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.



collinear points in 3D (R3)


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 \( \vec{AG} \).

cube



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

\( \dfrac{1}{|| \vec{AG} ||} \vec{AG} = \dfrac{1}{2\sqrt{3}}<2,2,2> = <1/\sqrt{3} ,1/\sqrt{3} ,1/\sqrt{3} > \)

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Updated: 20 January 2017 (A Dendane)

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