Q.8) The value of $$p$$ for which $$p(\hat{i}+\hat{j}+\hat{k})$$ is a unit vector is

(A) 0
(B) $$\frac{1}{\sqrt{3}}$$
(C ) 1
(D) $$\sqrt{3}$$

Solution:

Finding the value of $$p$$ for a unit vector

Step 1: Define a unit vector
A vector $$\mathbf{V}$$ is a unit vector if its magnitude, denoted as $$|\mathbf{V}|$$, is equal to $$1 .|\mathbf{V}|=1$$

Step 2: Express the given vector and calculate its magnitude
The given vector is $$\mathbf{V}=p(\hat{i}+\hat{j}+\hat{k})=p \hat{i}+p \hat{j}+p \hat{k}$$.
The magnitude of a vector $$\mathbf{V}=a \hat{i}+b \hat{j}+c \hat{k}$$ is given by $$|\mathbf{V}|=\sqrt{a^2+b^2+c^2}$$.
For the given vector, the magnitude is:
$$
|\mathbf{V}|=\sqrt{p^2+p^2+p^2}=\sqrt{3 p^2}
$$

Since the magnitude must be a positive value, we take the positive square root of $$p^2$$, which is $$|p|$$.
$$
|\mathbf{V}|=|p| \sqrt{3}
$$

Step 3: Solve for p
We set the magnitude equal to 1 :
$$
\begin{aligned}
& |p| \sqrt{3}=1 \\
& |p|=\frac{1}{\sqrt{3}}
\end{aligned}
$$
Answer:
The correct option is (B) $$\frac{1}{\sqrt{3}}$$.