Question
The distance between the line $\frac{x-1}{3}=\frac{y+2}{-2}=\frac{z-1}{2}$ and the plane $2 x+2 y-z-6=0$ is(a) 5(b) 4(c) 3(d) 0
Step 1
We can see that the line is parallel to the plane. Show more…
Show all steps
Your feedback will help us improve your experience
Ankur S and 87 other Calculus 3 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
The distance between the planes $2 \mathrm{x}-\mathrm{y}+2 \mathrm{z}+1=0$ and $4 \mathrm{x}-2 \mathrm{y}+4 \mathrm{z}+11=0$ is (a) $\frac{9}{2}$ (b) 3 (c) $\frac{7}{2}$ (d) $\frac{3}{2}$
Find the distance between the point and the plane (see figure). The distance $D$ between a point $\left(x_{0}, y_{0}, z_{0}\right)$ and the plane $a x+b y+c z+d=0$ is $$ D=\frac{\left|a x_{0}+b y_{0}+c z_{0}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}} $$ $$ (1,5,-4), 3 x-y+2 z=6 $$
Functions of Several Variables
Surfaces in Space
Find the distance between the point and the plane (see figure). The distance $D$ between a point $\left(x_{0}, y_{0}, z_{0}\right)$ and the plane $a x+b y+c z+d=0$ is $$ D=\frac{\left|a x_{0}+b y_{0}+c z_{0}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}} $$ $$ (2,-1,0), 3 x+3 y+2 z=6 $$
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD