30. Find all vectors $$ \boldsymbol{v} $$ such that $$ (-\boldsymbol{i}+2 \boldsymbol{j}+3 \boldsymbol{k}) imes \boldsymbol{v}=\boldsymbol{i}+5 \boldsymbol{j} $$. Otherwise, show that it is not possible.

Question

30. Find all vectors $$ \boldsymbol{v} $$ such that $$ (-\boldsymbol{i}+2 \boldsymbol{j}+3 \boldsymbol{k}) 	imes \boldsymbol{v}=\boldsymbol{i}+5 \boldsymbol{j} $$. Otherwise, show that it is not possible.

UpStudy AI · Accepted Answer

To find the vector $$ \boldsymbol{v} $$ such that

$$
(-\boldsymbol{i}+2 \boldsymbol{j}+3 \boldsymbol{k}) \times \boldsymbol{v}=\boldsymbol{i}+5 \boldsymbol{j},
$$

let's denote $$ \boldsymbol{v} $$ as $$ \boldsymbol{v} = x \boldsymbol{i} + y \boldsymbol{j} + z \boldsymbol{k} $$.

We can compute the cross product $$ (-\boldsymbol{i}+2 \boldsymbol{j}+3 \boldsymbol{k}) \times (x \boldsymbol{i} + y \boldsymbol{j} + z \boldsymbol{k}) $$ using the determinant of a matrix formed by the unit vectors and the components of the vectors involved.

The cross product can be expressed as:

$$
\begin{vmatrix}
\boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \
-1 & 2 & 3 \
x & y & z
\end{vmatrix}
$$

Calculating this determinant, we have:

$$
\boldsymbol{i} \begin{vmatrix} 2 & 3 \ y & z \end{vmatrix} - \boldsymbol{j} \begin{vmatrix} -1 & 3 \ x & z \end{vmatrix} + \boldsymbol{k} \begin{vmatrix} -1 & 2 \ x & y \end{vmatrix}
$$

Calculating each of these 2x2 determinants:

1. For $$ \boldsymbol{i} $$:
   $$
   2z - 3y
   $$

2. For $$ -\boldsymbol{j} $$:
   $$
   -(-1z - 3x) = z + 3x
   $$

3. For $$ \boldsymbol{k} $$:
   $$
   -1y - 2x = -y - 2x
   $$

Putting it all together, we have:

$$
(2z - 3y) \boldsymbol{i} - (z + 3x) \boldsymbol{j} + (-y - 2x) \boldsymbol{k}
$$

Setting this equal to $$ \boldsymbol{i} + 5 \boldsymbol{j} $$, we can equate the coefficients:

1. $$ 2z - 3y = 1 $$ (coefficient of $$ \boldsymbol{i} $$)
2. $$ -(z + 3x) = 5 $$ (coefficient of $$ \boldsymbol{j} $$)
3. $$ -y - 2x = 0 $$ (coefficient of $$ \boldsymbol{k} $$)

Now we can solve this system of equations step by step.

From the third equation, we can express $$ y $$ in terms of $$ x $$:

$$
y = -2x
$$

Substituting $$ y = -2x $$ into the first equation:

$$
2z - 3(-2x) = 1 \implies 2z + 6x = 1 \implies 2z = 1 - 6x \implies z = \frac{1 - 6x}{2}
$$

Now substituting $$ z $$ into the second equation:

$$
-(\frac{1 - 6x}{2} + 3x) = 5
$$

Multiplying through by -1:

$$
\frac{1 - 6x}{2} + 3x = -5
$$

Multiplying through by 2 to eliminate the fraction:

$$
1 - 6x + 6x = -10 \implies 1 = -10
$$

This is a contradiction, which means there are no solutions for $$ \boldsymbol{v} $$.

Thus, it is not possible to find such a vector $$ \boldsymbol{v} $$.
There is no vector $$ \boldsymbol{v} $$ that satisfies the equation $$ (-\boldsymbol{i}+2 \boldsymbol{j}+3 \boldsymbol{k}) \times \boldsymbol{v}=\boldsymbol{i}+5 \boldsymbol{j} $$.

Find all vectors such that .
Otherwise, show that it is not possible.

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