High School
SAT
ACT
University
MCAT
LSAT
DAT

Let L_1 be a line through the point (1,3,1) with direction vector 112^Tand let …

Related Topics

Wize University Linear Algebra Textbook > Equations of Lines and Planes

Distances

5 Activities

Let L1L_1 be a line through the point (1,3,1)(1,3,1) with direction vector
[112]T\begin{bmatrix}1&1&2\end{bmatrix}^Tand let L2L_2 be a line through the point (2,1,2)(2,1,2) with direction vector [113]T\begin{bmatrix}1&1&3\end{bmatrix}^T . Find the shortest distance from L1L_1 to L2L_2.
More Distances Questions:
Find the distance between following planes:
P1:2x+y+z=5,   P2:−6x−3y−3z=25P1:2x+y+z=5,\ \ \ P2:-6x-3y-3z=25
Find a point on the plane defined by 3x−y+z=8,3x-y+z=8, which is closest to (2,0,5)
Prove formula for distance between point and plane
Let a,b,c,d,p,q,r∈Ra,b,c,d,p,q,r\in\mathbb{R} be constants. Prove that the formula for the distance between the point Q⃗=(p,q,r)\vec Q=(p,q,r) and the plane ax+by+cz=dax+by+cz=d is ∣d−ap−bq−cr∣a2+b2+c2\frac{|d-ap-bq-cr|}{\sqrt{a^2+b^2+c^2}} .
Find the closest distance of point (2,2,2) to the line L by deriving and solving system of linear equations.
L:{𝑥1+x2+𝑥3=2𝑥1−𝑥2−2x3=5L:\begin{cases} 𝑥_1+ x_2 + 𝑥_3 = 2\\ 𝑥_1 − 𝑥_2 − 2x_3 = 5 \end{cases}
133 - FML 3 - 18.1W e.g. 41
Let P=(−3,−3,−4)\bcb{P = (-3, -3, -4)}, Q=(−5,−6,−3)\bcb{Q = (-5,-6,-3)}, R=(−6,−4,−6)\bcb{R = (-6,-4,-6)} and S=(−2,−3,−4)\bcb{S = (-2, -3, -4)}. Let l1\bcb{l_1} be the line passing through P\bcb{P} and Q\bcb{Q}, and let l2\bcb{l_2} be the line passing through R\bcb{R} and S\bcb{S}. Find the distance between R\bcb{R} and l1\bcb{l_1}. Find the distance between l1\bcb{l_1} and l2\bcb{l_2}.
$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 60.8
Given the planes
π1 : 4x−3y+z=2\bcb{\pi_1 \, : \, 4x - 3y + z = 2} and π2 : x+2y+2z=0\bcb{\pi_2 \, : \, x + 2y + 2z = 0}, and the lines
L1 : {x=3t+2y=−2t+1z=t, L2 : {x=−2t+1y=t+1z=t+3, L3 : {x=t+3y=−tz=2t+1\bcb{L_1 \, : \, \left\{ \begin{matrix} x & = & 3t + 2 \\ y & = & -2t + 1 \\ z & = & t \end{matrix} \right., ~ L_2 \, : \, \left\{ \begin{matrix} x & = & -2t + 1 \\ y & = & t + 1 \\ z & = & t + 3 \end{matrix} \right., ~ L_3 \, : \, \left\{ \begin{matrix} x & = & t + 3 \\ y & = & -t \\ z & = & 2t +1 \end{matrix} \right.},
$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 58.2
Given the vector v⃗=<1,1,2>\bcb{\vec{v} = \left< 1, 1, 2 \right>} and the points P1=(0,−1,0)\bcb{P_1 = (0,-1,0)} and P2=(−1,1,0)\bcb{P_2 = (-1,1,0)}, find the minimum distance of P2\bcb{P_2} to the line L\bcb{L} that is parallel to v⃗\bcb{\vec{v}} that passes through P1\bcb{P_1}.
Shortest Distance Between Two Lines, Parallel Edition
Find the shortest distance between the following two lines in R3\mathbb{R}^3:
L1={x=1+2ty=3−tz=2+3t,L2={1−4t2+2t−6tL_1 = \begin{cases} x = 1+2t \\ y = 3-t \\ z = 2+3t \end{cases}, L_2 = \begin{cases} 1-4t \\ 2+2t \\ -6t \end{cases}
Shortest Distance Between Point and Plane
Find the shortest distance between the point (1,2,3)(1,2,3) and the plane 2x−3y+z=42x-3y+z=4.
Shortest Distance Between Point and Line
Find the shortest distance between the point (1,2,3)(1,2,3) and the line L⃗(t)=<2t+3,t+4,−t−2>\vec{L}(t)=\left<2t+3,t+4,-t-2\right>.
Shortest Distance Between Two Planes
Find the shortest distance between the following two planes in R3\mathbb{R}^3:
P1:2x−3y+z=1P2:−6x+9y−3z=5P_1: 2x - 3y + z = 1\\ P_2: -6x + 9y - 3z = 5
Shortest Distance Between Two Lines, Parallel Edition
Find the shortest distance between the following two lines in R3\mathbb{R}^3:
L1={x=1+2ty=3−tz=2+3t,L2={1−4t2+2t−6tL_1 = \begin{cases} x = 1+2t \\ y = 3-t \\ z = 2+3t \end{cases}, L_2 = \begin{cases} 1-4t \\ 2+2t \\ -6t \end{cases}
Shortest Distance Between Two Lines
Find the shortest distance between the following two lines in R3\mathbb{R}^3:
L1={x=1+2ty=3−tz=2+3t,L2={−t2+t−3tL_1 = \begin{cases} x = 1+2t \\ y = 3-t \\ z = 2+3t \end{cases}, L_2 = \begin{cases} -t \\ 2+t \\ -3t \end{cases}
Shortest Distance Between Plane and Line
Find the shortest distance between the following plane and line in R3\mathbb{R}^3:
P:2x−3y+z=1,L:{x=2t+3y=t+4z=−t−2P: 2x-3y+z=1, L: \begin{cases} x = 2t+3 \\ y = t + 4 \\ z = -t - 2 \end{cases}
Shortest Distance
Find the shortest distance from the point (3, −1, 2)\left(3,\ -1,\ 2\right) to the plane 2x−y+z=102x-y+z=10.