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$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 58.3
Related Topics
Wize University Linear Algebra Textbook > Products of Vectors
Projection (Proj) and Perpendicular (Perp)
6 Activities
Given the vector
v
⃗
=
<
1
,
1
,
2
>
\bcb{\vec{v} = \left< 1, 1, 2 \right>}
v
=
⟨
1
,
1
,
2
⟩
and the points
P
1
=
(
0
,
−
1
,
0
)
\bcb{P_1 = (0,-1,0)}
P
1
=
(
0
,
−
1
,
0
)
and
P
2
=
(
−
1
,
1
,
0
)
\bcb{P_2 = (-1,1,0)}
P
2
=
(
−
1
,
1
,
0
)
, find the point on
the line
L
\bcb{L}
L
(which is parallel to
v
⃗
\bcb{\vec{v}}
v
that passes through
P
1
\bcb{P_1}
P
1
) that is closest to the point
P
2
\bcb{P_2}
P
2
.
<
1
6
,
(
1
6
+
1
)
,
−
2
3
>
\ds\left< \tfrac{1}{\sqrt{6}},\big( \tfrac{1}{\sqrt{6}}+1\big),-\sqrt{\tfrac{2}{3}}\right>
⟨
6
1
,
(
6
1
+
1
)
,
−
3
2
⟩
<
−
1
6
,
(
1
6
+
1
)
,
2
3
>
\ds\left< -\tfrac{1}{\sqrt{6}},\big( \tfrac{1}{\sqrt{6}}+1\big),\sqrt{\tfrac{2}{3}}\right>
⟨
−
6
1
,
(
6
1
+
1
)
,
3
2
⟩
<
1
6
,
−
5
6
,
2
6
>
\displaystyle \left< \frac{1}{{6}}, \;\frac{-5}{{6}},\;\frac{2}{6}\right>
⟨
6
1
,
6
−
5
,
6
2
⟩
<
−
1
6
,
(
1
6
−
1
)
,
−
2
3
>
\ds\left< -\tfrac{1}{\sqrt{6}},\big( \tfrac{1}{\sqrt{6}}-1\big),-\sqrt{\tfrac{2}{3}}\right>
⟨
−
6
1
,
(
6
1
−
1
)
,
−
3
2
⟩
I don't know
Check Submission
More Projection (Proj) and Perpendicular (Perp) Questions:
Example: Dot Product (2)
Find the scalar and vector projection of
b
⃗
=
⟨
2
,
1
,
3
⟩
\vec{b}=\langle 2,1,3\rangle
b
=
⟨
2
,
1
,
3
⟩
onto
a
⃗
=
⟨
−
1
,
4
,
2
⟩
\vec{a}=\langle -1,4,2\rangle
a
=
⟨
−
1
,
4
,
2
⟩
.
Find the vector projection of
[
3
,
2
]
[3,2]
[
3
,
2
]
onto the vector
[
3
,
6
]
\left[3,6\right]
[
3
,
6
]
133 - FML 3 - 18.1W - e.g. 9
Find the vector-component
ı
^
∣
∣
d
⃗
\ihat_{||_{\vd}} \,
^
∣
∣
d
of
ı
^
\bcb{\boldsymbol{ \ihat}}
^
in the direction of the line given by
y
=
m
x
\bcb{\boldsymbol{ y=mx}}
y
=
mx
. Repeat for
ȷ
^
∣
∣
d
⃗
\jhat_{||{\vd}} \,
^
∣∣
d
, the vector-component of
ȷ
^
\bcb{\boldsymbol{ \jhat }}
^
in the direction of the line.
133 - FML 3 - 18.1W e.g. 58.3
Given the vector
v
⃗
=
<
1
,
1
,
2
>
\bcb{\vec{v} = \left< 1, 1, 2 \right>}
v
=
⟨
1
,
1
,
2
⟩
and the points
P
1
=
(
0
,
−
1
,
0
)
\bcb{P_1 = (0,-1,0)}
P
1
=
(
0
,
−
1
,
0
)
and
P
2
=
(
−
1
,
1
,
0
)
\bcb{P_2 = (-1,1,0)}
P
2
=
(
−
1
,
1
,
0
)
Find the point on the line
L
\bcb{L}
L
closest to the point
P
2
\bcb{P_2}
P
2
.
