Question

Consider the vectors
u
=
(
−
1
−
1
−
2
)
u=
​

−1
−1
−2
​

​
and
u
×
v
=
(
5
1
−
3
)
u×v=
​

5
1
−3
​

​
.



Give one vector
v
v which is consistent with the expressions above.

Write your vector as a
3
×
1
3×1 column matrix.

Type none if you think such a vector does not exist.

Answer 1

To find a vector \( v \) such that the cross product \( u \times v = (5, 1, -3) \), we can use the formula for the cross product in terms of the components of the vectors.

Let \( u = \begin{pmatrix} -1 \\ -1 \\ -2 \end{pmatrix} \) and \( v = \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} \). The cross product \( u \times v \) is given by the determinant of the following matrix:

\[
u \times v = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-1 & -1 & -2 \\
v_1 & v_2 & v_3
\end{vmatrix}
\]

Calculating the determinant, we have:

\[
u \times v = \mathbf{i} \begin{vmatrix} -1 & -2 \\ v_2 & v_3 \end{vmatrix} - \mathbf{j} \begin{vmatrix} -1 & -2 \\ v_1 & v_3 \end{vmatrix} + \mathbf{k} \begin{vmatrix} -1 & -1 \\ v_1 & v_2 \end{vmatrix}
\]

This leads us to the components:

1. The \( x \)-component:
\[
-1 v_3 + 2 v_2 = 5
\]

2. The \( y \)-component:
\[
-(-1 v_3 + 2 v_1) = 1 \implies v_3 - 2 v_1 = 1 \implies v_3 = 1 + 2 v_1
\]

3. The \( z \)-component:
\[
-(-1 v_2 + -1 v_1) = -3 \implies v_2 - v_1 = -3 \implies v_2 = v_1 - 3
\]

Now, we can express \( v \) in terms of \( v_1 \):

Let \( v_1 = t \) (a parameter):
- \( v_2 = t - 3 \)
- \( v_3 = 1 + 2t \)

Thus, we have:

\[
v = \begin{pmatrix} t \\ t - 3 \\ 1 + 2t \end{pmatrix}
\]

To find a specific \( v \), let’s choose \( t = 0 \):
\[
v = \begin{pmatrix} 0 \\ -3 \\ 1 \end{pmatrix}
\]

So, one vector \( v \) that is consistent with the given expressions is:

\[
v = \begin{pmatrix} 0 \\ -3 \\ 1 \end{pmatrix}
\]