Question

if \[\hat n = a\hat i + b\hat j\] is perpendicular to the vector \[\left( {\hat i + \hat j} \right)\]. Then the value of \[a\] and \[b\] may be:
(A) \[1, - 1\]
(B) \[\dfrac{1}{{\sqrt 2 }},\dfrac{1}{{\sqrt 2 }}\]
(C) \[1,0\]
(D) \[\dfrac{1}{{\sqrt 2 }}, - \dfrac{1}{{\sqrt 2 }}\]

Answer 1

Hint: The cap on the \[n\] vector signifies that \[n\] is a unit vector, hence it has a magnitude equal to 1. Two vectors which are perpendicular must have a dot product equal to zero.

Formula used: In this solution we will be using the following formulae;
\[A \cdot B = {A_x}{B_x} + {A_y}{B_y}\] where \[A\] and \[B\] are vectors, \[{A_x}\] is the x component of the vector \[A\] while \[{A_y}\] is the y component. Similarly for the vector \[B\].
\[\left| A \right| = \sqrt {A_x^2 + A_y^2} \] where \[\left| A \right|\] signifies the magnitude of a vector \[A\].

Complete Step-by-Step Solution:
We have a particular vector with unknown components. This vector however is perpendicular to a vector of known components. We are to determine the component of the first vector
It is necessary to note that the first vector \[\hat n = a\hat i + b\hat j\] is a unit vector signified by the cap on the \[n\]. Hence, the magnitude of the vector is equal to 1.
This unit vector is perpendicular to the vector \[r = \hat i + \hat j\], the dot product of the two vectors is zero. Hence,
\[\hat n \cdot r = \left( {a\hat i + b\hat j} \right) \cdot \left( {\hat i + \hat j} \right) = a + b = 0\]
\[ \Rightarrow a = - b\]
Now, recall the unit vector has a magnitude of 1, hence
\[\left| {\hat n} \right| = \sqrt {{a^2} + {b^2}} = 1\]
\[ \Rightarrow \sqrt {{a^2} + {{\left( { - a} \right)}^2}} = \sqrt 2 a = 1\]
Then by making \[a\] subject, we get
\[a = \dfrac{1}{{\sqrt 2 }}\]
Now since, \[a = - b\]
Then
\[b = - a = - \dfrac{1}{{\sqrt 2 }}\]
Hence, the values of a and b may be \[\left( {\dfrac{1}{{\sqrt 2 }}, - \dfrac{1}{{\sqrt 2 }}} \right)\]

Hence, the correct option is D
Note: For clarity, observe that the values \[a = \dfrac{1}{{\sqrt 2 }}\] or \[b = - \dfrac{1}{{\sqrt 2 }}\] is peculiar to either of the variables as any of them can take any of the values (based on the calculations), as proven below;
At
\[\left| {\hat n} \right| = \sqrt {{a^2} + {b^2}} = 1\] we could say that since \[a = - b\] then
\[\sqrt {{{\left( { - b} \right)}^2} + {b^2}} = \sqrt 2 b = 1\]
Hence, by making \[b\] subject of the formula, we get
\[b = \dfrac{1}{{\sqrt 2 }}\]
And similarly, from \[a = - b\], we have
\[a = - \dfrac{1}{{\sqrt 2 }}\]
Hence, we see that the two variables have switched positions. What is important is that when one takes one value, the other must take the other value.