Master 3D Vectors (R³): 12 Practice Problems

Given points \( A(x_1 , y_1 ,z_1) = A(2,6,7) \) and \( B(x_2 , y_2 ,z_2) = B(0,-3,1) \), we use the component formula:

\[ \vec{AB} = \langle x_2-x_1, y_2-y_1, z_2-z_1 \rangle = \langle 0-2, -3-6, 1-7 \rangle = \langle -2, -9, -6 \rangle \] \[ \vec{BA} = \langle x_1-x_2, y_1-y_2, z_1-z_2 \rangle = \langle 2-0, 6-(-3), 7-1 \rangle = \langle 2, 9, 6 \rangle \]

To show the relationship:

\[ \vec{AB} = \langle -2, -9, -6 \rangle = -1 \langle 2, 9, 6 \rangle = -1\vec{BA} \]

a) Use the sum formula:

\[ \vec{v_1} + \vec{v_2} = \langle 0,-3,2 \rangle + \langle -3,4,5 \rangle = \langle 0+(-3), -3+4, 2+5 \rangle = \langle -3, 1, 7 \rangle \]

b) Use the difference formula:

\[ \vec{v_1} - \vec{v_2} = \langle 0,-3,2 \rangle - \langle -3,4,5 \rangle = \langle 0-(-3), -3-4, 2-5 \rangle = \langle 3, -7, -3 \rangle \]

c) Use scalar multiplication:

\[ -3\vec{v_1} = -3 \langle 0,-3,2 \rangle = \langle -3(0), -3(-3), -3(2) \rangle = \langle 0, 9, -6 \rangle \]

d) Combination of scalar and sum:

\[ -2\vec{v_1} + 3\vec{v_2} = -2\langle 0,-3,2 \rangle + 3\langle -3,4,5 \rangle = \langle 0,6,-4 \rangle + \langle -9,12,15 \rangle = \langle -9, 18, 11 \rangle \]

e) Find the vector \( \vec{v_1} + k\vec{v_2} \):

\[ \vec{v_1} + k\vec{v_2} = \langle 0,-3,2 \rangle + k \langle -3,4,5 \rangle = \langle -3k, -3+4k, 2+5k \rangle \]

Find the magnitude and set it equal to \( \sqrt{67} \):

\[ \sqrt{(-3k)^2 + (-3+4k)^2 + (2+5k)^2} = \sqrt{67} \]

Square both sides, expand, and group:

\[ 9k^2 + (9 - 24k + 16k^2) + (4 + 20k + 25k^2) = 67 \] \[ 50k^2 - 4k + 13 = 67 \implies 50k^2 - 4k - 54 = 0 \]

Solve the quadratic equation to obtain: \( k = -1 \) and \( k = 27/25 \).

The unit vector \( \vec{u} \) is given by \( \vec{u} = \dfrac{1}{|| \vec{v} ||} \vec{v} \).

\[ || \vec{v} || = \sqrt{0^2 + (-3)^2 + 2^2} = \sqrt{13} \] \[ \vec{u} = \dfrac{1}{\sqrt{13}} \langle 0,-3,2 \rangle = \left\langle 0, \frac{-3}{\sqrt{13}}, \frac{2}{\sqrt{13}} \right\rangle \]

Calculate the magnitude of \( \vec{u} \) to verify:

\[ || \vec{u} || = \sqrt{ 0^2 + \left(\frac{-3}{\sqrt{13}}\right)^2 + \left(\frac{2}{\sqrt{13}}\right)^2 } = \sqrt{ \frac{9}{13} + \frac{4}{13} } = \sqrt{ \frac{13}{13} } = 1 \]

Find the components using terminal minus initial coordinates:

\[ \vec{AB} = \langle 0-2, -3-6, 1-7 \rangle = \langle -2, -9, -6 \rangle \] \[ \vec{BC} = \langle 0-0, 3-(-3), 4-1 \rangle = \langle 0, 6, 3 \rangle \] \[ \vec{AC} = \langle 0-2, 3-6, 4-7 \rangle = \langle -2, -3, -3 \rangle \]

Now, verify the sum:

\[ \vec{AB} + \vec{BC} = \langle -2, -9, -6 \rangle + \langle 0, 6, 3 \rangle = \langle -2+0, -9+6, -6+3 \rangle = \langle -2, -3, -3 \rangle \]

This result exactly matches the components of \( \vec{AC} \).

\[ \vec{AB} = \langle 2-(-1), 4-2, 2-1 \rangle = \langle 3, 2, 1 \rangle \] \[ \vec{BC} = \langle 5-2, 6-4, 3-2 \rangle = \langle 3, 2, 1 \rangle \] \[ \vec{AC} = \langle 5-(-1), 6-2, 3-1 \rangle = \langle 6, 4, 2 \rangle \]

\( \vec{AB} \) and \( \vec{BC} \) have equal components and are therefore equivalent.

Notice the relationship with \( \vec{AC} \):

\[ \vec{AC} = \langle 6, 4, 2 \rangle = 2 \langle 3, 2, 1 \rangle = 2\vec{BC} \]

Hence, \( \vec{AC} \) is a scalar multiple of \( \vec{BC} \) (and \( \vec{AB} \)), meaning all three vectors are parallel.

