New year, and I am now in year 12, learning more vectors… In tutoring today, I did some 3D vector problems containing parametric equations. I’m going to post my solutions here.
Q. Use the shortest distance formula to find the shortest distance from the point P ( − 1 , 2 , 3 ) P(-1, 2, 3) P ( − 1 , 2 , 3 ) x = 1 + 2 t x = 1 + 2t x = 1 + 2 t y = − 4 + 3 t y = -4 + 3t y = − 4 + 3 t z = 3 + t z = 3 + t z = 3 + t
∵ [ 1 + 2 t − 4 + 3 t 3 + t ] = [ a 1 a 2 a 3 ] + t ⋅ [ d 1 d 2 d 3 ] \because \begin{bmatrix}1 + 2t \\ -4 + 3t \\ 3 + t\end{bmatrix} = \begin{bmatrix}a_1 \\ a_2 \\ a_3\end{bmatrix} + t \cdot \begin{bmatrix}d_1 \\ d_2 \\ d_3\end{bmatrix}
∵ ⎣ ⎢ ⎡ 1 + 2 t − 4 + 3 t 3 + t ⎦ ⎥ ⎤ = ⎣ ⎢ ⎡ a 1 a 2 a 3 ⎦ ⎥ ⎤ + t ⋅ ⎣ ⎢ ⎡ d 1 d 2 d 3 ⎦ ⎥ ⎤
Match the coefficients
∴ r ( t ) = [ 1 − 4 3 ] + t ⋅ [ 2 3 1 ] \therefore \mathbf{r}(t) = \begin{bmatrix}1 \\ -4 \\ 3\end{bmatrix} + t \cdot \begin{bmatrix}2 \\ 3 \\ 1\end{bmatrix}
∴ r ( t ) = ⎣ ⎢ ⎡ 1 − 4 3 ⎦ ⎥ ⎤ + t ⋅ ⎣ ⎢ ⎡ 2 3 1 ⎦ ⎥ ⎤
∣ A P ⃗ × v ^ ∣ |\vec{AP} \times \hat{\mathbf{v}}|
∣ A P × v ^ ∣
A P ⃗ \vec{AP} A P O A ⃗ = [ 1 − 4 3 ] O P ⃗ = [ − 1 2 3 ] \vec{OA} = \begin{bmatrix}1 \\ -4 \\ 3\end{bmatrix} \ \ \vec{OP} = \begin{bmatrix}-1 \\ 2 \\ 3\end{bmatrix}
O A = ⎣ ⎢ ⎡ 1 − 4 3 ⎦ ⎥ ⎤ O P = ⎣ ⎢ ⎡ − 1 2 3 ⎦ ⎥ ⎤
A P ⃗ = O P ⃗ − O A ⃗ = [ − 1 2 3 ] − [ 1 − 4 3 ] = [ − 2 6 0 ] \begin{aligned}
\vec{AP} &= \vec{OP} - \vec{OA} \\
&=\begin{bmatrix}-1 \\ 2 \\ 3\end{bmatrix} - \begin{bmatrix}1 \\ -4 \\ 3\end{bmatrix} \\
&=\begin{bmatrix}-2 \\ 6 \\ 0\end{bmatrix}
\end{aligned}
A P = O P − O A = ⎣ ⎢ ⎡ − 1 2 3 ⎦ ⎥ ⎤ − ⎣ ⎢ ⎡ 1 − 4 3 ⎦ ⎥ ⎤ = ⎣ ⎢ ⎡ − 2 6 0 ⎦ ⎥ ⎤
∣ A P ⃗ × v ^ ∣ = ∣ ∣ i ^ j ^ k ^ − 2 6 0 2 3 1 ∣ ∣ = ∣ [ 6 2 − 18 ] ∣ = 2 91 \begin{aligned}
|\vec{AP} \times \hat{\mathbf{v}}| &= \left|\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ -2 & 6 & 0 \\ 2 & 3 & 1\end{vmatrix}\right| \\
&= \left|\begin{bmatrix}6 \\ 2 \\ -18\end{bmatrix}\right| \\
&= 2\sqrt{91} \\
\end{aligned}
∣ A P × v ^ ∣ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ i ^ − 2 2 j ^ 6 3 k ^ 0 1 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ⎣ ⎢ ⎡ 6 2 − 1 8 ⎦ ⎥ ⎤ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = 2 9 1
∣ v ∣ = 14 |\mathbf{v}| = \sqrt{14}
∣ v ∣ = 1 4
∴ ∣ A P ⃗ × v ^ ∣ = 2 91 14 = 26 \begin{aligned}
\therefore |\vec{AP} \times \hat{\mathbf{v}}| &= \frac{2\sqrt{91}}{\sqrt{14}} \\
&= \sqrt{26} \\
\end{aligned}
∴ ∣ A P × v ^ ∣ = 1 4 2 9 1 = 2 6
Q.
