Vectors Joining Two Points - Practice Questions & MCQ

Let the points A(1, 0, 0), B(0, 1, 0) and C(0, 0, 1) on the x-axis, y-axis and z-axis, respectively. Then, clearly,$|\overrightarrow{\mathrm{OA}}|=1,|\overrightarrow{\mathrm{OB}}|=1$ and $|\overrightarrow{\mathrm{OC}}|=1$

The vectors,  $\overrightarrow{O A}, \overrightarrow{O B}$ and $\overrightarrow{O C}$ each having magnitude 1, are called unit vectors along the axes OX, OY, and OZ, respectively, and denoted by , , and respectively.

Now consider any point P(x, y, z) with position vector OP.  Let P₁ be the foot of the perpendicular from P on the plane XOY.  As we observe that P₁P is parallel to the z-axis. Also, , , and  are the unit vectors along the x, y, and z-axes, respectively, thus,  by the definition of the coordinates of P, we have $\overrightarrow{\mathrm{P}_1 \mathrm{P}}=\overrightarrow{\mathrm{OR}}=z \hat{\mathbf{k}}$.

Similarly, $\overrightarrow{\mathrm{QP}_1}=\overrightarrow{\mathrm{OS}}=y \hat{\mathbf{j}}$ and $\overrightarrow{\mathrm{OQ}}=x \hat{\mathbf{i}}$

Therefore,
$$
\begin{aligned}
& \overrightarrow{\mathrm{OP}_1}=\overrightarrow{\mathrm{OQ}}+\overrightarrow{\mathrm{QP}_1}=x \hat{i}+y \hat{j} \\
& \overrightarrow{\mathrm{OP}}=\overrightarrow{\mathrm{OP}_1}+\overrightarrow{\mathrm{P}_1 \mathrm{P}}=x \hat{i}+y \hat{j}+z \hat{k}
\end{aligned}
$$
and,
Hence, the position vector of P with reference to O is given by
$$
\overrightarrow{\mathrm{OP}}(\text { or } \vec{r})=x \hat{i}+y \hat{j}+z \hat{k}
$$

And, the length of any vector $\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}$ is given by
$$
|\vec{r}|=|x \hat{i}+y \hat{j}+z \hat{k}|=\sqrt{x^2+y^2+z^2}
$$

Vector Joining Two Points

If $\mathrm{A}\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)$ and $\mathrm{B}\left(\mathrm{x}_2, \mathrm{y}_2, \mathrm{z}_2\right)$ are any two points in three - dimensional system, then vector joining point A and B is the vector $\overrightarrow{A B}$.
Joining the point A and B with the origin, O , we get position vector of point A and B . i.e.
$$
\begin{aligned}
& \overrightarrow{O A}=x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k} \\
& \overrightarrow{O B}=x_2 \hat{i}+y_2 \hat{j}+z_2 \hat{k}
\end{aligned}
$$

Applying the triangle law of addition on the triangle $O A B$
$$
\overrightarrow{O A}+\overrightarrow{A B}=\overrightarrow{O B}
$$

Using the properties of vector addition, the above equation becomes
$$
\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}
$$
i.e.
$$
\begin{aligned}
\overrightarrow{A B} & =\left(x_2 \hat{i}+y_2 \hat{j}+z_2 \hat{k}\right)-\left(x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k}\right) \\
& =\left(x_2-x_1\right) \hat{i}+\left(y_2-y_1\right) \hat{j}+\left(z_2-z_1\right) \hat{k}
\end{aligned}
$$

The magnitude of vector $\overrightarrow{A B}$ is given by
$$
|\overrightarrow{A B}|=\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2+\left(\mathrm{z}_2-\mathrm{z}_1\right)^2}
$$