Chapter 9, Problem 86RE

To determine

To calculate:The vector orthogonal to both $v$ and $w$ .To use the vectors $v=3i+j-2k$ and $w=-3i+2j-k$ .

Answer to Problem 86RE

The vectors orthogonal to both $v$ and $w$ are: $0$ so the both vectors are orthogonal.

Explanation of Solution

Given information:

The vectors:

  $v=3i+j-2k$ and $w=-3i+2j-k$

Formula used:

The cross product of the vectors $v=a_{1}i+b_{1}j+c_{1}k$ and $w=a_{2}i+b_{2}j+c_{2}k$ that is,

  $\begin{array}{c}v\times w=\left(\begin{array}{ccc}i& j& k\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right)\\ =\left(\begin{array}{cc}{b}_1& c_1\\ b_2& c_2\end{array}\right)i-\left(\begin{array}{cc}{a}_1& c_1\\ a_2& c_2\end{array}\right)j+\left(\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right)k\\ =\left(b_1{c}_2-b_2{c}_1\right)i-\left(a_1{c}_2-a_2{c}_1\right)j+\left(a_1{b}_2-a_2{b}_1\right)k\end{array}$

The dot product of the vectors are solved by the formula: $v\cdot w=a_1{a}_2+b_1{b}_2+c_1{c}_2$

Calculation:

Consider the vectors:

  $v=3i+j-2k$ and $w=-3i+2j-k$

Now to find unit vector orthogonal to both $v$ and $w$ ,

First find cross product of the vectors,

Now to find the value of $v\times w$ .

Apply the Cross Product formula of matrix:

The cross product of the vectors $v=a_{1}i+b_{1}j+c_{1}k$ and $w=a_{2}i+b_{2}j+c_{2}k$ that is,

  $\begin{array}{c}v\times w=\left(\begin{array}{ccc}i& j& k\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right)\\ =\left(\begin{array}{cc}{b}_1& c_1\\ b_2& c_2\end{array}\right)i-\left(\begin{array}{cc}{a}_1& c_1\\ a_2& c_2\end{array}\right)j+\left(\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right)k\\ =\left(b_1{c}_2-b_2{c}_1\right)i-\left(a_1{c}_2-a_2{c}_1\right)j+\left(a_1{b}_2-a_2{b}_1\right)k\end{array}$

Apply the values in the formula:

  $\begin{array}{c}v\times w=\left(\begin{array}{ccc}i& j& k\\ 3& 1& -2\\ -3& 2& -1\end{array}\right)\\ =\left(\begin{array}{cc}1& -2\\ 2& -1\end{array}\right)i-\left(\begin{array}{cc}3& -2\\ -3& -1\end{array}\right)j+\left(\begin{array}{cc}3& 1\\ -3& 2\end{array}\right)k\\ =\left(-1+4\right)i-\left(-3-6\right)j+\left(6+3\right)k\\ =3i+9j+9k\end{array}$

Thus, Take a dot product with the result of $v\times w$ to check if the computed value is orthogonal on both $v$ and $w$ .

So apply the values in the $v\cdot \left(v\times w\right)$ .

As from the above result, the value of $v\times w$ is $3i+9j+9k$ .

So,

  $v\cdot \left(v\times w\right)=\left(3i+j-2k\right)\cdot \left(3i+9j+9k\right)$

Now apply the Dot product formula of the matrix:

  $\begin{array}{c}v\cdot w=a_1{a}_2+b_1{b}_2+c_1{c}_2\\ v\cdot \left(v\times w\right)=\left(3i+j-2k\right)\cdot \left(3i+9j+9k\right)\\ =3i\cdot 3i+j\cdot 9j+\left(-2k\right)\cdot 9k\\ =9+9-18\\ =0\end{array}$

Now check with the help of Dot product of the $v\times w$ to $w$ ,

  $\begin{array}{c}v\cdot w=a_1{a}_2+b_1{b}_2+c_1{c}_2\\ w\cdot \left(v\times w\right)=\left(-3i+2j-k\right)\cdot \left(3i+9j+9k\right)\\ =-3i\cdot 3i+2j\cdot 9j+\left(-k\right)\cdot 9k\\ =-9+18-9\\ =0\end{array}$

Thus, the cross product of the given vector is $0$ , so the both vectors are orthogonal.