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

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.