Detailed solutions to questions on 3D vectors are presented.

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 \( \vec{AG} \).

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