Question

What is the component of \[\left( {3\hat i + 4\hat j} \right)\] along $\left( {\hat i + \hat j} \right)$?
A) \[\dfrac{1}{2}\left( {\hat j + \hat i} \right)\]
B) $\dfrac{3}{2}\left( {\hat j + \hat i} \right)$
C) $\dfrac{5}{2}\left( {\hat j + \hat i} \right)$
D) $\dfrac{7}{2}\left( {\hat j + \hat i} \right)$

Answer 1

Hint: For any two vectors, their dot product i.e. scalar product of their unit vector provides the cosine of the angle between them. Scalar product of one vector with the unit vector along another vector’s direction gives the magnitude of the component of the first vector along the second one. Here, first find the cosine of the angle between the vectors by taking the dot product then use that to get the component along the second vector.

Formula used: For two vectors the cosine of the angle between them is given as:
\[\cos \theta = \dfrac{{\vec A.\vec B}}{{\left| {\vec A} \right|\left| {\vec B} \right|}}\]......................(1)
Where,
\[\theta \]is the angle between two vectors,
\[\vec A\]and \[\vec B\]are the two vectors,
\[\left| {\vec A} \right|\]and \[\left| {\vec B} \right|\]are the magnitude of each vector.
Component of vector \[\vec A\]along vector \[\vec B\]is given by:
\[{\vec A_{\vec B}} = \left| {\vec A} \right|\cos \theta \dfrac{{\vec B}}{{\left| {\vec B} \right|}}\]......................(2)

Complete step-by-step answer:
Given:
Here, the first vector is denoted as $\vec A = \left( {3\hat i + 4\hat j} \right)$.
Here, the second vector is denoted as $\vec B = \left( {\hat i + \hat j} \right)$.
To find: Component of vector \[\vec A\] along vector \[\vec B\].
Step 1
Put the relation from eq.(1) into relation of eq.(2) to get the expression of component as:
\[
  {{\vec A}_{\vec B}} = \left| {\vec A} \right|\cos \theta \dfrac{{\vec B}}{{\left| {\vec B} \right|}} = \left| {\vec A} \right| \times \dfrac{{\vec A.\vec B}}{{\left| {\vec A} \right|\left| {\vec B} \right|}} \times \dfrac{{\vec B}}{{\left| {\vec B} \right|}} \\
  \therefore {{\vec A}_{\vec B}} = \dfrac{{\vec A.\vec B}}{{{{\left| {\vec B} \right|}^2}}}\vec B \\
 \].............................(3)
Step 2
Now, substitute the values of \[\vec A\]and \[\vec B\]in eq.(3) to get the component as:
\[
  {{\vec A}_{\vec B}} = \dfrac{{\left( {3\hat i + 4\hat j} \right).\left( {\hat i + \hat j} \right)}}{{{{\left| {\left( {\hat i + \hat j} \right)} \right|}^2}}}\left( {\hat i + \hat j} \right) \\
  \therefore {{\vec A}_{\vec B}} = \dfrac{{(3 \times 1 + 4 \times 1)}}{{\left( {{1^2} + {1^2}} \right)}}\left( {\hat i + \hat j} \right) = \dfrac{7}{2}\left( {\hat i + \hat j} \right) \\
 \]

So, the component of \[\left( {3\hat i + 4\hat j} \right)\] along $\left( {\hat i + \hat j} \right)$ is (d) $\dfrac{7}{2}\left( {\hat j + \hat i} \right)$.

Note: While solving this problem students make a common mistake. First they find the \[\cos \theta \] term using dot product and then they take \[\vec A\cos \theta \] to get the answer. But this is actually wrong because when you are multiplying \[\vec A\] with \[\cos \theta \] then its unit vector’s direction is along \[\vec A\] not along \[\vec B\]. So, you need to find the magnitude of the projection of \[\vec A\] along \[\vec B\] first and that is given by \[\left| {\vec A} \right|\cos \theta \]. Now, this is the magnitude of \[\vec A\] along \[\vec B\], and its direction must be in the same direction of \[\vec B\]. So you need to multiply it with the unit vector along \[\vec B\] as given in eq.(2).