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Find the value(s) of k(if any) such that the vectors u=(1,k,1,0,-1)and v=(k,2,0…
Related Topics
Wize University Linear Algebra Textbook > Products of Vectors
Dot Product
3 Activities
Find the value(s) of
k
k
k
(if any) such that the vectors
u
→
=
(
1
,
k
,
1
,
0
,
−
1
)
\overrightarrow{u}=\left(1,k,1,0,-1\right)
u
=
(
1
,
k
,
1
,
0
,
−
1
)
and
v
→
=
(
k
,
2
,
0
,
3
,
−
4
)
\overrightarrow{v}=\left(k,2,0,3,-4\right)
v
=
(
k
,
2
,
0
,
3
,
−
4
)
are orthogonal.
k
=
−
1
k=-1
k
=
−
1
k
=
−
4
3
k=-\frac{4}{3}
k
=
−
3
4
k
=
±
2
k=\pm\sqrt{2}
k
=
±
2
k
=
0
k=0
k
=
0
No values of
k
k
k
I don't know
Check Submission
More Dot Product Questions:
133 - FML 3 - 18.1W e.g. 20
For two vectors
v
⃗
\bcb{\vv}
v
and
w
⃗
\bcb{\vw}
w
∈
I
R
n
\bcb{\in}\; \R{n}
∈
I
R
n
,
v
⃗
⋅
w
⃗
:
=
‾
\bcb{\vv \cdot \vw :=\underline{\qquad\qquad\qquad}}
v
⋅
w
:=
.
$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 40_$\tkcth{Mock F}\tkct{?}$_$\tkct{no soln / vid}$
If
u
⃗
=
<
1
,
−
2
,
3
>
\bcb{\vec{u} = \left< 1, -2, 3 \right>}
u
=
⟨
1
,
−
2
,
3
⟩
and
v
⃗
=
<
2
,
−
1
,
3
>
\bcb{\vec{v} = \left< 2,\, -1,\, 3 \right>}
v
=
⟨
2
,
−
1
,
3
⟩
, find
(
2
u
⃗
)
⋅
(
3
v
⃗
)
\bcb{(2\vec{u})\cdot (3\vec{v})}
(
2
u
)
⋅
(
3
v
)
.
Example: Orthogonal Vectors
Example:
Find all values of
k
k
k
such that the vectors
v
⃗
=
(
k
,
1
)
\vec{v}=\left(k,\ 1\right)
v
=
(
k
,
1
)
and
u
⃗
=
(
k
,
−
1
)
\vec{u}=\left(k,-1\right)
u
=
(
k
,
−
1
)
are orthogonal.
Practice Question: Orthogonal Vectors
Practice Question: Orthogonal Vectors
Find the value(s) of
c
c
c
such that
u
⃗
=
(
−
1
,
c
,
2
)
\vec{u}=(-1,c,2)\
u
=
(
−
1
,
c
,
2
)
and
v
⃗
=
(
−
7
,
c
,
−
4
)
\vec{v}=\left(-7,\ c,\ -4\right)
v
=
(
−
7
,
c
,
−
4
)
are orthogonal.
Practice Question: Orthogonal Vectors
Practice Question: Orthogonal Vectors
Find the value(s) of
c
c
c
such that
u
⃗
=
(
−
1
,
c
,
2
)
\vec{u}=(-1,c,2)\
u
=
(
−
1
,
c
,
2
)
and
v
⃗
=
(
−
7
,
c
,
−
4
)
\vec{v}=\left(-7,\ c,\ -4\right)
v
=
(
−
7
,
c
,
−
4
)
are orthogonal.
$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 52
If
∣
u
⃗
∣
=
2
\bcb{|{ \vec{u}| } = 2}
∣
u
∣
=
2
and
∣
v
⃗
∣
=
3
\bcb{|{ \vec{v} |} = 3}
∣
v
∣
=
3
and
u
⃗
\bcb{\vec{u}}
u
has the same direction as
v
⃗
\bcb{\vec{v}}
v
then
(
3
u
⃗
)
⋅
(
2
v
⃗
)
\bcb{(3\vec{u}) \cdot (2\vec{v})}
(
3
u
)
⋅
(
2
v
)
is:
$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | $\tkco{duplicate to mock ✓}$ 133 - FML 3 - 18.1W e.g. 46_$\tkcth{Mock F}\tkct{?}$_$\tkct{vid soln only}$
If
∣
u
⃗
∣
=
2
\bcb{|{\vec{u}|} = 2}
∣
u
∣
=
2
and
∣
v
⃗
∣
=
3
\bcb{|{ \vec{v} |} = 3}
∣
v
∣
=
3
, where
u
⃗
⋅
v
⃗
=
2
\bcb{\vec{u} \cdot \vec{v} = 2}
u
⋅
v
=
2
, find
∣
u
⃗
−
v
⃗
∣
\bcb{|{\vec{u} - \vec{v} }|}
∣
u
−
v
∣
.
