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) $${\displaystyle \frac{1}{\sqrt{2}}},{\displaystyle \frac{1}{\sqrt{2}}}$$
(C) $$1,0$$
(D) $${\displaystyle \frac{1}{\sqrt{2}}},-{\displaystyle \frac{1}{\sqrt{2}}}$$

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
$$a={\displaystyle \frac{1}{\sqrt{2}}}$$
Now since, $$a=-b$$
Then
$$b=-a=-{\displaystyle \frac{1}{\sqrt{2}}}$$
Hence, the values of a and b may be $$\left({\displaystyle \frac{1}{\sqrt{2}}},-{\displaystyle \frac{1}{\sqrt{2}}}\right)$$

Hence, the correct option is D
Note: For clarity, observe that the values $$a={\displaystyle \frac{1}{\sqrt{2}}}$$ or $$b=-{\displaystyle \frac{1}{\sqrt{2}}}$$ 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
$$b={\displaystyle \frac{1}{\sqrt{2}}}$$
And similarly, from $$a=-b$$, we have
$$a=-{\displaystyle \frac{1}{\sqrt{2}}}$$
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.