Topic: Distance between a point and a line
Question: Find the distance between the point and the line.
Point (1, −1, −1)
Line x= 1 −t y= 2t z=−1
Answer choices:
A 1
5
B 1
25
C 5
D 5
!
!
144
Solution: A
We have to start by converting the parametric equations to a vector
equation. Since we have x= 1 −t, y= 2t, and z=−1, we get
r= (1 −t)i+ (2t)j+ (−1)k
r= (1 −t)i+ 2tj−k
Now we’ll rearrange the vector equation until it matches the format
r=r0+tv.
r=i−ti+ 2tj−k
r= (i−k) + (−ti+ 2tj)
r= (i−k) + t(−i+ 2j)
Matching this to r=r0+tv gives us r0(1,0, −1) and v⟨−1,2,0⟩. We’ll rename
the vector v⟨−1,2,0⟩ to a⟨−1,2,0⟩. We’ll set a aside for a moment and work
on the vector b, which connects the given point (1, −1, −1) to the point on
the line, r0(1,0, −1).
b⟨1−1, −1−0, −1−(−1)⟩
b⟨0, −1,0⟩
Now we’ll find the cross product of a and b.
a×b=
i j k
a1a2a3
b1b2b3
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a×b=ia2a3
b2b3
−ja1a3
b1b3
+ka1a2
b1b2
a×b=(a2b3−a3b2)i−(a1b3−a3b1)j+(a1b2−a2b1)k
a×b=[(2)(0) −(0)(−1)]i−[(−1)(0) −(0)(0)]j+[(−1)(−1) −(2)(0)]k
a×b= (0 −0)i−(0 −0)j+(1−0)k
a×b= 0i−0j+ 1k
a×b=⟨0,0,1⟩
Then we need the magnitude of the cross product of a and b.
a×b= (0)2+ (0)2+ (1)2
a×b= 1
a×b= 1
We also need the magnitude of a⟨−1,2,0⟩.
|a|= (−1)2+ (2)2+ (0)2
|a|= 1 + 4
|a|= 5
Finally, we’ll use the distance formula to find the distance from the point to
the line.
146
d=
a×b
|a|
d=1
5
147
Topic: Distance between a point and a line
Question: Find the distance between the point and the line.
Point (1,1,1)
Line x=2+t y= 1 −2t z= 3t
Answer choices:
A 4
3
B 7
12
C 12
7
D 3
4
!
!
148
Solution: C
We have to start by converting the parametric equations to a vector
equation. Since we have x=2+t, y= 1 −2t, and z= 3t, we get
r= (2 + t)i+ (1 −2t)j+ (3t)k
r= (2 + t)i+ (1 −2t)j+ 3tk
Now we’ll rearrange the vector equation until it matches the format
r=r0+tv.
r= 2i+ti+j−2tj+ 3tk
r= (2i+j) + (ti−2tj+ 3tk)
r= (2i+j) + t(i−2j+ 3k)
Matching this to r=r0+tv gives us r0(2,1,0) and v⟨1, −2,3⟩. We’ll rename the
vector v⟨1, −2,3⟩ to a⟨1, −2,3⟩. We’ll set a aside for a moment and work on
the vector b, which connects the given point (1,1,1) to the point on the line,
r0(2,1,0).
b⟨1−2,1 −1,1 −0⟩
b⟨−1,0,1⟩
Now we’ll find the cross product of a and b.
a×b=
i j k
a1a2a3
b1b2b3
149
a×b=ia2a3
b2b3
−ja1a3
b1b3
+ka1a2
b1b2
a×b=(a2b3−a3b2)i−(a1b3−a3b1)j+(a1b2−a2b1)k
a×b=[(−2)(1) −(3)(0)]i−[(1)(1) −(3)(−1)]j+[(1)(0) −(−2)(−1)]k
a×b= (−2−0)i−(1 + 3)j+(0−2)k
a×b=−2i−4j−2k
a×b=⟨−2, −4, −2⟩
Then we need the magnitude of the cross product of a and b.
a×b= (−2)2+ (−4)2+ (−2)2
a×b= 4 + 16 + 4
a×b= 24
We also need the magnitude of a⟨1, −2,3⟩.
|a|= (1)2+ (−2)2+ (3)2
|a|= 1 + 4 + 9
|a|= 14
Finally, we’ll use the distance formula to find the distance from the point to
the line.
150
d=
a×b
|a|
d=24
14
d=24
14
d=12
7
151
Topic: Distance between a point and a line
Question: Find the distance between the point and the line.
Point (2, −4,5)
Line x=−3+2t y=3+t z= 2 −5t
Answer choices:
A 19
2
B 391
5
C 2
19
D 5
391
!
!
152
Solution: B
We have to start by converting the parametric equations to a vector
equation. Since we have x=−3+2t, y=3+t, and z= 2 −5t, we get
r= (−3 + 2t)i+ (3 + t)j+(2−5t)k
Now we’ll rearrange the vector equation until it matches the format
r=r0+tv.
r=−3i+ 2ti+ 3j+tj+ 2k−5tk
r= (−3i+ 3j+ 2k) + (2ti+tj−5tk)
r= (−3i+ 3j+ 2k)+t(2i+j−5k)
Matching this to r=r0+tv gives us r0(−3,3,2) and v⟨2,1, −5⟩. We’ll rename
the vector v⟨2,1, −5⟩ to a⟨2,1, −5⟩. We’ll set a aside for a moment and work
on the vector b, which connects the given point (2, −4,5) to the point on
the line, r0(−3,3,2).
b⟨2−(−3), −4−3,5 −2⟩
b⟨5, −7,3⟩
Now we’ll find the cross product of a and b.
a×b=
i j k
a1a2a3
b1b2b3
153
a×b=ia2a3
b2b3
−ja1a3
b1b3
+ka1a2
b1b2
a×b=(a2b3−a3b2)i−(a1b3−a3b1)j+(a1b2−a2b1)k
a×b=[(1)(3) −(−5)(−7)]i−[(2)(3) −(−5)(5)]j+[(2)(−7) −(1)(5)]k
a×b= (3 −35)i−(6 + 25)j+ (−14 −5)k
a×b=−32i−31j−19k
a×b=⟨−32, −31, −19⟩
Then we need the magnitude of the cross product of a and b.
a×b= (−32)2+ (−31)2+ (−19)2
a×b= 1,024 + 961 + 361
a×b= 2,346
We also need the magnitude of a⟨2,1, −5⟩.
|a|= (2)2+ (1)2+ (−5)2
|a|= 4 + 1 + 25
|a|= 30
Finally, we’ll use the distance formula to find the distance from the point to
the line.
154
d=
a×b
|a|
d=2,346
30
d=2,346
30
d=391
5
155
