# Solutions to Questions on 3D Vectors (R3)

Detailed solutions to questions on 3D vectors are presented.

Detailed Solutions Questions on 3D Vectors

1) Find the components of the vectors $\vec{AB}$ and $\vec{BA}$ where A and B are points given by their coordinates A(2,6,7) and B(0,-3,1) and show that $\vec{AB} = -1 \vec{BA}$.
Solution
Given points A and B are defined by their coordinates: $A (x_1 , y_1 ,z_1) = A(2,6,7)$ and B $(x_2 , y_2 ,z_2) = B(0,-3,1)$ , we use the formula
$\vec{AB} = < x_2-x_1,y_2-y_1,z_2-z_1> = <0-2,-3-6,1-7> = <-2,-9,-6>$
$\vec{BA} = < x_1-x_2,y_1-y_2,z_1-z_2> = <2-0,6-(-3),7-1> = <2,9,6>$
$\vec{AB} = <-2,-9,-6> = -1 <2,9,6> = -1\vec{BA}$

2) Given vectors $\vec{v_1} = <0,-3,2>$ and $\vec{v_2} = <-3,4,5>$, find:
a) $\vec{v_1} + \vec{v_2}$
b) $\vec{v_1} - \vec{v_2}$
c) $-3\vec{v_1}$
d) $-2\vec{v_1} + 3\vec{v_2}$
e) $k$such that $||\vec{v_1} + k\vec{v_2}|| = \sqrt{67}$.
Solution
Given vectors $\vec{v_1} = < a_1,b_1,c_1> = <0,-3,2>$ and $\vec{v_2} = < a_2,b_2,c_2> = <-3,4,5>$, the sum $\vec{v_1} + \vec{v_2}$ , the difference $\vec{v_1} - \vec{v_2}$ and scalar multiplication $k \vec{v_1}$, k a real number, are given by
sum: $\vec{v_1} + \vec{v_2} = < a_1+a_2,b_1+b_2,c_1+c_2>$
difference: $\vec{v_1} - \vec{v_2} = < a_1 - a_2,b_1 - b_2,c_1 - c_2>$
multiply by a scalar: $k \vec{v_1} = < k a_1,k b_1,k c_1>$.
a) Use the sum formula
$\vec{v_1} + \vec{v_2} = <0,-3,2> + <-3,4,5> = <0+(-3) , -3+4 , 2 + 5 > = <-3 , 1 , 7>$
b) Use the difference formula
$\vec{v_1} - \vec{v_2} = <0,-3,2> - <-3,4,5> = <0 -(-3) , -3 - 4 , 2 - 5 > = <3 , -7 , -3>$
c) Use the scalar multiplication formula
$-3\vec{v_1} = -3 <0,-3,2> = <-3\cdot0 , -3\cdot(-3) , -3\cdot2> = <0,9,-6>$
d) Use a combination of scalar and sum formula
$-2\vec{v_1} + 3\vec{v_2} = -2<0,-3,2> + 3<-3,4,5> = <0,6,-4> + <-9,12,15> = <-9,18,11>$
e) Find the vector $\vec{v_1} + k\vec{v_2}$.
$\vec{v_1} + k\vec{v_2} = <0,-3,2> + k<-3,4,5> = <-3k ,-3+4k ,2+5k>$
Find the magnitude of $\vec{v_1} + k\vec{v_2}$ and make it equal to $\sqrt{67}$.
$\sqrt{(-3k)^2 + (-3+4k)^2 + (2+5k)^2} = \sqrt{67}$
Square both sides of the above equation.
$(-3k)^2 + (-3+4k)^2 + (2+5k)^2 = 67$
Expand and group.
$50k^2-4k+13 = 67$
Solve the above quadratic equation for k to obtain.
$k = -1$ and $k = 27/25$.

