Suggested languages for you:

Americas

Europe

Q11E

Expert-verified
Found in: Page 564

### Essential Calculus: Early Transcendentals

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

# Find the resultant vector of $$(j - k) \times (k - i)$$ using cross product.

The cross product $$(j - k) \times (k - i)$$ is$$i + j + k$$.

See the step by step solution

## Step 1: Formula used

Consider the following property

$$\begin{array}{l}{\rm{j}} \times {\rm{k}} = {\rm{i}}\\{\rm{k}} \times {\rm{k}} = 0\\{\rm{j}} \times {\rm{i}} = - {\rm{k}}\\{\rm{k}} \times {\rm{i}} = {\rm{j}}\end{array}$$

## Step 2: Find the resultant vector

As given,$$(j - k) \times (k - i) \ldots \ldots \ldots (1)$$

Modify equation (1)

$$\begin{array}{l}({\rm{j}} - {\rm{k}}) \times (k - i) = ({\rm{j}} - {\rm{k}}) \times {\rm{k}} + ({\rm{j}} - {\rm{k}}) + ({\rm{j}} - {\rm{k}}) \times ( - {\rm{i}})\\({\rm{j}} - {\rm{k}}) \times (k - i) = ({\rm{j}} \times {\rm{k}}) + ( - {\rm{k}}) \times {\rm{k}} + {\rm{j}} \times ( - {\rm{i}}) + ( - {\rm{k}}) \times ( - {\rm{i}})\\({\rm{j}} - {\rm{k}}) \times (k - i) = ({\rm{j}} \times {\rm{k}}) + ( - 1)({\rm{k}} \times {\rm{k}}) + ( - 1)({\rm{j}} \times {\rm{i}}) + {( - 1)^2}({\rm{k}} \times {\rm{i}})\end{array}$$

Substitute $$i$$ for $$j \times k,0$$ for $$k \times k, - k$$ for $$j \times i$$ and $$j$$ for $$k \times i$$

$$\begin{array}{l}({\rm{j}} - {\rm{k}}) \times ({\rm{k}} - {\rm{i}}) = {\rm{i}} + ( - 1)(0) + ( - 1)( - {\rm{k}}) + {( - 1)^2}({\rm{j}})\\({\rm{j}} - {\rm{k}}) \times ({\rm{k}} - {\rm{i}}) = {\rm{i}} - 0 + {\rm{k}} + {\rm{j}}\\({\rm{j}} - {\rm{k}}) \times ({\rm{k}} - {\rm{i}}) = {\rm{i}} + {\rm{j}} + {\rm{k}}\end{array}$$

Thus, the cross product $$(j - k) \times (k - i)$$ is $$i + j + k$$.