Question

How do you determine whether u and v are orthogonal, parallel or neither given $u = < 3,15 >$ and $v = < 1,5 >$ .

Answer 1

The concept behind orthogonal vectors and their definition is used in order to find if the given vectors are orthogonal to each other by computing the dot product between the vectors. Similarly, the concept of parallel vectors is applied where the magnitude and the angle between them is found out in order to find out if the given vectors are parallel.

Complete Step By Step Answer:
The above problem revolves around the concept of orthogonality between two vectors and also we are required to find if the two given vectors are parallel to each other. In order to find out if the two vectors are orthogonal we need to know the concept of orthogonality.
Orthogonal vectors are vectors that are defined to be perpendicular or at right angles to each other. Since, we are given two vectors we need to find out whether the angle between them is ${90^ \circ }$ . The angle between them is calculated by using the concept of dot product between two vectors.
The dot product between the two vectors is defined as the product of the magnitudes of the two vectors and the cosine of the angle between them. Since, $\cos {90^ \circ } = 0$ we regard that for two vectors to be orthogonal to each other the dot product between the two vectors must be zero.
The question gives two vectors and their X and Y coordinates:
$u = < 3,15 >$ $v = < 1,5 >$
The above are two vectors represented by variables u and v respectively where u has its X and Y coordinates as and v has its X and Y coordinates as respectively. The dot product of these two vectors can be found out by computing the product between their corresponding X coordinates and their corresponding Y coordinates and summing them up. Hence, the formula is as follows:
$u \cdot v = ac + bd$
Where, a and c here represent the x-coordinates of the two given vectors and b and d represent the y-coordinates of the two given vectors. Thus from the given data we have:
$a = 3$ , $b = 15$ , $c = 1$ and $d = 5$
We know that the dot product between them must be zero and hence the above equation becomes:
Since, for orthogonal vectors: $u \cdot v = 0$
$\Rightarrow ac + bd = 0$
The above equation is the condition for the two vectors to be orthogonal and hence the above equation must be proved in order to know if they are orthogonal and hence to do this we need to compute the dot product. From the known values by substituting into equation one we get:
Since, $u \cdot v = ac + bd$
$\Rightarrow (3)(1) + (15)(5)$
On solving out we get:
$\Rightarrow 3 + 75 = 78$
$\Rightarrow u \cdot v = 78$
We can see that the dot product of the two vectors is 78 which contracts the property that the dot product of orthogonal vectors must be zero. Hence, we see that:
$78 \ne 0$
Hence, the two given vectors u and v are not orthogonal as their dot product is not equal to zero.
Next, in order to find out if the two vectors are parallel we again need to find the dot product between them but the angle between them for the vectors to be parallel must be equivalent to either ${0^ \circ }$ or ${180^ \circ }$ . As mentioned earlier the dot product is defined as the product between their corresponding magnitudes and the cosine of the angle between them. Hence the dot product between the two given vectors can be written as:
$u \cdot v = \left| u \right|\left| v \right|\cos \theta$
By rearranging the terms in order to find the angle between the two vectors we get:
$\cos \theta = \dfrac{{u \cdot v}}{{\left| u \right|\left| v \right|}}$
$\Rightarrow \theta = {\cos ^{ - 1}}\left( {\dfrac{{u \cdot v}}{{\left| u \right|\left| v \right|}}} \right)$ -------( $1$ )
Here, $\left| u \right|$ and $\left| v \right|$ represent the magnitudes of the two vectors respectively. Thus, in order to find the angle between them the magnitudes must be found out. The magnitude of vectors is computed by using the formula:
$\left| r \right| = \sqrt {{x^2} + {y^2}}$
Here, for vector u:
$x = 3$ and $y = 15$ [ $\because$ Given: $u = < 3,15 >$ ]
Thus, by substituting the values we have:
$\left| u \right| = \sqrt {{{(3)}^2} + {{(15)}^2}}$
On solving further we get:
$\left| u \right| = \sqrt {9 + 225}$
$\Rightarrow \left| u \right| = 3\sqrt {26}$
Here, for vector v:
$x = 1$ and $y = 5$ [ $\because$ Given: $v = < 1,5 >$ ]
Thus, by substituting the values we have:
$\left| v \right| = \sqrt {{{(1)}^2} + {{(5)}^2}}$
On solving further we get:
$\left| v \right| = \sqrt {1 + 25}$
$\Rightarrow \left| v \right| = \sqrt {26}$
By substituting the values into the equation ( $1$ ) we get:
$\Rightarrow \theta = {\cos ^{ - 1}}\left( {\dfrac{{u \cdot v}}{{3\sqrt {26} \sqrt {26} }}} \right)$
From equation ( $2$ ) we know that:
$u \cdot v = 78$
By substituting into the above equation we get:
$\Rightarrow \theta = {\cos ^{ - 1}}\left( {\dfrac{{78}}{{3\sqrt {26} \sqrt {26} }}} \right)$
On solving further we have:
$\Rightarrow \theta = {\cos ^{ - 1}}\left( {\dfrac{{78}}{{3 \times 26}}} \right)$ [ $\because$ $\sqrt a \times \sqrt a = a$ ]
$\Rightarrow \theta = {\cos ^{ - 1}}\left( {\dfrac{{78}}{{78}}} \right)$
$\Rightarrow \theta = {\cos ^{ - 1}}\left( 1 \right)$
Hence, $\theta = {0^ \circ }$ which means that the two vectors are parallel to each other.

Note:
Vectors are quantities which have a magnitude as well as direction wherein the magnitude would often be the distance of the vectors and the direction is given by the angle that it makes. A common mistake which is made is that the formula for finding the orthogonality may not be applied or the method to find the dot product given the x and y coordinates of the vectors may be unknown.