Question

Determine the value of $k$ for which the direction vectors of the lines $\frac{x-1}{k}=\frac{y-2}{2}=\frac{z+1}{k-1}$ and $\frac{x+3}{-2}=\frac{z}{1}, y=-1$ are perpendicular.

Name: Determine the value of k for which the direction vectors of the lines (x-1)/(k)=(y-2)/(2)=(z+1)/(k-1) and (x+3)/(-2)=(z)/(1), y=-1 are perpendicular.
Uploaded: 2023-06-27T12:01:33-08:00
Duration: 1 min 48 s
Channel: Adi S
Description: Determine the value of k for which the direction vectors of the lines (x-1)/(k)=(y-2)/(2)=(z+1)/(k-1) and (x+3)/(-2)=(z)/(1), y=-1 are perpendicular.

          Determine the value of $k$ for which the direction vectors of the lines $\frac{x-1}{k}=\frac{y-2}{2}=\frac{z+1}{k-1}$ and $\frac{x+3}{-2}=\frac{z}{1}, y=-1$ are perpendicular.

Added by Michael G.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

06/27/2023

Step 1

For the first line, we have the direction vector $\vec{d_1} = \begin{pmatrix} k \\ 2 \\ k-1 \end{pmatrix}$. For the second line, we have the direction vector $\vec{d_2} = \begin{pmatrix} -2 \\ 0 \\ 1 \end{pmatrix}$. Now, we know that two vectors are Show more…

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Determine the value of $k$ for which the direction vectors of the lines $\frac{x-1}{k}=\frac{y-2}{2}=\frac{z+1}{k-1}$ and $\frac{x+3}{-2}=\frac{z}{1}, y=-1$ are perpendicular.