8.4 Expression of a Vector as the Linear Combination of a Few Vectors

8.4 Expression of a Vector as the Linear Combination of a Few Vectors

1. Polygon Law for Vectors

\vec{P Q} = \vec{P U} + \vec{U T} + \vec{T S} + \vec{S R} + \vec{R Q}

2. To prove that two vectors are parallel, we must express one of the vectors as a scalar multiple of the other vector.

For example,

\vec{A B} = k \vec{C D} or \vec{C D} = h \vec{A B} .

3. To prove that points P, Q and R are collinear, prove one of the following.

\begin{array}{l} • \vec{P Q} = k \vec{Q R} or \vec{Q R} = h \vec{P Q} \\ • \vec{P R} = k \vec{P Q} or \vec{P Q} = h \vec{P R} \\ • \vec{P R} = k \vec{Q R} or \vec{Q R} = h \vec{P R} \end{array}

Example:

Diagram below shows a parallelogram ABCD. Point Q lies on the straight line AB and point S lies on the straight line DC. The straight line AS is extended to the point T such that AS = 2ST.

It is given that AQ : QB = 3 : 1, DS : SC = 3 : 1,

\vec{A Q} = 6 \underset{˜}{a} and \vec{A D} = \underset{˜}{b}

(a) Express, in terms of

\underset{˜}{a} and \underset{˜}{b} :

(i) \vec{A S} (ii) \vec{Q C}

(b) Show that the points Q, C and T are collinear.

Solution:

\begin{array}{l} (a)(i) \vec{A S} = \vec{A D} + \vec{D S} \\ = \vec{A D} + \vec{A Q} \leftarrow \begin{array}{l} A Q : Q B = 3 : 1 and \\ D S : S C = 3 : 1 \\ ∴ \vec{A Q} = \vec{D S} \end{array} \\ = \underset{˜}{b} + 6 \underset{˜}{a} \\ = 6 \underset{˜}{a} + \underset{˜}{b} \end{array}

\begin{array}{l} (a)(ii) \vec{Q C} = \vec{Q B} + \vec{B C} \\ = \frac{1}{3} \vec{A Q} + \vec{A D} \leftarrow \begin{array}{l} A Q : Q B = 3 : 1 \\ \frac{A Q}{Q B} = \frac{3}{1} \Rightarrow Q B = \frac{1}{3} A Q \\ and for parallelogram, \\ B C / / A D, B C = A D \end{array} \\ = \frac{1}{3} (6 \underset{˜}{a}) + \underset{˜}{b} \\ = 2 \underset{˜}{a} + \underset{˜}{b} \end{array}

\begin{array}{l} (b) \vec{Q T} = \vec{Q A} + \vec{A T} \\ = \vec{Q A} + \frac{3}{2} \vec{A S} \leftarrow \begin{array}{l} A S = 2 S T \\ A T = 3 S T = \frac{3}{2} A S \end{array} \\ = - 6 \underset{˜}{a} + \frac{3}{2} (6 \underset{˜}{a} + \underset{˜}{b}) \\ = 3 \underset{˜}{a} + \frac{3}{2} \underset{˜}{b} \\ = \frac{3}{2} (2 \underset{˜}{a} + \underset{˜}{b}) \\ = \frac{3}{2} \vec{Q C} \\ ∴ Points Q, C and T are collinear . \end{array}

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