Question

If \[\vec a = 2\hat i - \hat j + 5\hat k\] and \[\vec b = 4\hat i - 2\hat j + \lambda \hat k\] such that \[\vec a\parallel \vec b\], find the value of \[\lambda \].

Answer 1

Hint:
Here, we will use the concept of the properties of the vectors . We will use the identity of the parallel vectors and substitute the given values in the condition. Then we will equate the similar terms to get 3 different equations. Then by solving the equations, we will get the value of \[\lambda \].

Complete step by step solution:
Given vectors are \[\vec a = 2\hat i - \hat j + 5\hat k\] and \[\vec b = 4\hat i - 2\hat j + \lambda \hat k\] such that \[\vec a\parallel \vec b\].
We know that if the two vectors are parallel to each other they satisfies the basic condition that \[\vec a - \mu \vec b = 0\]. Therefore, putting the values of the vectors in the equation of the condition, we get
\[\begin{array}{l}2\hat i - \hat j + 5\hat k - \mu \left( {4\hat i - 2\hat j + \lambda \hat k} \right) = 0\\ \Rightarrow 2\hat i - 4\hat i\mu - \hat j + 2\hat j\mu + 5\hat k - \lambda \hat k\mu = 0\end{array}\]
Rewriting like terms together, we get
\[ \Rightarrow \left( {2 - 4\mu } \right)\hat i + \left( { - 1 + 2\mu } \right)\hat j + \left( {5 - \lambda \mu } \right)\hat k = 0\hat i + 0\hat j + 0\hat k\]
Now by equate each term to 0, we will get the basic equations. Therefore, we get
\[\begin{array}{l}2 - 4\mu = 0\\ - 1 + 2\mu = 0\\5 - \lambda \mu = 0\end{array}\]
Now by solving the equation \[2 - 4\mu = 0\] we will get the value of \[\mu \]. Therefore,
\[\mu = \dfrac{2}{4} = \dfrac{1}{2}\]
Now by solving the equation \[5 - \lambda \mu = 0\] we will get the value of \[\lambda \].
We will put the value of \[\mu \] in the above equation, we get
\[ \Rightarrow 5 - \lambda \dfrac{1}{2} = 0\]
\[ \Rightarrow \dfrac{\lambda }{2} = 5\]
Multiplying both sides by 2, we get
\[ \Rightarrow \lambda = 10\]

Hence, the value of \[\lambda \] is equal to 10.

Note:
Here we have to keep in mind that while expanding the dot product of vectors is related to cos function not the sin function. Sin function is related to the cross product of the vectors. Also we have to remember that the dot product of the perpendicular vectors is always zero as the angle between them is \[90^\circ \] and the dot product is related to cos function. Also the cross product of the parallel vectors is always zero as the angle between the parallel vectors is \[0^\circ \] or \[{\rm{180}}^\circ \] and cross product is related to the sin function.
Dot product of two vectors is \[a.b = \left| a \right|\left| b \right|\cos \theta \]
Cross product of two vectors is \[a \times b = \left| a \right|\left| b \right|\sin \theta \]
Where, \[\theta \] is the angle between the vectors.