Let \( \vec{v} = \langle x,y,z \rangle \) and rewrite the equation using components:

\[ \langle -4,0,2 \rangle - 2 \langle x,y,z \rangle = 3 \langle x,y,z \rangle - 3 \langle -1,-4,2 \rangle \]

Multiply and simplify each side:

\[ \langle -4-2x, -2y, 2-2z \rangle = \langle 3x - 3(-1), 3y - 3(-4), 3z - 3(2) \rangle \] \[ \langle -4-2x, -2y, 2-2z \rangle = \langle 3x + 3, 3y + 12, 3z - 6 \rangle \]

Set corresponding components equal and solve:

• \( -4 - 2x = 3x + 3 \implies -5x = 7 \implies x = -7/5 \)
• \( -2y = 3y + 12 \implies -5y = 12 \implies y = -12/5 \)
• \( 2 - 2z = 3z - 6 \implies -5z = -8 \implies z = 8/5 \)

Final Vector: \( \vec{v} = \langle -7/5, -12/5, 8/5 \rangle \)

A vector \( \vec{u} \) in the same direction with twice the length is simply the scalar multiple of \( \vec{v} \) by 2:

\[ \vec{u} = 2 \vec{v} = 2 \langle -4,2,2 \rangle = \langle -8, 4, 4 \rangle \]

First, find the unit vector in the opposite direction of \( \vec{v} \), which is \( -\dfrac{1}{||\vec{v}||} \vec{v} \).

\[ ||\vec{v}|| = \sqrt{(-1)^2 + 2^2 + 2^2} = \sqrt{9} = 3 \]

The unit vector in the opposite direction is \( -\frac{1}{3} \langle -1, 2, 2 \rangle \).

To give it a length of 5, multiply by the scalar 5:

\[ \vec{u} = 5 \left( -\frac{1}{3} \langle -1, 2, 2 \rangle \right) = -\frac{5}{3} \langle -1, 2, 2 \rangle = \left\langle \frac{5}{3}, -\frac{10}{3}, -\frac{10}{3} \right\rangle \]

We use the scalar magnitude property: \( ||k \vec{v} || = |k| \cdot || \vec{v} || \).

First, find the magnitude of \( \vec{v} \):

\[ || \vec{v} || = \sqrt{(-1)^2 + 2^2 + 2^2} = 3 \]

Substitute this into the equation:

\[ 3|k| = \frac{1}{5} \implies |k| = \frac{1}{15} \]

This yields two possible solutions for \( k \):

\[ k = \frac{1}{15} \quad \text{and} \quad k = -\frac{1}{15} \]

Vectors are parallel if one is a scalar multiple of the other: \( \vec{v_1} = k \vec{v_2} \).

\[ \langle -4,6,2 \rangle = k \langle 2,b,c \rangle = \langle 2k, kb, kc \rangle \]

Set up equations for each component:

• \( -4 = 2k \implies k = -2 \)
• \( 6 = kb \implies 6 = -2b \implies b = -3 \)
• \( 2 = kc \implies 2 = -2c \implies c = -1 \)

Result: \( b = -3 \) and \( c = -1 \).

For points A, B, and C to be collinear, the vectors \( \vec{AB} \) and \( \vec{AC} \) must be parallel (meaning \( \vec{AC} = k \vec{AB} \)).

\[ \vec{AC} = \langle 0-2, 2-6, 3-7 \rangle = \langle -2, -4, -4 \rangle \] \[ \vec{AB} = \langle 1-2, 4-6, 5-7 \rangle = \langle -1, -2, -2 \rangle \]

Check for a scalar multiplier:

\[ \langle -2, -4, -4 \rangle = 2 \langle -1, -2, -2 \rangle \]

Since \( \vec{AC} = 2 \vec{AB} \), the vectors are parallel and share the point A. Therefore, the points are collinear.

a) First, define the coordinates of the vertices based on a side length of 2 and A at the origin:

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).

• \( \vec{AB} = \langle 2-0, 0-0, 0-0 \rangle = \langle 2,0,0 \rangle \)
• \( \vec{EF} = \langle 2-0, 0-0, 2-2 \rangle = \langle 2,0,0 \rangle \)
• \( \vec{DC} = \langle 2-0, 2-2, 0-0 \rangle = \langle 2,0,0 \rangle \)
• \( \vec{HG} = \langle 2-0, 2-2, 2-2 \rangle = \langle 2,0,0 \rangle \)
• \( \vec{AC} = \langle 2-0, 2-0, 0-0 \rangle = \langle 2,2,0 \rangle \)
• \( \vec{AG} = \langle 2-0, 2-0, 2-0 \rangle = \langle 2,2,2 \rangle \)

b) The vectors \( \vec{AB} \), \( \vec{EF} \), \( \vec{DC} \), and \( \vec{HG} \) have equal components \( \langle 2,0,0 \rangle \) and are therefore equivalent.

c) Calculate the left and right sides algebraically:

Left side: \( \vec{AB} + \vec{BF} + \vec{FG} = \langle 2,0,0 \rangle + \langle 0,0,2 \rangle + \langle 0,2,0 \rangle = \langle 2,2,2 \rangle \)

Right side: \( \vec{AC} + \vec{CG} = \langle 2,2,0 \rangle + \langle 0,0,2 \rangle = \langle 2,2,2 \rangle \)

Hence, \( \vec{AB} + \vec{BF} + \vec{FG} = \vec{AC} + \vec{CG} \).

d) Find the magnitude of \( \vec{AG} \):

\[ || \vec{AG} || = \sqrt{2^2 + 2^2 + 2^2} = \sqrt{12} = 2\sqrt{3} \]

e) The unit vector in the direction of \( \vec{AG} \):

\[ \frac{1}{|| \vec{AG} ||} \vec{AG} = \frac{1}{2\sqrt{3}} \langle 2,2,2 \rangle = \left\langle \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right\rangle \]