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Q.
The unit vector in the direction of vector $P Q$, where $P$ and $Q$ are the points $(1,2,3)$ and $(4,5,6)$ respectively, is
Vector Algebra
Solution:
A vector with initial point $\left(x_1, y_1, z_1\right)$ and terminal point $\left(x_2, y_2, z_2\right)$ is given by $\left(x_2-x_1\right) \hat{i}+\left(y_2-y_1\right) \hat{j}+\left(z_2-z_1\right) \hat{k}$
The given points are $P(1,2,3)$ and $Q(4,5,6)$.
$\therefore x_1=1, y_1=2, z_1=3$ and $x_2=4, y_2=5, z_2=6$
So, vector $P Q=\left(x_2-x_1\right) \hat{i}+\left(y_2-y_1\right) \hat{j}+\left(z_2-z_1\right) \hat{k}$
$=(4-1) \hat{i}+(5-2) \hat{j}+(6-3) \hat{k}=3 \hat{i}+3 \hat{j}+3 \hat{k}$
$\therefore$ Magnitude of given vector
$| PQ |=\sqrt{3^2+3^2+3^2}=\sqrt{9+9+9}=\sqrt{27}=3 \sqrt{3}$
Hence, the unit vector in the direction of $P Q$,
$\frac{P Q}{|P Q|} =\frac{3 \hat{i}+3 \hat{j}+3 \hat{k}}{3 \sqrt{3}}=\frac{3}{3 \sqrt{3}}(\hat{i}+\hat{j}+\hat{k}) $
$ =\frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}+\frac{1}{\sqrt{3}} \hat{k}$