$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 54
Find the value of
k
\bcb{k}
k
such that the vector
u
⃗
=
<
1
,
k
,
3
>
\bcb{\vec{u} = \left< 1, k, 3 \right>}
u
=
⟨
1
,
k
,
3
⟩
is perpendicular to the vector
v
⃗
=
<
2
,
−
1
,
k
>
\bcb{\vec{v} = \left< 2, -1, k\right>}
v
=
⟨
2
,
−
1
,
k
⟩
.
If
𝑢
⃗
=
(
2
,
0
,
−
3
)
,
𝑣
⃗
=
(
−
2
,
2
,
−
4
)
,
𝑢⃗ = (2, 0, −3), \vec𝑣 = (−2, 2, −4),
u
⃗
=
(
2
,
0
,
−
3
)
,
v
=
(
−
2
,
2
,
−
4
)
,
and
𝑤
⃗
=
(
1
,
2
,
1
)
𝑤⃗= (1,2,1)
w
⃗
=
(
1
,
2
,
1
)
, compute
u
⃗
∙
v
⃗
\vec{u}\ \bullet\ \vec{v}
u
∙
v
.
Practice Question: Vector Product
Practice Question: Vector Product
Given that
𝑢
⃗
=
(
1
,
1
,
−
1
)
𝑢⃗=(1,1,−1)
u
⃗
=
(
1
,
1
,
−
1
)
and
𝑣
⃗
=
(
0
,
−
2
,
2
)
,
𝑣⃗=(0,−2,2),
v
⃗
=
(
0
,
−
2
,
2
)
,
compute
(
𝑢
⃗
⋅
𝑣
⃗
)
(
𝑣
⃗
×
𝑢
⃗
)
.
(𝑢⃗\cdot𝑣⃗)(𝑣⃗\times𝑢⃗).
(
u
⃗
⋅
v
⃗
)
(
v
⃗
×
u
⃗
)
.
$\tkcth{Moved }\bcth{\to} \key{ Mid} \tkco{ S } \tkcth{ch 3 quiz ✓}$ | Basic dot product.
The dot product
v
⃗
⋅
w
⃗
\vv \cdot \vw
v
⋅
w
, for
v
⃗
=
<
−
1
,
1
,
−
1
>
\vv = \rowvecth{-1}{1}{-1}
v
=
⟨
−
1
,
1
,
−
1
⟩
and
w
⃗
=
<
−
2
,
−
2
,
3
>
\vw = \rowvecth{-2}{-2}{3}
w
=
⟨
−
2
,
−
2
,
3
⟩
, is equal to:
$\tkcth{Moved }\bcth{\to} \key{ Mid} \tkco{ S } \tkcth{ch 3 quiz ✓}$ |
A unit vector perpendicular to the vector
w
⃗
=
<
2
,
5
>
\vw = \rowvec{2}{5}
w
=
⟨
2
,
5
⟩
is:
$\tkcth{Moved }\bcth{\to} \key{ Mid} \tkco{ S } \tkcth{ch 3 quiz ✓}$ |
A vector perpendicular to the vector connecting the points
P
1
(
2
,
2
,
2
)
P_1(2,2,2)
P
1
(
2
,
2
,
2
)
and
P
2
(
1
,
−
1
,
1
)
P_2(1,-1,1)
P
2
(
1
,
−
1
,
1
)
is:
$\tkcth{Moved }\bcth{\to} \key{ Mid} \tkco{ S } \tkcth{ch 3 quiz ✓}$ |
A vector perpendicular to the vector connecting the points
P
1
(
0
,
1
,
−
2
)
P_1(0,1,-2)
P
1
(
0
,
1
,
−
2
)
and
P
2
(
−
2
,
0
,
1
)
P_2(-2,0,1)
P
2
(
−
2
,
0
,
1
)
is:
Dot and Cross Products: Vector Norm
If
u
⃗
,
v
⃗
,
w
⃗
∈
R
3
\vec{u},\ \vec{v},\ \vec{w}\ \ \in\ R^3
u
,
v
,
w
∈
R
3
, which of the following operations is/are defined?
(i.e. which of the following can be calculated?)
Select all that are correct.
Unit Vectors: Dot Product
Consider the vectors
u
=
i
−
4
k
\bm{u}=\bm{i}-4\bm{k}
u
=
i
−
4
k
and
v
=
3
i
−
2
j
+
k
\bm{v}=3\bm{i}-2\bm{j}+\bm{k}
v
=
3
i
−
2
j
+
k
in
R
3
\reals^3
R
3
. Find (if possible)
3
v
⋅
(
k
−
2
u
)
3\bm{v}\cdot(\bm{k}-2\bm{u})
3
v
⋅
(
k
−
2
u
)
.