Question

For the following questions, consider A (7, -9) and B (2,3)
a) Determine $\overrightarrow {AB}$
b) Determine $|\overrightarrow {AB} |$
c) A unit vector in the direction of $\overrightarrow {AB}$
d) A vector three times as long as $\overrightarrow {AB}$ in the opposite direction

Answer 1

A vector from point A to B can be found by subtracting the coordinates of the former from the latter. For example a vector from (2,5) to (7,8) would be $5\hat i + 3\hat j$ . Mod of this vector would be $\sqrt {{5^2} + {3^2}}$.

Complete step by step answer:
First, we would know how a vector is defined in the coordinate plane. For any point with coordinates (a,b), a vector from origin is said to be $a\hat i + b\hat j$. Let there be two points P and Q with coordinates (a,b) and (c,d) respectively. Corresponding vectors from origin would be
$\overrightarrow {OP} = a\hat i + b\hat j$ and $\overrightarrow {OQ} = c\hat i + d\hat j$ .
So here, according to Triangle law of vector addition (that states that if two vectors are represented by two sides of a triangle in sequence, then third closing side of the triangle, in the opposite direction of the sequence, represents the sum (or resultant) of the two vectors in both magnitude and direction) we can say that,

$$\overrightarrow {OP} + \overrightarrow {PQ} = \overrightarrow {OQ} \\\ \Rightarrow \overrightarrow {PQ} = \overrightarrow {OQ} - \overrightarrow {OP} \\\ \Rightarrow \overrightarrow {PQ} = \left( {c\hat i + d\hat j} \right) - (a\hat i + b\hat j) \\\ \Rightarrow \overrightarrow {PQ} = \left( {c - a} \right)\hat i + (d - b)\hat j \\\$$

Also the mod of vector equals,
$|\overrightarrow {PQ|} = \sqrt {{{\left( {c - a} \right)}^2} + {{\left( {d - b} \right)}^2}}$
Using the above formulas, we can find
a) $\overrightarrow {AB}$

$$\overrightarrow {AB} = \overrightarrow {OB} - \overrightarrow {OA} \\\ \Rightarrow \overrightarrow {AB} = \left( {2\hat i + 3\hat j} \right) - (7\hat i - 9\hat j) \\\ \Rightarrow \overrightarrow {AB} = (2 - 7)\hat i + \left( {3 + 9} \right)\hat j \\\ \Rightarrow \overrightarrow {AB} = - 5\hat i + 12\hat j \\\$$

b) $|\overrightarrow {AB} |$
$|\overrightarrow {AB} | = \sqrt {{{\left( { - 5} \right)}^2} + {{\left( {12} \right)}^2}} \\\ \Rightarrow |\overrightarrow {AB} | = \sqrt {25 + 144} = \sqrt {169} \\\ \Rightarrow |\overrightarrow {AB} | = 13 \\\$
c) A unit vector in the direction of $\overrightarrow {AB}$
Unit vector in any direction is a vector in the given direction having a magnitude of 1.
Therefore, a unit vector in the direction of AB vector would be AB divided by its mod. Let the vector be $\overrightarrow U$.
$\overrightarrow U = \dfrac{{\overrightarrow {AB} }}{{|\overrightarrow {AB} |}} = \dfrac{{ - 5\hat i + 12\hat j}}{{13}} \\\ \Rightarrow \overrightarrow U = \dfrac{{ - 5}}{{13}}\hat i + \dfrac{{12}}{{13}}\hat j \\\$

d) A vector three times as long as $\overrightarrow {AB}$ in the opposite direction
A vector in the opposite direction is simply denoted by a negative sign. Let the required vector be $\overrightarrow T$ .So, a vector in the opposite direction of $\overrightarrow {AB}$ and 3 times as long would be
$\overrightarrow T = 3 \times \left( { - \overrightarrow {AB} } \right) \\\ \Rightarrow \overrightarrow T = 3 \times (5\hat i - 12\hat j) \\\ \Rightarrow \overrightarrow T = 15\hat i - 36\hat j \\\$

Note: Here we can also solve the latter two parts simply in mind. Also, in these kinds of questions it is always recommended to make a figure before coming to the numerical part as it helps greatly in solving the question.