q. Thus, the points  $$\vec a,\,\,\vec b\,\,{\text{and}}\,\vec c$$are collinear.

Now, we prove the other way implication, i.e, we first assume that points $$\vec a,\,\,\vec b\,\,{\text{and}}\,\vec c$$ are collinear:

\begin{align}& \Rightarrow \quad \vec b - \vec a\, = \lambda \left( {\vec c - \vec a} \right) \hfill \\\\ & \Rightarrow \quad \left( {1 - \lambda } \right)\vec a + \left( { - 1} \right)\vec b\, + \left( \lambda \right)\vec c = \vec 0 \hfill \\\\&\Rightarrow \quad p\vec a + q\vec b + r\vec c = \vec 0 \hfill \\\\& where\quad \quad p + q + r = \left( {1 - \lambda } \right) + \left( { - 1} \right) + \left( \lambda \right) = 0 \hfill \\ \end{align}

This completes our proof.

(b) Again, first assume the existence of scalars p, q, r, s such that

$p + q + r + s = 0$

$\Rightarrow \qquad s = - \left( {p + q + r} \right)$

Now,

\begin{align}& p\vec a + q\vec b + r\vec c + s\vec d = \vec 0 \hfill \\\\&\Rightarrow \quad p\vec a + q\vec b + r\vec c - \left( {p + q + r} \right)\vec d = \vec 0 \hfill \\\\& \Rightarrow \quad p\left( {\vec a - \vec d} \right) + q\left( {\vec b - \vec d} \right) + r\left( {\vec c - \vec d} \right) = \vec 0 \hfill \\ \end{align}

Assuming  $$p \ne 0,$$ we have

\begin{align}& \left( {\vec a - \vec d} \right) = \left( { - \frac{q}{p}} \right)\left( {\vec b - \vec d} \right) + \left( { - \frac{r}{p}} \right)\left( {\vec c - \vec d} \right) \hfill \\\\&\quad\qquad\;\;= \lambda \left( {\vec b - \vec d} \right) + \mu \left( {\vec c - \vec d} \right) \hfill \\ \end{align}

That $$\left( {\vec a - \vec d} \right)$$  can be written as a linear combination of the vectors $$\left( {\vec b - \vec d} \right)$$ and $$\left( {\vec c - \vec d} \right)$$ implies that $$\left( {\vec a - \vec d} \right)$$ , $$\left( {\vec b - \vec d} \right)$$and $$\left( {\vec c - \vec d} \right)$$ are coplanar.

$\Rightarrow \quad Points{\text{ }}with{\text{ }}position{\text{ }}vectors\,\vec a,\vec b,\vec c,\vec d{\text{ }}are{\text{ }}coplanar.$

Now lets prove the other way implication. Assume that $$A\left( {\vec a} \right),B\left( {\vec b} \right),C\left( {\vec c} \right){\text{and}}\,D\left( {\vec d} \right)$$ are coplanar points. Thus, there must exist scalars $$\lambda ,\mu$$ such that

\begin{align}&\quad\quad\quad\overrightarrow {AB} = \lambda \overrightarrow {AC} + \mu \overrightarrow {AD} \hfill \\\\&\Rightarrow \quad\ \left( {\vec b - \vec a} \right) = \lambda \left( {\vec c - \vec a} \right) + \mu \left( {\vec d - \vec a} \right) \hfill \\\\&\Rightarrow \quad \left( {\lambda + \mu - 1} \right)\vec a + \left( 1 \right)\vec b + \left( { - \lambda } \right)\vec c + \left( { - \mu } \right)\vec d = \vec 0 \hfill \\\\ &\Rightarrow \quad\ p\vec a + q\vec b + r\vec c + s\vec d = \vec 0 \hfill \\\end{align}

where $$p + q + r + s = \left( {\lambda + \mu - 1} \right) + \left( 1 \right) + \left( { - \lambda } \right) + \left( { - \mu } \right)$$

= 0

This completes the second proof.

Example – 16

If any point O inside or outside a tetrahedron ABCD is joined to the vertices and AO, BO, CO, DO are produced so as to cut the planes of the opposite faces in P, Q, R, S respectively, prove that

$\frac{{OP}}{{AP}} + \frac{{OQ}}{{BQ}} + \frac{{OR}}{{CR}} + \frac{{OS}}{{DS}} = 1$

Solution: Assume O to be the origin, and the position vectors of A, B, C, D to be $$\vec a,\,\,\vec b,\,\,\vec c,\,\,\vec d$$ respectively.

