Question

How do you find the angle between the vectors \[\overset{\to }{\mathop{u}}\,=2\overset{\wedge }{\mathop{i}}\,+4\overset{\wedge }{\mathop{j}}\,\] and \[\overset{\to }{\mathop{v}}\,=-7\overset{\wedge }{\mathop{i}}\,+5\overset{\wedge }{\mathop{j}}\,\]?

Answer 1

Hint: Take the dot product of the two vectors by assuming that the angle between them is $\theta $. To find \[\overset{\to }{\mathop{u}}\,.\overset{\to }{\mathop{v}}\,\], that is the dot product of vectors u and v, multiply the coefficients of \[\overset{\wedge }{\mathop{i}}\,,\overset{\wedge }{\mathop{j}}\,\] of vector u with the corresponding coefficients of \[\overset{\wedge }{\mathop{i}}\,,\overset{\wedge }{\mathop{j}}\,\] of vector v. Use the fact that the dot product of two perpendicular vectors is 0, so \[\overset{\wedge }{\mathop{i}}\,.\overset{\wedge }{\mathop{j}}\,=\overset{\wedge }{\mathop{j}}\,.\overset{\wedge }{\mathop{i}}\,=0\], to simplify the answer. Find the magnitude of the vector given as \[\overset{\to }{\mathop{a}}\,=x\overset{\wedge }{\mathop{i}}\,+y\overset{\wedge }{\mathop{j}}\,\] by using the formula: $a=\sqrt{{{x}^{2}}+{{y}^{2}}}$.

Complete step by step solution:
Here, we have been provided with two vectors, \[\overset{\to }{\mathop{u}}\,=2\overset{\wedge }{\mathop{i}}\,+4\overset{\wedge }{\mathop{j}}\,\] and \[\overset{\to }{\mathop{v}}\,=-7\overset{\wedge }{\mathop{i}}\,+5\overset{\wedge }{\mathop{j}}\,\], and we are asked to find the angle between the two vectors. Here, we have to take the dot product of the two vectors.
Now, the dot product of two vectors is defined as the product of the magnitude of two vectors and the cosine of the angle between them. For example: - let us consider two vectors, \[\overset{\to }{\mathop{x}}\,\] and \[\overset{\to }{\mathop{y}}\,\] and the angle between them is \[\theta \]. So, the dot product of these vectors is given as: -

\[\Rightarrow \overrightarrow{x}.\overrightarrow{y}=xy\cos \theta \]
Here, x denotes the magnitude of \[\overrightarrow{x}\] and similarly y denotes the magnitude of \[\overrightarrow{y}\]. Now, one thing you may note is that when the two vectors are perpendicular, i.e., \[\theta ={{90}^{\circ }}\], then the dot product of two vectors will become zero. This is because \[\cos {{90}^{\circ }}=0\].
Now, we know that \[\overset{\wedge }{\mathop{i}}\,,\overset{\wedge }{\mathop{j}}\,\] are perpendicular to each other as they represent unit vectors along the x, y axis respectively. So, we have,
\[\Rightarrow \overset{\wedge }{\mathop{i}}\,.\overset{\wedge }{\mathop{j}}\,=\overset{\wedge }{\mathop{j}}\,.\overset{\wedge }{\mathop{i}}\,=0\]
That means we have to take the product of \[\overset{\wedge }{\mathop{i}}\,,\overset{\wedge }{\mathop{j}}\,\] of one vector with the corresponding \[\overset{\wedge }{\mathop{i}}\,,\overset{\wedge }{\mathop{j}}\,\] of the second vector. Here, \[\overset{\wedge }{\mathop{i}}\,.\overset{\wedge }{\mathop{i}}\,=\overset{\wedge }{\mathop{j}}\,.\overset{\wedge }{\mathop{j}}\,=1\] because in these two cases \[\theta ={{0}^{\circ }}\] and we know that \[\cos {{0}^{\circ }}=1\].
Let us come to the question, considering the dot product of vectors u and v by assuming that the angle between them is $\theta $, we get,
\[\begin{align}
  & \Rightarrow \overset{\to }{\mathop{u}}\,.\overset{\to }{\mathop{v}}\,=\left( 3\overset{\wedge }{\mathop{i}}\,+4\overset{\wedge }{\mathop{j}}\, \right).\left( -7\overset{\wedge }{\mathop{i}}\,+5\overset{\wedge }{\mathop{j}}\, \right) \\
 & \Rightarrow uv\cos \theta =\left( 3\times \left( -7 \right) \right)+\left( 4\times 5 \right) \\
 & \Rightarrow uv\cos \theta =-1 \\
\end{align}\]
Using the formula for finding the magnitude of \[\overset{\to }{\mathop{a}}\,=x\overset{\wedge }{\mathop{i}}\,+y\overset{\wedge }{\mathop{j}}\,\] given as: $a=\sqrt{{{x}^{2}}+{{y}^{2}}}$, we get,
\[\begin{align}
  & \Rightarrow \sqrt{{{3}^{2}}+{{4}^{2}}}\times \sqrt{{{\left( -7 \right)}^{2}}+{{5}^{2}}}\times \cos \theta =-1 \\
 & \Rightarrow 5\times \sqrt{74}\times \cos \theta =-1 \\
 & \Rightarrow \cos \theta =\dfrac{-1}{5\sqrt{74}} \\
 & \Rightarrow \theta ={{\cos }^{-1}}\left( \dfrac{-1}{5\sqrt{74}} \right) \\
\end{align}\]
Hence, the above relation is required answer

Note: One may note that we cannot find the value of the angle $\theta $ obtained above without using the cosine table. You can also solve the above question by using the formula for magnitude of cross product of two vectors. In that case we have \[\overset{\wedge }{\mathop{i}}\,.\overset{\wedge }{\mathop{i}}\,=\overset{\wedge }{\mathop{j}}\,.\overset{\wedge }{\mathop{j}}\,=0\] because the relation is given as \[\left| \overrightarrow{u}\times \overrightarrow{v} \right|=uvsin\theta \] and we know that $\sin {{0}^{\circ }}=0$. But it will be easier to use the dot product to solve the above question.