Let N be the foot of the perpendicular of length P, from the origin to a plane and l, m, n be th

step 1 Before solving that question, let us discuss this concept here. So if we have the plane here, th

Let N be the foot of the perpendicular of length P, from the...

Key Concepts

Point-Normal Form of a Plane

The point?normal form of a plane's equation uses a known point on the plane and a normal vector to the plane. It is formulated by taking the dot product of the normal vector with the vector difference between any point on the plane and the known point, and setting this equal to zero. This form provides a direct method to derive the plane's equation when the normal and a point are known.

Foot of a Perpendicular

The foot of a perpendicular from a point to a plane is the unique point on the plane where the line drawn at a right angle (i.e. perpendicular) from the given point first intersects the plane. This point minimizes the distance between the point and the plane and is used to determine the shortest distance (or the perpendicular distance) from the point to the plane.

Direction Cosines

Direction cosines are the cosines of the angles that a given vector makes with the positive directions of the coordinate axes. When the vector is a unit vector, its components are exactly these direction cosines. They are crucial in expressing the orientation of a line or the normal to a plane in three-dimensional space.

Vector Dot Product

The vector dot product is an operation that multiplies two vectors to yield a scalar, and it is defined in terms of the magnitudes of the vectors and the cosine of the angle between them. In the context of finding the equation of a plane, the dot product is used to express the condition of perpendicularity between the plane’s normal vector and any vector lying in the plane.

Transcript

00:01 Before solving that question, let us discuss this concept here.

00:03 So if we have the plane here, this is the plane here we have.

00:05 This is here we have x1, y, 1, z1, a point in the plane.

00:08 This is x, y, z, a general point in the plane.

00:10 And this is the normal to the plane given by a normal vector a, b, c.

00:14 So we get this vector here we have that is given as x negative x1, x, negative x1, then we have y negative y1, then we have b negative z1.

00:29 Which we have here.

00:29 So this is, we have this is a, this is b.

00:34 This is cv this is equal to we have we just when you just take the the we have got product so we just solve this that is coming out to be that is equal to zero here we have got product is going to be zero because two vectors are perpendicular so we solve this from here we get a x plus b y plus c z and this here we get negative a x1 negative b y 1 negative c that is a constant so we just put in on the other side that is equal to d this is the equation of the plane.

01:07 Now, if we have the point, any point here we have, just to divide this by, let's say we have a x over we have square root a square plus b square plus c square, positive b, y over square root a square plus b square plus c square.

01:23 Positive, we have tz over square root a square plus b square plus b square.

01:29 That equals d over root a square plus b square plus c square.

01:35 This is coming out to be a over root a square plus b square that's our direction cosine uh we have with respect to x x and same with respect to y x is and that we get here we get here lx plus vy plus c z that is coming i'm sorry we will take here lm and here we have this coming out to be m and this will be n here so we get lx plus my plus n z that is coming out to be equal to k okay, now moving forward, now we'll find out the length of perpendicular from the origin to the plane we have...

Please give Ace some feedback