Line 1 has equation x = − 1 + 2 s x = -1 + 2s x = − 1 + 2 s y = 1 − 2 s y = 1 - 2s y = 1 − 2 s z = 1 + 4 s z = 1 + 4s z = 1 + 4 s
Line 2 has equation x = 1 − t x = 1 - t x = 1 − t y = t y = t y = t z = 3 − 2 t z = 3 - 2t z = 3 − 2 t
Line 3 has equation x − 1 2 = − y − 1 = z − 4 3 \frac{x - 1}{2} = -y - 1 = \frac{z - 4}{3} 2 x − 1 = − y − 1 = 3 z − 4
L 1 ( s ) = [ − 1 1 1 ] + s ⋅ [ 2 − 2 4 ] L 2 ( t ) = [ 1 0 3 ] + t ⋅ [ − 1 1 − 2 ] \begin{aligned}
L_1(s) &= \begin{bmatrix}-1 \\ 1 \\ 1\end{bmatrix} + s \cdot \begin{bmatrix}2 \\ -2 \\ 4\end{bmatrix} \\
L_2(t) &= \begin{bmatrix}1 \\ 0 \\ 3\end{bmatrix} + t \cdot \begin{bmatrix}-1 \\ 1 \\ -2\end{bmatrix}
\end{aligned}
L 1 ( s ) L 2 ( t ) = ⎣ ⎢ ⎡ − 1 1 1 ⎦ ⎥ ⎤ + s ⋅ ⎣ ⎢ ⎡ 2 − 2 4 ⎦ ⎥ ⎤ = ⎣ ⎢ ⎡ 1 0 3 ⎦ ⎥ ⎤ + t ⋅ ⎣ ⎢ ⎡ − 1 1 − 2 ⎦ ⎥ ⎤
For Line 3, define a new variable k, and set: \text{For Line 3, define a new variable k, and set:} \\
For Line 3, define a new variable k, and set:
x − 1 2 = k − y − 1 = k z − 4 3 = k \begin{aligned}
\frac{x - 1}{2} &= k \\
-y - 1 &= k \\
\frac{z - 4}{3} &= k
\end{aligned}
2 x − 1 − y − 1 3 z − 4 = k = k = k
from that, we get: \text{from that, we get:}
from that, we get:
x = 2 k + 1 y = − k − 1 z = 3 k + 4 \begin{aligned}
x &= 2k + 1 \\
y &= -k - 1 \\
z &= 3k + 4
\end{aligned}
x y z = 2 k + 1 = − k − 1 = 3 k + 4
∴ L 3 ( k ) = [ 1 − 1 4 ] + k ⋅ [ 2 − 1 3 ] \therefore L_3(k) = \begin{bmatrix}1 \\ -1 \\ 4\end{bmatrix} + k \cdot \begin{bmatrix}2 \\ -1 \\ 3\end{bmatrix}
∴ L 3 ( k ) = ⎣ ⎢ ⎡ 1 − 1 4 ⎦ ⎥ ⎤ + k ⋅ ⎣ ⎢ ⎡ 2 − 1 3 ⎦ ⎥ ⎤
a) Show that line 1 and line 2 are parallel.
direction of L 1 ( t ) = [ 2 − 2 4 ] \text{direction of} \ L_1(t) = \begin{bmatrix}2 \\ -2 \\ 4\end{bmatrix}
direction of L 1 ( t ) = ⎣ ⎢ ⎡ 2 − 2 4 ⎦ ⎥ ⎤
direction of L 2 ( s ) = [ − 1 1 − 2 ] \text{direction of} \ L_2(s) = \begin{bmatrix}-1 \\ 1 \\ -2\end{bmatrix}
direction of L 2 ( s ) = ⎣ ⎢ ⎡ − 1 1 − 2 ⎦ ⎥ ⎤
∵ [ 2 − 2 4 ] = − 2 ⋅ [ − 1 1 − 2 ] \because \begin{bmatrix}2 \\ -2 \\ 4\end{bmatrix} = -2 \cdot \begin{bmatrix}-1 \\ 1 \\ -2\end{bmatrix}
∵ ⎣ ⎢ ⎡ 2 − 2 4 ⎦ ⎥ ⎤ = − 2 ⋅ ⎣ ⎢ ⎡ − 1 1 − 2 ⎦ ⎥ ⎤
∴ L 1 ( t ) ∥ L 2 ( s ) \therefore L_1(t) \parallel L_2(s)
∴ L 1 ( t ) ∥ L 2 ( s )
b) Show that line 2 and line 3 intercept and find the angle between them.