3) Given vector $\vec{v} = <0,-3,2>$, find the unit vector in the same direction as $\vec{v}$ and check that its magnitude is equal to 1.
Solution
The unit vector $\vec{u}$ in the same direction as vector $\vec{v}$ is given by.
$\vec{u} = \dfrac{1}{|| \vec{v} ||} \vec{v} = \dfrac{1}{\sqrt{0^2+(-3)^2+2^2}} <0,-3,2> = <0,-3/\sqrt{13} , 2/\sqrt{13}>$
Calculate magnitude of $\vec{u}$ and check that it is equal to 1.
$|| \vec{u} || = \sqrt{ 0^2 + (-3/\sqrt{13})^2 + (2/\sqrt{13})^2 } = \sqrt{ 9/13 + 4/13} = 1$

4) Given the points A(2,6,7), B(0,-3,1) and C(0,3,4), find the components of the vectors $\vec{AB}$, $\vec{AC}$ and $\vec{BC}$ and show that $\vec{AB} + \vec{BC} = \vec{AC}$.
Solution
The components of a vector defined by two points are given by the difference between the coordinates of the terminal and the initial points.
$\vec{AB} = < x_2-x_1,y_2-y_1,z_2-z_1> = <0-2,-3-6,1-7> = <-2,-9,-6>$
$\vec{BC} = <0-0,3-(-3),4-1> = <0,6,3>$
$\vec{AC} = <0-2,3-6,4-7> = <-2,-3,-3>$
$\vec{AB} + \vec{BC} = <-2,-9,-6> + <0,6,3> = <-2+0,-9+6,-6+3> = <-2,-3,-3> = \vec{AC}$

5) Given the points A(-1,2,1), B(2,4,2) and C(5,6,3), find the components of the vectors $\vec{AB}$, $\vec{BC}$ and $\vec{AC}$ and determine which of these vectors are equivalent and which are parallel.
Solution
The components of a vector defined by two points are given by the difference between the coordinates of the terminal and the initial points.
$\vec{AB} = <2-(-1),4-2,2-1> = <3,2,1>$
$\vec{BC} = <5-2,6-4,3-2> = <3,2,1>$
$\vec{AC} = <5-(-1),6-2,3-1> = <6,4,2>$
$\vec{AB}$ and $\vec{BC}$ has equal components and are therefore equivalent.
Note that
$\vec{AC} = <6,4,2> = 2 <3,2,1> = 2\vec{BC}$
Hence, $\vec{AC}$ is parallel to $\vec{BC}$ and $\vec{AB}$.

6) Given vectors $\vec{v_1} = <-4,0,2>$ and $\vec{v_2} = <-1,-4,2>$, find vector $\vec{v}$ such that $\vec{v_1} - 2 \vec{v} = 3 \vec{v} - 3 \vec{v_2}$
Solution
Let $\vec{v} = \lt x,y,z \gt$ and rewrite the vector equation $\vec{v_1} - 2 \vec{v} = 3 \vec{v} - 3 \vec{v_2}$ using the components.
$<-4,0,2> - 2 \lt x,y,z \gt = 3 \lt x,y,z \gt - 3 <-1,-4,2>$
Multiply and subtract and group each side of the vector equation.
$<-4-2x,0-2y,2-2z> = <3x -3(-1) , 3y - 3(-4) , 3z - 3(2)>$
$<-4-2x,-2y,2-2z> = <3x + 3 , 3y + 12 , 3z - 6>$
Two vectors are equal ( or equivalent) if their components are equal. Hence the equations:
$-4-2x = 3x + 3 \;\; , \;\; solution: x = -7/5$
$-2y = 3y + 12 \;\; , \;\; solution: y = -12/5$
$2-2z = 3z - 6 \;\; , \;\; solution: z = 8/5$
$\vec{v} = \lt -7/5,-12/5,8/5 \gt$

7) Find a vector $\vec{u}$ in the same direction as vector $\vec{v} = <-4,2,2>$ but with twice the length of $\vec{v}$.
Solution
$\vec{u}$ is twice vector $\vec{v}$. Hence
$\vec{u} = 2 \vec{v} = 2<-4,2,2> = <-8 , 4 , 4>$

8) Find a vector $\vec{u}$ in the opposite direction of vector $\vec{v} = <-1,2,2>$ but with a length of 5 units.
Solution
The unit vector in the opposite direction of $\vec{v}$ is given by
$-\dfrac{1}{||\vec{v}||} \vec{v}$
where $||\vec{v} ||$ is the magnitude of $\vec{v}$ and is given by
$||\vec{u} || =\sqrt{(-1)^2+2^2+2^2} = 3$
$\vec{u}$ is given by
$5(-\dfrac{1}{3} \vec{v}) = (-5/3)<-1 , 2 , 2> = <5/3 , -10/3 , -10/3>$

