Converting vector-form to cartesian-form line equations in 3D

Convert \frac{x-5}2=y+1=\frac{z+3}6 to vector form

You can see that the first vector corresponds to the negatives of the constant added to each position (e.g. x-5 becomes 5 in the vector), and the second vector corresponds to the denominators of each fraction (e.g. \frac{x-5}2 becomes 2 in the vector).

Convert \frac{x+2}4=\frac{y-3}5=z+1 to vector form

flashcards

QuestionAnswer
What is the first step to convert \frac{x-5}{2}=y+1=\frac{z+3}{6} into vector form?Let \lambda = \frac{x-5}{2} = y+1 = \frac{z+3}{6}.
How do you find the x component from \lambda = \frac{x-5}{2}?x = 5 + 2\lambda.
How do you find the y component from \lambda = y+1?y = -1 + \lambda.
How do you find the z component from \lambda = \frac{z+3}{6}?z = -3 + 6\lambda.
What is the vector form of the line \frac{x-5}{2}=y+1=\frac{z+3}{6}?\vec r = \begin{pmatrix}5\\-1\\-3\end{pmatrix} + \lambda \begin{pmatrix}2\\1\\6\end{pmatrix}.
In the vector form from the symmetric form, what do the constants in the first vector correspond to?The negatives of the constant added to each position (e.g. x-5 becomes 5).
In the vector form from the symmetric form, what does the second vector correspond to?The denominators of each fraction (e.g. \frac{x-5}{2} becomes 2).
How do you convert \frac{x+2}{4}=\frac{y-3}{5}=z+1 to vector form?\vec r = \begin{pmatrix}-2\\3\\-1\end{pmatrix} + \lambda \begin{pmatrix}4\\5\\1\end{pmatrix}.
In the conversion of \frac{x+2}{4}=\frac{y-3}{5}=z+1, what is the x component?x = -2 + 4\lambda.
In the conversion of \frac{x+2}{4}=\frac{y-3}{5}=z+1, what is the y component?y = 3 + 5\lambda.
In the conversion of \frac{x+2}{4}=\frac{y-3}{5}=z+1, what is the z component?z = -1 + \lambda.