Question

Two points P and Q have cordinates (2, 3, −1) and (4, −3, 2) respectively. Find a unit vector parallel to the vector joining P with Q.​

Answer 1

Answer:

Two points P and Q have coordinates (2, 3, −1) and (4, −3, 2) respectively, the unit vector parallel to the vector joining P with Q. is $\frac{2i-6j+3k}{7}$

Explanation:

• A unit vector has a magnitude of 1 unit

• If A and B are two points the line joining these two points is vector AB

• The unit vector along this vector is $n=\frac{AB}{/AB/}$ where $/AB/$ is the magnitude of the vector which can be calculated as $/AB/=\sqrt{A^{2} +B^{2} }$ (the root of the sum of squares of its components)

From the question, we have

two points are $A=(2,3,-1)$ and $B=(4,-3,2)$

the vector joining these two points is $AB=(4-2)i+(-3-3)j+(2--1)k=2i-6j+3k$

the magnitude of the vector is $/AB/=\sqrt{2^{2} +6^{2} +3^{2} } =7$ units

therefore the unit vector is $\frac{2i-6j+3k}{7}$