L 2 ( s ) = L 3 ( k ) L_2(s) = L_3(k)
L 2 ( s ) = L 3 ( k )
{ 1 − t = 1 + 2 k ( 1 ) t = − 1 − k ( 2 ) 3 − 2 t = 4 + 3 k ( 3 ) \begin{cases}
\begin{aligned}
1 - t = 1 + 2k \quad &(1)\\
t = -1 - k \quad &(2)\\
3 - 2t = 4 + 3k \quad &(3)
\end{aligned}
\end{cases}
⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 1 − t = 1 + 2 k t = − 1 − k 3 − 2 t = 4 + 3 k ( 1 ) ( 2 ) ( 3 )
S u b ( 2 ) i n t o ( 1 ) Sub \ (2) \ into \ (1)
S u b ( 2 ) i n t o ( 1 )
1 − ( − 1 − k ) = 1 + 2 k 2 + k = 1 + 2 k k = 1 \begin{aligned}
1 - (-1 - k) &= 1 + 2k \\
2 + k &= 1 + 2k \\
k &= 1
\end{aligned}
1 − ( − 1 − k ) 2 + k k = 1 + 2 k = 1 + 2 k = 1
S u b k = 1 i n t o ( 2 ) Sub \ k = 1\ into \ (2)
S u b k = 1 i n t o ( 2 )
t = − 1 − 1 t = − 2 \begin{aligned}
t &= -1 - 1 \\
t &= -2
\end{aligned}
t t = − 1 − 1 = − 2
T e s t w i t h ( 3 ) Test \ with \ (3)
T e s t w i t h ( 3 )
3 − 2 ⋅ − 2 = 7 4 + 3 ⋅ 1 = 7 \begin{aligned}
3 - 2 \cdot -2 &= 7 \\
4 + 3 \cdot 1 &= 7
\end{aligned}
3 − 2 ⋅ − 2 4 + 3 ⋅ 1 = 7 = 7
7 = 7 7 = 7
7 = 7
The solutions are consistent, hence, they are intersecting at t = − 2 and k = 1 \text{The solutions are consistent, hence, they are intersecting at} \ t = -2 \ \text{and} \ k = 1
The solutions are consistent, hence, they are intersecting at t = − 2 and k = 1
The angle between two vectors can be found by: \text{The angle between two vectors can be found by:}
The angle between two vectors can be found by:
θ = a r c c o s ( a ⋅ b ∣ a ∣ ⋅ ∣ b ∣ ) \theta = arccos(\frac{\mathbf{a}\cdot\mathbf{b}}{\lvert\mathbf{a}\rvert\cdot\lvert\mathbf{b}\rvert})
θ = a r c c o s ( ∣ a ∣ ⋅ ∣ b ∣ a ⋅ b )
direction of L 2 ( s ) = [ − 1 1 − 2 ] direction of L 3 ( k ) = [ 2 − 1 3 ] \begin{aligned}
\text{direction of} \ L_2(s) &= \begin{bmatrix}-1 \\ 1 \\ -2\end{bmatrix} \\
\text{direction of} \ L_3(k) &= \begin{bmatrix}2 \\ -1 \\ 3\end{bmatrix}
\end{aligned}
direction of L 2 ( s ) direction of L 3 ( k ) = ⎣ ⎢ ⎡ − 1 1 − 2 ⎦ ⎥ ⎤ = ⎣ ⎢ ⎡ 2 − 1 3 ⎦ ⎥ ⎤
θ = a r c c o s ( [ − 1 1 − 2 ] ⋅ [ 2 − 1 3 ] ∣ [ − 1 1 − 2 ] ∣ ⋅ ∣ [ 2 − 1 3 ] ∣ ) = a r c c o s ( − 9 6 ⋅ 14 ) = a r c c o s ( − 9 84 ) = a r c c o s ( 7 14 ) + π 2 (from CAS) \begin{aligned}
\theta &= arccos(\frac{\begin{bmatrix}-1 & 1 & -2\end{bmatrix}\cdot\begin{bmatrix}2 & -1 & 3\end{bmatrix}}{\left|\begin{bmatrix}-1 & 1 & -2\end{bmatrix}\right|\cdot\left|\begin{bmatrix}2 & -1 & 3\end{bmatrix}\right|}) \\
&= arccos(\frac{-9}{\sqrt{6}\cdot\sqrt{14}}) \\
&=arccos(\frac{-9}{\sqrt{84}}) \\
&=arccos(\frac{\sqrt{7}}{14}) + \frac{\pi}{2} \text{ (from CAS)}
\end{aligned}
θ = a r c c o s ( ∣ ∣ ∣ [ − 1 1 − 2 ] ∣ ∣ ∣ ⋅ ∣ ∣ ∣ [ 2 − 1 3 ] ∣ ∣ ∣ [ − 1 1 − 2 ] ⋅ [ 2 − 1 3 ] ) = a r c c o s ( 6 ⋅ 1 4 − 9 ) = a r c c o s ( 8 4 − 9 ) = a r c c o s ( 1 4 7 ) + 2 π (from CAS)
