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Found in: Page 565

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Find the two unit vectors orthogonal to both $${\rm{j}} - {\rm{k}}$$ and $${\rm{i}} + {\rm{j}}$$.

The unit vectors orthogonal to both $${\rm{j}} - {\rm{k}}$$ and $${\rm{i}} + {\rm{j}}$$ are $$\frac{1}{{\sqrt 3 }}{\rm{i}} - \frac{1}{{\sqrt 3 }}{\rm{j}} - \frac{1}{{\sqrt 3 }}{\rm{k}}$$ and$$- \frac{1}{{\sqrt 3 }}{\rm{i}} + \frac{1}{{\sqrt 3 }}{\rm{j}} + \frac{1}{{\sqrt 3 }}{\rm{k}}$$.

See the step by step solution

## Step 1: Theorem used

The cross product of two vectors is orthogonal to both vectors, when their dot product is equal to zero.

Write the expression to find unit vector of vector$${\bf{a}}$$

Unit vector $$= \frac{{\rm{a}}}{{|{\rm{a}}|}}$$

## Step 2: Find the cross product between $${\rm{j}} - {\rm{k}}$$ and $${\rm{i}} + {\rm{j}}$$

As given, $${\rm{j}} - {\rm{k}}$$ and $${\rm{i}} + {\rm{j}}$$.

Find the cross product between $${\rm{j}} - {\rm{k}}$$ and $${\rm{i}} + {\rm{j}}$$

$$\begin{array}{l}(j - k) \times (i + j) = \left| {\begin{array}{*{20}{c}}i&j&k\\0&1&{ - 1}\\1&1&0\end{array}} \right|\\(j - k) \times (i + j) = \left| {\begin{array}{*{20}{c}}1&{ - 1}\\1&0\end{array}} \right|{\rm{i}} - \left| {\begin{array}{*{20}{c}}0&{ - 1}\\1&0\end{array}} \right|j + \left| {\begin{array}{*{20}{c}}0&1\\1&1\end{array}} \right|k\\(j - k) \times (i + j) = (0 + 1){\rm{i}} - (0 + 1){\rm{j}} + (0 - 1){\rm{k}}\\(j - k) \times (i + j) = {\rm{i}} - {\rm{j}} - {\rm{k}}\end{array}$$

## Step 3: Find unit vectors

$$\begin{array}{l}{\rm{Unit vectors }} = \frac{{{\rm{i}} - {\rm{j}} - {\rm{k}}}}{{|{\rm{i}} - {\rm{j}} - {\rm{k}}|}}\\{\rm{Unit vectors}} = \pm \frac{{{\rm{i}} - {\rm{j}} - {\rm{k}}}}{{\sqrt {{{(1)}^2} + {{( - 1)}^2} + {{( - 1)}^2}} }}\\{\rm{Unit vectors}} = \pm \frac{{{\rm{i}} - {\rm{j}} - {\rm{k}}}}{{\sqrt {1 + 1 + 1} }}\\{\rm{Unit vectors}} = \pm \frac{{{\rm{i}} - {\rm{j}} - {\rm{k}}}}{{\sqrt 3 }}\end{array}$$

The unit vectors are $$\frac{1}{{\sqrt 3 }}{\rm{i}} - \frac{1}{{\sqrt 3 }}{\rm{j}} - \frac{1}{{\sqrt 3 }}{\rm{k}}$$ and$$- \frac{1}{{\sqrt 3 }}{\rm{i}} + \frac{1}{{\sqrt 3 }}{\rm{j}} + \frac{1}{{\sqrt 3 }}{\rm{k}}$$.

Thus, the two unit vectors orthogonal to both $${\rm{j}} - {\rm{k}}$$ and $${\rm{i}} + {\rm{j}}$$.are $$\frac{1}{{\sqrt 3 }}{\rm{i}} - \frac{1}{{\sqrt 3 }}{\rm{j}} - \frac{1}{{\sqrt 3 }}{\rm{k}}$$ and$$- \frac{1}{{\sqrt 3 }}{\rm{i}} + \frac{1}{{\sqrt 3 }}{\rm{j}} + \frac{1}{{\sqrt 3 }}{\rm{k}}$$.