Different Representation of Curves – Switch from Cartesian to vector – Exercise 3807 Post category:Different Representations of Curves Post comments:0 Comments Exercise A line is given as an intersection of the plains 2x−y−z−1=02x-y-z-1=02x−y−z−1=0 x−y+z−2=0x-y+z-2=0x−y+z−2=0 Present the curve in the vector presentation. Final Answer Show final answer r⃗(t)=(2t−1)i⃗+(3t−3)j⃗+tk⃗\vec{r}(t)=(2t-1)\vec{i}+(3t-3)\vec{j}+t\vec{k}r(t)=(2t−1)i+(3t−3)j+tk Solution Coming soon… Share with Friends Read more articles Previous PostDifferent Representation of Curves – Switch from Cartesian to vector – Exercise 3809 Next PostDifferent Representation of Curves – Switch from Cartesian to vector – Exercise 3796 You Might Also Like Different Representation of Curves – Switch from parametric to Cartesian – Exercise 3704 March 2, 2019 Different Representation of Curves – Switch from parametric to Cartesian – Exercise 3722 March 2, 2019 Different Representation of Curves – Switch from parametric to Cartesian – Exercise 3707 March 2, 2019 Different Representation of Curves – Switch from parametric to Cartesian – Exercise 3715 March 2, 2019 Different Representation of Curves – Switch from parametric to Cartesian – Exercise 3732 March 2, 2019 הDifferent Representation of Curves – Switch from parametric to Cartesian – Exercise 3742 March 2, 2019 Leave a Reply Cancel replyCommentEnter your name or username to comment Enter your email address to comment Enter your website URL (optional) Δ
Different Representation of Curves – Switch from parametric to Cartesian – Exercise 3704 March 2, 2019
Different Representation of Curves – Switch from parametric to Cartesian – Exercise 3722 March 2, 2019
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