133 - FML 3 - 18.1W e.g. 58.2
Given the vector
v
⃗
=
<
1
,
1
,
2
>
\bcb{\vec{v} = \left< 1, 1, 2 \right>}
v
=
⟨
1
,
1
,
2
⟩
and the points
P
1
=
(
0
,
−
1
,
0
)
\bcb{P_1 = (0,-1,0)}
P
1
=
(
0
,
−
1
,
0
)
and
P
2
=
(
−
1
,
1
,
0
)
\bcb{P_2 = (-1,1,0)}
P
2
=
(
−
1
,
1
,
0
)
Find the
minimum
distance of
P
2
\bcb{P_2}
P
2
to the line
L
\bcb{L}
L
found above.
133 - FML 3 - 18.1W e.g. 48
Find a unit vector in the direction of the projection of
u
⃗
=
<
102
,
112
,
−
56
>
\bcb{\vec{u} = \left< 102, 112, -56 \right>}
u
=
⟨
102
,
112
,
−
56
⟩
onto
v
⃗
=
<
0
,
3
,
4
>
\bcb{\vec{v} = \left< 0, 3, 4 \right>}
v
=
⟨
0
,
3
,
4
⟩
.
133 - FML 3 - 18.1W e.g. 43
Given
u
⃗
=
<
8
,
2
,
2
>
\bcb{\vec{u} = \left< 8, 2, 2 \right>}
u
=
⟨
8
,
2
,
2
⟩
and
v
⃗
=
<
2
,
−
1
,
−
1
>
\bcb{\vec{v} = \left< 2, -1, -1 \right>}
v
=
⟨
2
,
−
1
,
−
1
⟩
, find the vector projection of
u
⃗
\bcb{\vec{u}}
u
onto
v
⃗
\bcb{\vec{v}}
v
.
$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 41
Let
P
=
(
−
3
,
−
3
,
−
4
)
\bcb{P = (-3, -3, -4)}
P
=
(
−
3
,
−
3
,
−
4
)
,
Q
=
(
−
5
,
−
6
,
−
3
)
\bcb{Q = (-5,-6,-3)}
Q
=
(
−
5
,
−
6
,
−
3
)
,
R
=
(
−
6
,
−
4
,
−
6
)
\bcb{R = (-6,-4,-6)}
R
=
(
−
6
,
−
4
,
−
6
)
and
S
=
(
−
2
,
−
3
,
−
4
)
\bcb{S = (-2, -3, -4)}
S
=
(
−
2
,
−
3
,
−
4
)
. Let
l
1
\bcb{l_1}
l
1
be the line passing through
P
\bcb{P}
P
and
Q
\bcb{Q}
Q
, and let
l
2
\bcb{l_2}
l
2
be the line passing through
R
\bcb{R}
R
and
S
\bcb{S}
S
. Find the distance between
R
\bcb{R}
R
and
l
1
\bcb{l_1}
l
1
.
133 - FML 3 - 18.1W e.g. 36
For two lines having directional vectors
v
1
\mathbf{v}_1
v
1
and
v
2
\mathbf{v}_2
v
2
, respectively, the shortest distance between them can be found by creating a vector that connects the two lines (by subtracting a point on one line from a point on the other),
i.e.
(
r
2
−
r
1
)
\left(\mathbf{r}_2 - \mathbf{r}_1\right)
(
r
2
−
r
1
)
, and projecting it onto the unit normal vector connecting the two lines. The unit normal is easily found by crossing their respective directionals and dividing by the magnitude of the resulting vector, as illustrated in Fig. 5.
i.e.
