
How do you determine whether \[u\] and \[v\] are orthogonal, parallel or neither? Given \[u = \left\langle {3,15} \right\rangle \] and \[v = \left\langle { - 1,5} \right\rangle \].
Answer
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Hint: In the given question, we have been given a pair of vectors whose type is to be determined – orthogonal, parallel, or neither. To check for orthogonal, we find the dot product – if it is equal to zero. Then, we check for parallel by finding the angle between them – if it is \[0^\circ \] or \[180^\circ \].
Formula Used:
We are going to find the angle between them by using the scalar-product formula:
\[\theta = {\cos ^{ - 1}}\left( {\dfrac{{\overrightarrow a .\overrightarrow b }}{{\left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|}}} \right)\].
Complete step by step answer:
The given two vectors are:
\[\overrightarrow u = \left\langle {3,15} \right\rangle \] and \[\overrightarrow v = \left\langle { - 1,5} \right\rangle \].
First, we find their dot-product:
\[\overrightarrow u .\overrightarrow v = 3\left( { - 1} \right) + 15\left( 5 \right) = - 3 + 75 = 72\].
Hence, the two vectors are not orthogonal because the dot-product of two such vectors is zero.
Now, we check for the angle between them.
\[\left| {\overrightarrow u } \right| = \sqrt {{3^2} + {{15}^2}} = \sqrt {234} = 3\sqrt {26} \]
\[\left| {\overrightarrow v } \right| = \sqrt {{{\left( { - 1} \right)}^2} + {5^2}} = \sqrt {26} \]
\[\theta = {\cos ^{ - 1}}\left( {\dfrac{{\overrightarrow u .\overrightarrow v }}{{\left| {\overrightarrow u } \right|.\left| {\overrightarrow v } \right|}}} \right) = {\cos ^{ - 1}}\left( {\dfrac{{72}}{{\sqrt {234} .\sqrt {26} }}} \right) = {\cos ^{ - 1}}\left( {\dfrac{{72}}{{78}}} \right) \approx 22.6^\circ \].
Hence, the two vectors are not parallel because the angle between two such vectors is \[0^\circ \] or \[180^\circ \].
Note: In the given question, we had to find the type of pair of vectors given. To do that, we check for orthogonal by checking their dot-product. For checking for parallel, we do that by checking the angle between the two vectors. So, it is really important that we know the formulae and where, when, and how to use them so that we can get the correct result.
Formula Used:
We are going to find the angle between them by using the scalar-product formula:
\[\theta = {\cos ^{ - 1}}\left( {\dfrac{{\overrightarrow a .\overrightarrow b }}{{\left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|}}} \right)\].
Complete step by step answer:
The given two vectors are:
\[\overrightarrow u = \left\langle {3,15} \right\rangle \] and \[\overrightarrow v = \left\langle { - 1,5} \right\rangle \].
First, we find their dot-product:
\[\overrightarrow u .\overrightarrow v = 3\left( { - 1} \right) + 15\left( 5 \right) = - 3 + 75 = 72\].
Hence, the two vectors are not orthogonal because the dot-product of two such vectors is zero.
Now, we check for the angle between them.
\[\left| {\overrightarrow u } \right| = \sqrt {{3^2} + {{15}^2}} = \sqrt {234} = 3\sqrt {26} \]
\[\left| {\overrightarrow v } \right| = \sqrt {{{\left( { - 1} \right)}^2} + {5^2}} = \sqrt {26} \]
\[\theta = {\cos ^{ - 1}}\left( {\dfrac{{\overrightarrow u .\overrightarrow v }}{{\left| {\overrightarrow u } \right|.\left| {\overrightarrow v } \right|}}} \right) = {\cos ^{ - 1}}\left( {\dfrac{{72}}{{\sqrt {234} .\sqrt {26} }}} \right) = {\cos ^{ - 1}}\left( {\dfrac{{72}}{{78}}} \right) \approx 22.6^\circ \].
Hence, the two vectors are not parallel because the angle between two such vectors is \[0^\circ \] or \[180^\circ \].
Note: In the given question, we had to find the type of pair of vectors given. To do that, we check for orthogonal by checking their dot-product. For checking for parallel, we do that by checking the angle between the two vectors. So, it is really important that we know the formulae and where, when, and how to use them so that we can get the correct result.
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