Detailed solutions to questions on 3D vectors are presented.
\( \vec{AB} = \lt x_2-x_1,y_2-y_1,z_2-z_1 > = \lt 0-2,-3-6,1-7 > = \lt -2,-9,-6> \)
\( \vec{BA} = \lt x_1-x_2,y_1-y_2,z_1-z_2 > = \lt 2-0,6-(-3),7-1 > = \lt 2,9,6>\)
\( \vec{AB} = \lt -2,-9,-6> = -1 \lt 2,9,6> = -1\vec{BA} \)
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 \( k = 27/25\).
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.
\( || \vec{u} || = \sqrt{ 0^2 + (-3/\sqrt{13})^2 + (2/\sqrt{13})^2 } = \sqrt{ 9/13 + 4/13} = 1\)
\( \vec{AB} = \lt x_2-x_1,y_2-y_1,z_2-z_1> = \lt 0-2,-3-6,1-7> = \lt -2,-9,-6>\)
\( \vec{BC} = \lt 0-0,3-(-3),4-1> = \lt 0,6,3>\)
\( \vec{AC} = \lt 0-2,3-6,4-7> = \lt -2,-3,-3>\)
\( \vec{AB} + \vec{BC} = \lt -2,-9,-6> + \lt 0,6,3> = \lt -2+0,-9+6,-6+3> = \lt -2,-3,-3> = \vec{AC} \)
\( \vec{AB} = \lt 2-(-1),4-2,2-1> = \lt 3,2,1>\)
\( \vec{BC} = \lt 5-2,6-4,3-2> = \lt 3,2,1>\)
\( \vec{AC} = \lt 5-(-1),6-2,3-1> = \lt 6,4,2>\)
\( \vec{AB} \) 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 \( \vec{AB} \).
\(\lt -4,0,2> - 2 \lt x,y,z \gt = 3 \lt x,y,z \gt - 3 \lt -1,-4,2> \)
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:
\( -4-2x = 3x + 3 \;\; , \;\; solution: x = -7/5 \)
\( -2y = 3y + 12 \;\; , \;\; solution: y = -12/5 \)
\( 2-2z = 3z - 6 \;\; , \;\; solution: z = 8/5 \)
\( \vec{v} = \lt -7/5,-12/5,8/5 \gt \)
\( \vec{u} = 2 \vec{v} = 2 \lt -4,2,2> = \lt -8 , 4 , 4> \)
\(-\dfrac{1}{||\vec{v}||} \vec{v} \)
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
\(5(-\dfrac{1}{3} \vec{v}) = (-5/3) \lt -1 , 2 , 2> = \lt 5/3 , -10/3 , -10/3>\)
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
\(\vec{v_1} = k \vec{v_2} \)
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 \( c = 1 \)
\( \vec{AC} = k \vec{AB} \) , 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.
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
\( \dfrac{1}{|| \vec{AG} ||} \vec{AG} = \dfrac{1}{2\sqrt{3}}\lt2,2,2> = \lt1/\sqrt{3} ,1/\sqrt{3} ,1/\sqrt{3} > \)