9) Given vector $\vec{v} = <-1,2,2>$, find a real number $k$ such that $||k \vec{v} || = 1/5$.
Solution
We first note that
$||k \vec{v} || = |k| || \vec{v} ||$
$|| \vec{v} ||$ is given by
$|| \vec{v} ||= \sqrt{(-1)^2+2^2+2^2} = 3$
Substituting in the equation $||k \vec{v} || = 1/5$, we obtain
$3 |k| = 1/5$
which gives
|k| = 1/15
Two solutions.
k = 1/15 and k = - 1/15

10) Find $b$ and $c$ such that vectors $\vec{v_1} = <-4,6,2>$ and $\vec{v_2} = <2,b,c>$ are parallel.
Solution
Vectors $\vec{v_1}$ and $\vec{v_2}$ are parallel if there exists k such that
$\vec{v_1} = k \vec{v_2}$
Hence the vector equation
$<-4,6,2> = k <2,b,c> = <2k , k b , k c >$
The above vector equation gives 3 components equations:
$-4 = 2 k$ , hence $k = -2$
$6 = k b = - 2 b$ , hence $b = -3$
$2 = k c = -2 c$ , hence $c = 1$

11) Are the three points A(2,6,7), B(1,4,5) and C(0,2,3) collinear?
Solution
For the points A, B and C to be collinear, we need to find k such that
$\vec{AC} = k \vec{AB}$ , vector AC and AB are collinear.
Find the components of vectors $\vec{AC}$ and $\vec{AB}$ using the coordinates of the points A, B and C.
$\vec{AC} = <0 - 2 , 2-6 , 3 - 7> = <-2 , -4 , -4>$
$\vec{AB} = <1-2 , 4- 6 , 5 - 7> = <-1 , -2 , -2>$
Note that.
$\vec{AC} = <-2 , -4 , -4> = 2 <-1 , -2 , -2> = 2 \vec{AB}$
Hence
$\vec{AC} = 2 \vec{AB}$ , k = 2
Hence vectors $\vec{AC}$ and $\vec{AB}$ are collinear and therefore the points A, B and C are collinear (on the same line) as shown below in the rectangular system of coordinates. 12) A cube of side 2 units is shown below.
a) Find the components of the vectors $\vec{AB}$, $\vec{EF}$, $\vec{DC}$, $\vec{HG}$, $\vec{AC}$ and $\vec{AG}$.
b) Which of the vectors in part a) are equivalent?
c) Prove algebraically that $\vec{AB} + \vec{BF} + \vec{FG} = \vec{AC} + \vec{CG}$.
d) Find $|| \vec{AG} ||$.
e) Find the unit vector in the same direction as vector $\vec{AG}$. Solution
a)
We first need to write the coordinates of points A, B, C, D, E, F, G and H.
A(0,0,0,), B(2,0,0), C(2,2,0), D(0,2,0), E(0,0,2), F(2,0,2), G(2,2,2), H(0,2,2).
$\vec{AB} = <2-0,0-0,0-0> = <2,0,0>$
$\vec{EF} = <2-0,0-0,2-2> = <2,0,0>$
$\vec{DC} = <2-0, 2-2,0-0> = <2,0,0>$
$\vec{HG} = <2- 0 ,2 -2, 2-2 > =<2,0,0>$
$\vec{AC} = <2 - 0 , 2 - 0, 0 - 0> = <2,2,0>$
$\vec{AG} = <2-0,2-2,2-0> = <2,2,2>$
b) The vectors $\vec{AB}$, $\vec{EF}$, $\vec{DC}$ and $\vec{HG}$ have equal components and are therefore equivalent (equal).
c)
Calculate $\vec{AB} + \vec{BF} + \vec{FG}$ and $\vec{AC} + \vec{CG}$ and compare.
$\vec{AB} + \vec{BF} + \vec{FG} = <2,0,0> + <0,0,2> + <0,2,0> = <2,2,2>$
$\vec{AC} + \vec{CG} = <2,2,0> + <0,0,2> = <2,2,2>$
Hence $\vec{AB} + \vec{BF} + \vec{FG}$ = $\vec{AC} + \vec{CG}$.
d) Find $|| \vec{AG} || = \sqrt{2^2+2^2+2^2} = 2\sqrt{3}$.
e) Unit vector in the same direction as vector $\vec{AG}$ is given by
$\dfrac{1}{|| \vec{AG} ||} \vec{AG} = \dfrac{1}{2\sqrt{3}}<2,2,2> = <1/\sqrt{3} ,1/\sqrt{3} ,1/\sqrt{3} >$