Question
A plane passes through a fixed point $(a, b, c)$. Show that the locus of the foot of the perpendicular to it from the origin is the surface $x^{2}+y^{2}+z^{2}-a x-b y-c z=0$
Step 1
The equation of the plane passing through the point $(a, b, c)$ and perpendicular to the line joining the origin and $(\alpha, \beta, \gamma)$ can be written as: \[\alpha(x - a) + \beta(y - b) + \gamma(z - c) = 0\] Show more…
Show all steps
Your feedback will help us improve your experience
Nida Wasiq and 93 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
A plane passes through a fixed point $(a, b, c)$. The locus of the foot of the perpendicular to it from the origin is a sphere of radius (A) $\sqrt{a^{2}+b^{2}+c^{2}}$ (B) $\frac{1}{2} \sqrt{a^{2}+b^{2}+c^{2}}$ (C) $a^{2}+b^{2}+c^{2}$ (D) none of these
If the foot of the perpendicular from the origin to a plane in $(a, b, c)$, the equation of the plane is (a) $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ (b) $a x+b y+c z=3$ (c) $a x+b y+c z=a^{2}+b^{2}+c^{2}$ (d) $a x+b y+c z=a+b+c$
Let $N$ be the foot of the perpendicular of length $P$, from the origin to a plane and $l, m, n$ be the direction cosines of $O N$, the equation of the plane is (a) $p x+m y+n z=1$ (b) $L x+p y+n=m$ (c) $k x+m y+p z=n$ (d) $\lfloor x+m y+n z=p$
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD