Answer :
To find a vector of magnitude [tex]\( S \)[/tex] unit that is perpendicular to both [tex]\(\vec{a} = -2\hat{i} + \hat{j} + \hat{k}\)[/tex] and [tex]\(\vec{b} = \hat{i} - 3\hat{j} - \hat{k}\)[/tex], we need to follow these steps:
1. Compute the Cross Product:
The cross product of two vectors [tex]\(\vec{a}\)[/tex] and [tex]\(\vec{b}\)[/tex] gives us a vector that is perpendicular to both.
Given:
[tex]\[ \vec{a} = -2\hat{i} + \hat{j} + \hat{k} \][/tex]
[tex]\[ \vec{b} = \hat{i} - 3\hat{j} - \hat{k} \][/tex]
The cross product [tex]\(\vec{a} \times \vec{b}\)[/tex] is determined by:
[tex]\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 1 & 1 \\ 1 & -3 & -1 \end{vmatrix} \][/tex]
Expanding this determinant, we find:
[tex]\[ \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} 1 & 1 \\ -3 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} -2 & 1 \\ 1 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} -2 & 1 \\ 1 & -3 \end{vmatrix} \][/tex]
Calculate each of the smaller [tex]\(2 \times 2\)[/tex] determinants:
[tex]\[ \begin{vmatrix} 1 & 1 \\ -3 & -1 \end{vmatrix} = (1 \times -1) - (1 \times -3) = -1 + 3 = 2 \][/tex]
[tex]\[ \begin{vmatrix} -2 & 1 \\ 1 & -1 \end{vmatrix} = (-2 \times -1) - (1 \times 1) = 2 - 1 = 1 \][/tex]
[tex]\[ \begin{vmatrix} -2 & 1 \\ 1 & -3 \end{vmatrix} = (-2 \times -3) - (1 \times 1) = 6 - 1 = 5 \][/tex]
Therefore:
[tex]\[ \vec{a} \times \vec{b} = 2\hat{i} - 1\hat{j} + 5\hat{k} \][/tex]
Thus, the cross product vector is:
[tex]\[ \vec{c} = 2\hat{i} - \hat{j} + 5\hat{k} \][/tex]
2. Compute the Magnitude of the Cross Product Vector:
[tex]\[ |\vec{c}| = \sqrt{(2)^2 + (-1)^2 + (5)^2} = \sqrt{4 + 1 + 25} = \sqrt{30} \approx 5.477225575051661 \][/tex]
3. Find the Unit Vector:
The unit vector in the direction of [tex]\(\vec{c}\)[/tex] is obtained by dividing each component by the magnitude:
[tex]\[ \hat{u} = \frac{1}{|\vec{c}|}\vec{c} = \frac{1}{\sqrt{30}}(2\hat{i} - \hat{j} + 5\hat{k}) \][/tex]
[tex]\[ = \left(\frac{2}{\sqrt{30}}\right)\hat{i} + \left(\frac{-1}{\sqrt{30}}\right)\hat{j} + \left(\frac{5}{\sqrt{30}}\right)\hat{k} \][/tex]
Numerically, this simplifies to:
[tex]\[ \hat{u} = 0.3651483716701107 \hat{i} - 0.18257418583505536 \hat{j} + 0.9128709291752769 \hat{k} \][/tex]
In summary:
- The vector perpendicular to both [tex]\(\vec{a}\)[/tex] and [tex]\(\vec{b}\)[/tex] is [tex]\(2\hat{i} - \hat{j} + 5\hat{k}\)[/tex].
- The magnitude of this vector is [tex]\(5.477225575051661\)[/tex].
- The unit vector in this direction is [tex]\(0.3651483716701107 \hat{i} - 0.18257418583505536 \hat{j} + 0.9128709291752769 \hat{k}\)[/tex].
1. Compute the Cross Product:
The cross product of two vectors [tex]\(\vec{a}\)[/tex] and [tex]\(\vec{b}\)[/tex] gives us a vector that is perpendicular to both.
Given:
[tex]\[ \vec{a} = -2\hat{i} + \hat{j} + \hat{k} \][/tex]
[tex]\[ \vec{b} = \hat{i} - 3\hat{j} - \hat{k} \][/tex]
The cross product [tex]\(\vec{a} \times \vec{b}\)[/tex] is determined by:
[tex]\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 1 & 1 \\ 1 & -3 & -1 \end{vmatrix} \][/tex]
Expanding this determinant, we find:
[tex]\[ \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} 1 & 1 \\ -3 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} -2 & 1 \\ 1 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} -2 & 1 \\ 1 & -3 \end{vmatrix} \][/tex]
Calculate each of the smaller [tex]\(2 \times 2\)[/tex] determinants:
[tex]\[ \begin{vmatrix} 1 & 1 \\ -3 & -1 \end{vmatrix} = (1 \times -1) - (1 \times -3) = -1 + 3 = 2 \][/tex]
[tex]\[ \begin{vmatrix} -2 & 1 \\ 1 & -1 \end{vmatrix} = (-2 \times -1) - (1 \times 1) = 2 - 1 = 1 \][/tex]
[tex]\[ \begin{vmatrix} -2 & 1 \\ 1 & -3 \end{vmatrix} = (-2 \times -3) - (1 \times 1) = 6 - 1 = 5 \][/tex]
Therefore:
[tex]\[ \vec{a} \times \vec{b} = 2\hat{i} - 1\hat{j} + 5\hat{k} \][/tex]
Thus, the cross product vector is:
[tex]\[ \vec{c} = 2\hat{i} - \hat{j} + 5\hat{k} \][/tex]
2. Compute the Magnitude of the Cross Product Vector:
[tex]\[ |\vec{c}| = \sqrt{(2)^2 + (-1)^2 + (5)^2} = \sqrt{4 + 1 + 25} = \sqrt{30} \approx 5.477225575051661 \][/tex]
3. Find the Unit Vector:
The unit vector in the direction of [tex]\(\vec{c}\)[/tex] is obtained by dividing each component by the magnitude:
[tex]\[ \hat{u} = \frac{1}{|\vec{c}|}\vec{c} = \frac{1}{\sqrt{30}}(2\hat{i} - \hat{j} + 5\hat{k}) \][/tex]
[tex]\[ = \left(\frac{2}{\sqrt{30}}\right)\hat{i} + \left(\frac{-1}{\sqrt{30}}\right)\hat{j} + \left(\frac{5}{\sqrt{30}}\right)\hat{k} \][/tex]
Numerically, this simplifies to:
[tex]\[ \hat{u} = 0.3651483716701107 \hat{i} - 0.18257418583505536 \hat{j} + 0.9128709291752769 \hat{k} \][/tex]
In summary:
- The vector perpendicular to both [tex]\(\vec{a}\)[/tex] and [tex]\(\vec{b}\)[/tex] is [tex]\(2\hat{i} - \hat{j} + 5\hat{k}\)[/tex].
- The magnitude of this vector is [tex]\(5.477225575051661\)[/tex].
- The unit vector in this direction is [tex]\(0.3651483716701107 \hat{i} - 0.18257418583505536 \hat{j} + 0.9128709291752769 \hat{k}\)[/tex].
