# 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
$$\vec{AB} = \lt x_2-x_1,y_2-y_1,z_2-z_1 > = \lt 0-2,-3-6,1-7 > = \lt -2,-9,-6>$$
$$\vec{BA} = \lt x_1-x_2,y_1-y_2,z_1-z_2 > = \lt 2-0,6-(-3),7-1 > = \lt 2,9,6>$$
$$\vec{AB} = \lt -2,-9,-6> = -1 \lt 2,9,6> = -1\vec{BA}$$

2)
Given vectors $$\vec{v_1} = \lt 0,-3,2>$$ and $$\vec{v_2} = \lt -3,4,5>$$, 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 $$\vec{v_1} = \lt a_1,b_1,c_1> = \lt 0,-3,2>$$ and $$\vec{v_2} = \lt a_2,b_2,c_2> = \lt -3,4,5>$$, 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: $$\vec{v_1} + \vec{v_2} = \lt a_1+a_2,b_1+b_2,c_1+c_2>$$
difference: $$\vec{v_1} - \vec{v_2} = \lt a_1 - a_2,b_1 - b_2,c_1 - c_2>$$
multiply by a scalar: $$k \vec{v_1} = \lt k a_1,k b_1,k c_1>$$.
a) Use the sum formula
$$\vec{v_1} + \vec{v_2} = \lt 0,-3,2> + \lt -3,4,5> = \lt 0+(-3) , -3+4 , 2 + 5 > = \lt -3 , 1 , 7>$$
b) Use the difference formula
$$\vec{v_1} - \vec{v_2} = \lt 0,-3,2> - \lt -3,4,5> = \lt 0 -(-3) , -3 - 4 , 2 - 5 > = \lt 3 , -7 , -3>$$
c) Use the scalar multiplication formula
$$-3\vec{v_1} = -3 \lt 0,-3,2> = \lt -3\cdot0 , -3\cdot(-3) , -3\cdot2> = \lt 0,9,-6>$$
d) Use a combination of scalar and sum formula
$$-2\vec{v_1} + 3\vec{v_2} = -2\lt 0,-3,2> + 3\lt-3,4,5> = \lt 0,6,-4> + \lt-9,12,15> = \lt -9,18,11>$$
e) Find the vector $$\vec{v_1} + k\vec{v_2}$$.
$$\vec{v_1} + k\vec{v_2} = \lt 0,-3,2> + k \lt -3,4,5> = \lt -3k ,-3+4k ,2+5k>$$
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 $$\vec{v} = \lt 0,-3,2>$$, 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.
$$\vec{u} = \dfrac{1}{|| \vec{v} ||} \vec{v} = \dfrac{1}{\sqrt{0^2+(-3)^2+2^2}} \lt 0,-3,2> = \lt 0,-3/\sqrt{13} , 2/\sqrt{13}>$$
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.
$$\vec{AB} = \lt x_2-x_1,y_2-y_1,z_2-z_1> = \lt 0-2,-3-6,1-7> = \lt -2,-9,-6>$$
$$\vec{BC} = \lt 0-0,3-(-3),4-1> = \lt 0,6,3>$$
$$\vec{AC} = \lt 0-2,3-6,4-7> = \lt -2,-3,-3>$$
$$\vec{AB} + \vec{BC} = \lt -2,-9,-6> + \lt 0,6,3> = \lt -2+0,-9+6,-6+3> = \lt -2,-3,-3> = \vec{AC}$$

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.
$$\vec{AB} = \lt 2-(-1),4-2,2-1> = \lt 3,2,1>$$
$$\vec{BC} = \lt 5-2,6-4,3-2> = \lt 3,2,1>$$
$$\vec{AC} = \lt 5-(-1),6-2,3-1> = \lt 6,4,2>$$
$$\vec{AB}$$ and $$\vec{BC}$$ has equal components and are therefore equivalent.
Note that
$$\vec{AC} = \lt 6,4,2> = 2 \lt 3,2,1> = 2\vec{BC}$$
Hence, $$\vec{AC}$$ is parallel to $$\vec{BC}$$ and $$\vec{AB}$$.

6)
Given vectors $$\vec{v_1} = \lt -4,0,2>$$ and $$\vec{v_2} = \lt -1,-4,2>$$, 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.
$$\lt -4,0,2> - 2 \lt x,y,z \gt = 3 \lt x,y,z \gt - 3 \lt -1,-4,2>$$
Multiply and subtract and group each side of the vector equation.
$$\lt -4-2x,0-2y,2-2z> = \lt 3x -3(-1) , 3y - 3(-4) , 3z - 3(2)>$$
$$\lt -4-2x,-2y,2-2z> = \lt 3x + 3 , 3y + 12 , 3z - 6>$$
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 $$\vec{v} = \lt -4,2,2>$$ but with twice the length of $$\vec{v}$$.
Solution
$$\vec{u}$$ is twice vector $$\vec{v}$$. Hence
$$\vec{u} = 2 \vec{v} = 2 \lt -4,2,2> = \lt -8 , 4 , 4>$$

8)
Find a vector $$\vec{u}$$ in the opposite direction of vector $$\vec{v} = \lt -1,2,2>$$ 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
$$5(-\dfrac{1}{3} \vec{v}) = (-5/3) \lt -1 , 2 , 2> = \lt 5/3 , -10/3 , -10/3>$$

9)
Given vector $$\vec{v} = \lt -1,2,2>$$, 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 $$\vec{v_1} = \lt -4,6,2>$$ and $$\vec{v_2} = \lt 2,b,c>$$ 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
$$\lt -4,6,2> = k \lt 2,b,c> = \lt 2k , k b , k c >$$
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.
$$\vec{AC} = \lt 0 - 2 , 2-6 , 3 - 7> = \lt -2 , -4 , -4>$$
$$\vec{AB} = \lt 1-2 , 4- 6 , 5 - 7> = \lt -1 , -2 , -2>$$
Note that.
$$\vec{AC} = \lt -2 , -4 , -4> = 2 \lt-1 , -2 , -2> = 2 \vec{AB}$$
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).
$$\vec{AB} = \lt 2-0,0-0,0-0> = \lt 2,0,0>$$
$$\vec{EF} = \lt2-0,0-0,2-2> = \lt2,0,0>$$
$$\vec{DC} = \lt2-0, 2-2,0-0> = \lt2,0,0>$$
$$\vec{HG} = \lt2- 0 ,2 -2, 2-2 > =\lt2,0,0>$$
$$\vec{AC} = \lt2 - 0 , 2 - 0, 0 - 0> = \lt2,2,0>$$
$$\vec{AG} = \lt2-0,2-2,2-0> = \lt2,2,2>$$
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.
$$\vec{AB} + \vec{BF} + \vec{FG} = \lt2,0,0> + \lt0,0,2> + \lt0,2,0> = \lt2,2,2>$$
$$\vec{AC} + \vec{CG} = \lt2,2,0> + \lt0,0,2> = \lt2,2,2>$$
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
$$\dfrac{1}{|| \vec{AG} ||} \vec{AG} = \dfrac{1}{2\sqrt{3}}\lt2,2,2> = \lt1/\sqrt{3} ,1/\sqrt{3} ,1/\sqrt{3} >$$