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If l, m, n are the direction of the normal to the plane and p be the perpendicular distance from the origin on it, then the equation of the plane is of the type
- a)x/l + y/m + z/n = p
- b)lx+my+nz=l/p
- c)lx=my+nz=p
- d)lx+my+nz=√p
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
If l, m, n are the direction of the normal to the plane and p be the p...
OP is perpendicular on the plane ABC
∴ OP = p and d.c.'s of OP are [1, m, n]
Let Q(x, y, z) be an arbitrary point on the plane.
Then
OP is the projection of OQ on OP.
⇒ OP = I(x- 0) + m(y- 0) + n(z- 0)
⇒ lx + my + nz = p
This is the required equation of the plane in normal form.
∴ OP = p and d.c.'s of OP are [1, m, n]
Let Q(x, y, z) be an arbitrary point on the plane.
Then
OP is the projection of OQ on OP.
⇒ OP = I(x- 0) + m(y- 0) + n(z- 0)
⇒ lx + my + nz = p
This is the required equation of the plane in normal form.
Most Upvoted Answer
If l, m, n are the direction of the normal to the plane and p be the p...
OP is perpendicular on the plane ABC
∴ OP = p and d.c.'s of OP are [1, m, n]
Let Q(x, y, z) be an arbitrary point on the plane.
Then
OP is the projection of OQ on OP.
⇒ OP = I(x- 0) + m(y- 0) + n(z- 0)
⇒ lx + my + nz = p
This is the required equation of the plane in normal form.
∴ OP = p and d.c.'s of OP are [1, m, n]
Let Q(x, y, z) be an arbitrary point on the plane.
Then
OP is the projection of OQ on OP.
⇒ OP = I(x- 0) + m(y- 0) + n(z- 0)
⇒ lx + my + nz = p
This is the required equation of the plane in normal form.
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