Examples On Basics Of Vectors Set-5

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.