Chapter 3.1, Problem 3.17P

A plane contains the vectors A and B. Determine the unit vector normal to the plane when A and B are equal to, respectively, (a) 7i + 8j – 2k and 9i – 4j – 5k, (b) 6i – 3j + 9k and –5i + 4j – 3k.

To determine

The unit vector perpendicular to the plane.

Answer to Problem 3.17P

The unit vector perpendicular to the plane is $\left(-0.428\right)i+\left(0.1515\right)j+\left(-0.891\right)k$.

Explanation of Solution

Write an expression to calculate the unit vector perpendicular to the plane.

$\lambda =\frac{A\times B}{\left|A\times B\right|}$ (I)

Here, $\lambda$ is the unit vector normal to the plane $A$ and $B$ are two vectors in the plane.

Write an expression to represent first vector.

$A=A_{x}i+A_{y}j+A_{z}k$ (II)

Here, $A_x$ is the x component of the vector, $A_y$ is the y component of the vector and $A_z$ is the z component of the vector.

Write an expression to represent first vector.

$B=B_{x}i+B_{y}j+B_{z}k$ (III)

Here, $B_x$ is the x component of the vector, $B_y$ is the y component of the vector and $B_z$ is the z component of the vector.

Write an expression to calculate the cross product of the vectors.

$A\times B=\left|\begin{array}{ccc}i& j& k\\ A_x& A_y& A_z\\ B_x& B_y& B_z\end{array}\right|$ (IV)

Write an expression to calculate the magnitude of the cross product of the vectors.

$\left|A\times B\right|=\sqrt{{\left(A\times B\right)}_x^2+{\left(A\times B\right)}_y^2+{\left(A\times B\right)}_z^2}$ (V)

Here, ${\left(A\times B\right)}_x$ is the x component of  $A\times B$, ${\left(A\times B\right)}_y$ is the y component of ${\left(A\times B\right)}_z$ and $A\times B$ is the z component of $A\times B$.

Conclusion:

Substitute $7i+8j-2k$ for $A$ and $9i-4j-5k$ for $B$ in equation (IV) to find $A\times B$.

$\begin{array}{c}A\times B=\left(7i+8j-2k\right)\times \left(9i-4j-5k\right)\\ =\left|\begin{array}{ccc}i& j& k\\ 7& 8& -2\\ 9& -4& -5\end{array}\right|\\ =\left(-40-8\right)i+\left(-18+35\right)j+\left(-28-72\right)k\\ =\left(-48\right)i+\left(17\right)j+\left(-100\right)k\end{array}$

Substitute $-48$ for ${\left(A\times B\right)}_x$, $17$ for ${\left(A\times B\right)}_y$, and $-100$ for ${\left(A\times B\right)}_z$ in equation (V) to find $\left|A\times B\right|$.

$\begin{array}{c}\left|A\times B\right|=\sqrt{{\left(-48\right)}^2+{\left(17\right)}^2+{\left(-100\right)}^2}\\ =\sqrt{12593}\end{array}$

Substitute $\left(-48\right)i+\left(17\right)j+\left(-100\right)k$ for $A\times B$, and $\sqrt{12593}$ for $\left|A\times B\right|$ in equation (I) to find $\lambda$.

$\begin{array}{c}\lambda =\frac{\left(-48\right)i+\left(17\right)j+\left(-100\right)k}{\sqrt{12593}}\\ =\left(-0.428\right)i+\left(0.1515\right)j+\left(-0.891\right)k\end{array}$

Thus, the unit vector perpendicular to the plane is $\left(-0.428\right)i+\left(0.1515\right)j+\left(-0.891\right)k$.

To determine

The unit vector perpendicular to the plane.

Answer to Problem 3.17P

The unit vector perpendicular to the plane is $\frac{\left(-3\right)i+\left(-3\right)j+\left(1\right)k}{\sqrt{19}}$.

Explanation of Solution

Write an expression to calculate the unit vector perpendicular to the plane.

$\lambda =\frac{A\times B}{\left|A\times B\right|}$ (VI)

Write an expression to calculate the cross product of the vectors.

$A\times B=\left|\begin{array}{ccc}i& j& k\\ A_x& A_y& A_z\\ B_x& B_y& B_z\end{array}\right|$ (VII)

Write an expression to calculate the magnitude of the cross product of the vectors.

$\left|A\times B\right|=\sqrt{{\left(A\times B\right)}_x^2+{\left(A\times B\right)}_y^2+{\left(A\times B\right)}_z^2}$ (VIII)

Conclusion:

Substitute $6i-3j+9k$ for $A$ and $-5i+4j-3k$ for $B$ in equation (VII) to find $A\times B$.

$\begin{array}{c}A\times B=\left(7i+8j-2k\right)\times \left(9i-4j-5k\right)\\ =\left|\begin{array}{ccc}i& j& k\\ 6& -3& 9\\ -5& 4& -3\end{array}\right|\\ =\left(9-36\right)i+\left(-45+18\right)j+\left(-24-15\right)k\\ =\left(-27\right)i+\left(-27\right)j+\left(9\right)k\end{array}$

Substitute $-27$ for ${\left(A\times B\right)}_x$, $-27$ for ${\left(A\times B\right)}_y$, and $9$ for ${\left(A\times B\right)}_z$ in equation (VIII) to find $\left|A\times B\right|$.

$\begin{array}{c}\left|A\times B\right|=\sqrt{{\left(-27\right)}^2+{\left(-27\right)}^2+{\left(9\right)}^2}\\ =9\sqrt{19}\end{array}$

Substitute $\left(-27\right)i+\left(-27\right)j+\left(9\right)k$ for $A\times B$, and $9\sqrt{19}$ for $\left|A\times B\right|$ in equation (VI) to find $\lambda$.

$\begin{array}{c}\lambda =\frac{\left(-27\right)i+\left(-27\right)j+\left(9\right)k}{9\sqrt{19}}\\ =\frac{\left(-3\right)i+\left(-3\right)j+\left(1\right)k}{\sqrt{19}}\end{array}$

Thus, the unit vector perpendicular to the plane is $\frac{\left(-3\right)i+\left(-3\right)j+\left(1\right)k}{\sqrt{19}}$.