Question

Points P and Q are located at (0, 2, 4) and (-3, 1,5). Calculate

a) The position vector of P

b) The distance vector from P to Q

c) The distance between P and Q

d) A vector parallel to PQ with magnitude of 10​

Answer 1

Answer:

a) The position vector of P is $\vec P =(0, 2,4)$.

b) The distance vector from P to Q is $\overrightarrow{PQ} = (-3,-1,1)$.

c) The distance between P and Q is $\|\overrightarrow{PQ}\|=\sqrt{11}$.

d) A vector parallel to PQ with magnitude of 10 is $\vec v = \left(-\frac{30\sqrt{11}}{11},-\frac{10\sqrt{11}}{11}, \frac{10\sqrt{11}}{11} \right)$.

Explanation:

a) The position vector of a point is the vector displacement from the origin to the location of the point. That is:

$\vec P = (0,2,4)-(0,0,0)$

$\vec P = (0-0, 2-0, 4-0)$

$\vec P =(0, 2,4)$

The position vector of P is $\vec P =(0, 2,4)$.

b) First, we calculate the position vector of point Q:

$\vec Q = (-3,1,5)-(0,0,0)$

$\vec Q = (-3-0,1-0,5-0)$

$\vec Q =(-3,1,5)$

The distance vector from P to Q is define by the following vectorial expression:

$\overrightarrow{PQ} = \vec Q - \vec P$ (1)

$\overrightarrow{PQ} = (-3,1,5)-(0,2,4)$

$\overrightarrow{PQ} =(-3-0,1-2,5-4)$

$\overrightarrow{PQ} = (-3,-1,1)$

The distance vector from P to Q is $\overrightarrow{PQ} = (-3,-1,1)$.

c) There are two approaches to calculate the distance between P and Q:

First Method - Pythagorean Theorem:

$\|\overrightarrow{PQ}\| = \sqrt{(-3)^{2}+(-1)^{2}+1^{2}}$

$\|\overrightarrow{PQ}\|=\sqrt{11}$

Second Method - Dot Product:

$\|\overrightarrow{PQ}\| = \sqrt{\overrightarrow{PQ}\,\bullet\,\overrightarrow{PQ}}$ (2)

$\|\overrightarrow{PQ}\| = \sqrt{(-3,-1,1)\,\bullet (-3,-1,1)}$

$\|\overrightarrow{PQ}\|=\sqrt{11}$

The distance between P and Q is $\|\overrightarrow{PQ}\|=\sqrt{11}$.

d) To determine a vector parallel to PQ with a given magnitude is determined by the following expression:

$\vec v = \frac{k}{\|\overrightarrow{PQ}\|} \cdot \overrightarrow{PQ}$ (3)

Where $k$ is the scale factor.

If we know that $\overrightarrow{PQ} = (-3,-1,1)$, $\|\overrightarrow{PQ}\|=\sqrt{11}$ and $k = 10$, then the vector is:

$\vec v = \frac{10}{\sqrt{11}}\cdot (-3,-1,1)$

$\vec v = \left(-\frac{30}{\sqrt{11}},-\frac{10}{\sqrt{11}},\frac{10}{\sqrt{11}}\right)$

$\vec v = \left(-\frac{30\sqrt{11}}{11},-\frac{10\sqrt{11}}{11}, \frac{10\sqrt{11}}{11} \right)$

A vector parallel to PQ with magnitude of 10 is $\vec v = \left(-\frac{30\sqrt{11}}{11},-\frac{10\sqrt{11}}{11}, \frac{10\sqrt{11}}{11} \right)$.