Question

Let \[\vec a = \hat i + \hat j + \hat k\], \[\vec b = \hat i - \hat j + \hat k\] and \[\vec c = \hat i - \hat j - \hat k\] be three vectors. A vector \[\vec v\] of the form \[\vec a + \lambda \vec b\] for some scalar \[\lambda \] whose projection on \[\vec c\] is \[\dfrac{1}{{\sqrt 3 }}\], is given by
A. \[\hat i - 3\hat j + \hat k\]
B. \[ - 3\hat i - 3\hat j - \hat k\]
C. \[3\hat i - \hat j + 3\hat k\]
D. \[\hat i + 3\hat j - 3\hat k\]

Answer 1

Hint: We should use the formula for the projection of one vector on the other vector. We can also use the formula for the dot product of the two vectors. Using the given values of the vectors \[\vec a\] and \[\vec b\], solve the vector \[\vec v\]. Then solve the formula for the projection of the vector \[\vec v\] on the vector \[\vec c\] and calculate the value of lambda. Substitute this value of lambda in the solved expression for.

Formulae used:
The projection of a vector \[\vec a\] on the vector \[\vec b\] is given by
\[\vec X = \dfrac{{\vec a \cdot \vec b}}{{\left| {\vec b} \right|}}\] …… (1)
The dot product of the two vectors \[\vec a\] and \[\vec b\] is given by
\[\vec a \cdot \vec b = {a_x}{b_x} + {a_y}{b_y} + {a_z}{b_z}\] …… (2)

Complete step by step answer:
We have given three vectors as
\[\vec a = \hat i + \hat j + \hat k\]
\[\Rightarrow \vec b = \hat i - \hat j + \hat k\]
\[\Rightarrow \vec c = \hat i - \hat j - \hat k\]
We have also given that the vector \[\vec v\] is given by
\[\vec v = \vec a + \lambda \vec b\]
Substitute \[\hat i + \hat j + \hat k\] for \[\vec a\] and \[\hat i - \hat j + \hat k\] for \[\vec b\] in the above equation.
\[\vec v = \left( {\hat i + \hat j + \hat k} \right) + \lambda \left( {\hat i - \hat j + \hat k} \right)\]
\[ \Rightarrow \vec v = \left( {\hat i + \hat j + \hat k} \right) + \left( {\lambda \hat i - \lambda \hat j + \lambda \hat k} \right)\]
\[ \Rightarrow \vec v = \left( {1 + \lambda } \right)\hat i + \left( {1 - \lambda } \right)\hat j + \left( {1 + \lambda } \right)\hat k\] …… (3)

We have given that the projection of the vector \[\vec v\] on the vector \[\vec c\] is \[\dfrac{1}{{\sqrt 3 }}\].
According to equation, we can write
\[\dfrac{{\vec v \cdot \vec c}}{{\left| {\vec c} \right|}} = \dfrac{1}{{\sqrt 3 }}\] …… (4)
Let us first calculate the dot product of the vectors \[\vec v\] and \[\vec c\].
\[\vec v \cdot \vec c = \left[ {\left( {1 + \lambda } \right)\hat i + \left( {1 - \lambda } \right)\hat j + \left( {1 + \lambda } \right)\hat k} \right] \cdot \left( {\hat i - \hat j - \hat k} \right)\]
\[ \Rightarrow \vec v \cdot \vec c = \left( {1 + \lambda } \right) - \left( {1 - \lambda } \right) - \left( {1 + \lambda } \right)\]
\[ \Rightarrow \vec v \cdot \vec c = \lambda - 1\]

Let now calculate the magnitude of the vector \[\vec c\].
\[\left| {\vec c} \right| = \sqrt {{{\left( 1 \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( { - 1} \right)}^2}} \]
\[ \Rightarrow \left| {\vec c} \right| = \sqrt {1 + 1 + 1} \]
\[ \Rightarrow \left| {\vec c} \right| = \sqrt 3 \]
Substitute \[\lambda - 1\] for \[\vec v \cdot \vec c\] and \[\sqrt 3 \] for \[\left| {\vec c} \right|\] in equation (4).
\[\dfrac{{\lambda - 1}}{{\sqrt 3 }} = \dfrac{1}{{\sqrt 3 }}\]
\[ \Rightarrow \lambda - 1 = 1\]
\[ \Rightarrow \lambda = 2\]
Let now determine the vector \[\vec v\].Substitute \[2\] for \[\lambda \] in equation (5).
\[ \Rightarrow \vec v = \left( {1 + 2} \right)\hat i + \left( {1 - 2} \right)\hat j + \left( {1 + 2} \right)\hat k\]
\[ \therefore \vec v = 3\hat i - \hat j + 3\hat k\]
Therefore, the value of the required vector is \[3\hat i - \hat j + 3\hat k\].

Hence, the correct option is C.

Note: The students should be careful while solving the expressions in the vector form because the mistake in only the sign of the components of the unit vectors in X, Y or Z direction can make the whole solution wrong and we will end with a wrong answer. Hence, the students should be careful while performing calculations.