Since $$\vec a,\,\,\vec b,\,\,\vec c,\,\,\vec d$$ are non-coplanar vectors, we must have scalars $${\lambda _1},{\lambda _2},{\lambda _3},{\lambda _4}$$ such that (Page 17)

${\lambda _1}\vec a + {\lambda _2}\vec b + {\lambda _3}\vec c + {\lambda _4}\vec d = \vec 0\quad\quad\quad...{\text{ }}\left( 1 \right)$

Since AO is produced to meet the plane of the opposite face in P, $$\overrightarrow {AO} \,\,{\text{and}}\,\,\overrightarrow {OP}$$ must be collinear vectors. Thus,

\begin{align}& \overrightarrow {AO} = \mu \overrightarrow {OP} for{\text{ }}some\mu \in \mathbb{R} \hfill \\\\& \Rightarrow \quad - \vec a = \mu \overrightarrow {OP} \hfill \\\\& \qquad\qquad = \mu \vec p\left( {{\text{ }}\vec p\,is{\text{ }}\,the\,{\text{ }}position{\text{ }}\,vector{\text{ }}\,of\,P} \right) \hfill \\ \end{align}

This when used in (1) gives

$\left( { - \mu {\lambda _1}} \right)\vec p + {\lambda _2}\vec b + {\lambda _3}\vec c + {\lambda _4}\vec d = \vec 0$

However, since B, C, D and P will be coplanar, we have, using the result of the last example,

\begin{align}& - \mu {\lambda _1} + {\lambda _2} + {\lambda _3} + {\lambda _4} = 0 \hfill \\\\& \Rightarrow \quad \mu = \frac{{{\lambda _1}}}{{{\lambda _2} + {\lambda _3} + {\lambda _4}}} \hfill \\\\& \Rightarrow \quad \frac{{OP}}{{AP}} = \frac{{\left| {\overrightarrow {OP} } \right|}}{{\left| {\overrightarrow {AP} } \right|}} = \frac{\mu }{{1 + \mu }}\left( {From{\text{ }}\left( 2 \right)} \right) \hfill \\\\& \quad \quad\quad\quad\qquad\qquad\;= \frac{{{\lambda _1}}}{{{\lambda _1} + {\lambda _2} + {\lambda _3} + {\lambda _4}}} \hfill \\\\& \quad \quad\quad\quad\qquad\qquad\;= \frac{{{\lambda _1}}}{{\sum {\lambda _i}}} \hfill \\\end{align}

We similarly have,

\begin{align}& \frac{{OQ}}{{BQ}} = \frac{{{\lambda _2}}}{{\sum {\lambda _i}}},\frac{{OR}}{{CR}} = \frac{{{\lambda _3}}}{{\sum {\lambda _i}}},\frac{{OS}}{{DS}} = \frac{{{\lambda _4}}}{{\sum {\lambda _i}}} \hfill \\\\& \Rightarrow\quad \frac{{OP}}{{AP}} + \frac{{OQ}}{{BQ}} + \frac{{OR}}{{CR}} + \frac{{OS}}{{DS}} = 1 \hfill \\ \end{align}

## TRY YOURSELF - I

Q. 1 If the points with position vectors $$60\hat i + 3\hat j,40\hat i - 8\hat j\,and\,a\hat i - 52\hat j$$ are collinear, find a.

Q. 2 Let$$\vec a,\vec b,\vec c,$$ be three non-coplanar vectors. Prove that the points with position vectors $$\vec a - 2\vec b + 3\vec c,$$,$$2\vec a + 3\vec b - 4\vec c$$ and $$- 7\vec b + 10\vec c$$are collinear.

Q. 3 Find a point P within a quadrilateral ABCD such that

$\overrightarrow {PA} + \overrightarrow {PB} + \overrightarrow {PC} + \overrightarrow {PD} = \vec 0$

Q. 4ABCD is a parallelogram. Let L and M be the mid-points of BC and CD respectively. Prove that

$\overrightarrow {AL} + \overrightarrow {AM} = \frac{3}{2}\overrightarrow {AC}$

Q. 5 Prove that the line segments joining the mid-points of the adjacent sides of a quadrilateral form a parallelogram.

Q. 6 Prove that the line segment joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides and equal to half of their difference.

Q. 7 Let ABCD be a quadrilateral and E and F be the mid-points of AC and BD respectively. Prove that

$\overrightarrow {AB} + \overrightarrow {AD} + \overrightarrow {CB} + \overrightarrow {CD} = 4\,\overrightarrow {EF}$

Q. 8 Let  $$\vec a,\vec b,\vec c,$$ be non-coplanar vectors. Prove that the three vectors $$3\vec a - 7\vec b - 4\vec c,\,\,\vec a + \vec b + 2\vec c$$ and $$3\vec a - 2\vec b + \vec c$$ are coplanar

Q. 9 Let $$\vec a,\vec b,\vec c,$$ from a linearly independent system of vectors. Show that the system of vectors $$m\vec a + \vec b + \vec c,$$ $$\vec a + m\vec b + \vec c$$ and $$\vec a + \vec b + m\vec c$$ form a linearly independent system iff m = –2.

Q. 10 In $$\Delta ABC,$$ the point D lies on AC such that AD : DC = 2 : 1. BD is produced to F such that DF = 2BD . Prove that AF is parallel to BC and is equal to 2BD.

Vectors
grade 11 | Questions Set 1
Vectors