s
=
∣
proj
n
^
(
r
2
−
r
1
)
∣
=
∣
(
r
2
−
r
1
)
⋅
v
1
×
v
2
∣
v
1
×
v
2
∣
∣
s = \left| \text{proj}_{\mathbf{\hat{n}}} \left(\mathbf{r}_2 - \mathbf{r}_1\right)\right| = \left| \left(\mathbf{r}_2 - \mathbf{r}_1\right) \cdot \frac{\mathbf{v}_1 \times \mathbf{v}_2}{\left|\mathbf{v}_1 \times \mathbf{v}_2\right|}\right|
s
=
∣
proj
n
^
(
r
2
−
r
1
)
∣
=
(
r
2
−
r
1
)
⋅
∣
v
1
×
v
2
∣
v
1
×
v
2
Find the equations of the line passing through the point
(
2
,
3
,
−
4
)
\bcb{(2, 3, -4)}
(
2
,
3
,
−
4
)
parallel to the vector
i
−
4
k
\bcb{\mathbf{i} - 4\mathbf{k}}
i
−
4k
; express it in vector-parametric, scalar-parametric and standard form.
$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 59.3
Given the points
P
1
=
(
0
,
0
,
1
)
\bcb{P_1 = (0,0,1)}
P
1
=
(
0
,
0
,
1
)
and
P
2
=
(
1
,
−
2
,
3
)
\bcb{P_2 = (1,-2,3)}
P
2
=
(
1
,
−
2
,
3
)
and the vector
v
⃗
=
<
2
,
−
3
,
1
>
\bcb{\vec{v} = \left< 2, -3, 1 \right>}
v
=
⟨
2
,
−
3
,
1
⟩
, find the point
P
o
\bcb{P_o}
P
o
on the plane
Π
:
2
x
−
3
y
+
z
=
1
\bcb{\Pi: \ 2x - 3y + z = 1}
Π
:
2
x
−
3
y
+
z
=
1
that is closest to the point
P
2
\bcb{P_2}
P
2
.
$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 48
Find a unit vector in the direction of the projection of
u
⃗
=
<
102
,
112
,
−
56
>
\bcb{\vec{u} = \left< 102, 112, -56 \right>}
u
=
⟨
102
,
112
,
−
56
⟩
onto
v
⃗
=
<
0
,
3
,
4
>
\bcb{\vec{v} = \left< 0, 3, 4 \right>}
v
=
⟨
0
,
3
,
4
⟩
.
$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 43
Given
u
⃗
=
<
8
,
2
,
2
>
\bcb{\vec{u} = \left< 8, 2, 2 \right>}
u
=
⟨
8
,
2
,
2
⟩
and
v
⃗
=
<
2
,
−
1
,
−
1
>
\bcb{\vec{v} = \left< 2, -1, -1 \right>}
v
=
⟨
2
,
−
1
,
−
1
⟩
, find the vector projection of
u
⃗
\bcb{\vec{u}}
u
onto
v
⃗
\bcb{\vec{v}}
v
.
$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 58.2
Given the vector
v
⃗
=
<
1
,
1
,
2
>
\bcb{\vec{v} = \left< 1, 1, 2 \right>}
v
=
⟨
1
,
1
,
2
⟩
and the points
P
1
=
(
0
,
−
1
,
0
)
\bcb{P_1 = (0,-1,0)}
P
1
=
(
0
,
−
1
,
0
)
and
P
2
=
(
−
1
,
1
,
0
)
\bcb{P_2 = (-1,1,0)}
P
2
=
(
−
1
,
1
,
0
)
, find the
minimum
distance of
P
2
\bcb{P_2}
P
2
to the line
L
\bcb{L}
L
that is parallel to
v
⃗
\bcb{\vec{v}}
v
that passes through
P
1
\bcb{P_1}
P
1
.
$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 59.2
Given the points
P
1
=
(
0
,
0
,
1
)
\bcb{P_1 = (0,0,1)}
P
1
=
(
0
,
0
,
1
)
and
P
2
=
(
1
,
−
2
,
3
)
\bcb{P_2 = (1,-2,3)}
P
2
=
(
1
,
−
2
,
3
)
and the vector
v
⃗
=
<
2
,
−
3
,
1
>
\bcb{\vec{v} = \left< 2, -3, 1 \right>}
v
=
⟨
2
,
−
3
,
1
⟩
Find the
minimum
distance between the point
P
2
\bcb{P_2}
P
2
and the plane
Π
:
2
x
−
3
y
+
z
=
1
\bcb{\Pi:\ 2x-3y + z = 1}
Π
:
2
x
−
3
y
+
z
=
1
.