Given the vectors A=0.6j+0.8k and B=30i−5j−k . A unit vector that is normal to the plane containing A and B is

To find a unit vector that is normal to the plane containing vectors \vec{A} = 0.6\vec{j} + 0.8\vec{k} and \vec{B} = 30\vec{i} - 5\vec{j} - \vec{k} , we need to find the cross product of vectors \vec{A} and \vec{B} .

The cross product of two vectors \vec{u} = u_x\vec{i} + u_y\vec{j} + u_z\vec{k} and \vec{v} = v_x\vec{i} + v_y\vec{j} + v_z\vec{k} is given by:
\vec{u} \times \vec{v} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \ u_x & u_y & u_z \ v_x & v_y & v_z \end{vmatrix}

Given \vec{A} = 0.6\vec{j} + 0.8\vec{k} and \vec{B} = 30\vec{i} - 5\vec{j} - \vec{k} , the cross product \vec{A} \times \vec{B} is:
\vec{A} \times \vec{B} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \ 0 & 0.6 & 0.8 \ 30 & -5 & -1 \end{vmatrix}
= -30\vec{i} - 24\vec{j} + 18\vec{k}

To get a unit vector that is normal to the plane containing \vec{A} and \vec{B} , we need to divide \vec{A} \times \vec{B} by its magnitude.

The magnitude of \vec{A} \times \vec{B} is given by:
| \vec{A} \times \vec{B} | = \sqrt{(-30)^2 + (-24)^2 + (18)^2}
= \sqrt{900 + 576 + 324}
= \sqrt{1800}
= 30

Therefore, the unit vector that is normal to the plane containing \vec{A} and \vec{B} is:
\hat{N} = \frac{\vec{A} \times \vec{B}}{| \vec{A} \times \vec{B} |} = \frac{-30\vec{i} - 24\vec{j} + 18\vec{k}}{30}
\hat{N} = -\vec{i} - 0.8\vec{j} + 0.6\vec{k}

\boxed{\hat{N} = -\vec{i} - 0.8\vec{j} + 0.6\vec{k}}