We are given with points,

Origin O(0, 0, 0) and point A(a, b, c).

Here, a, b, c are direction ratios.

We need to find:

(a). Direction cosines of line OA.

(b). Equation of plane through A at right angle to OA.

Let us start from (a).

(a). The given points are A(a, b, c) and O(0, 0, 0). Then,

We know that,

If (a, b, c) are direction ratios of a given vector, then its direction cosines are

As in the question,

Direction ratios are (a, b, c), then direction cosines of are

Now, let us solve (b).

(b). It is given in the question that the plane is perpendicular to OA.

And we know that,

A normal is an object such as a line or vector that is perpendicular to a given object.

So, we can say that,

[∵, ]

Also,

Vector equation of a plane where is the normal to the plane and passing through is,

Where,

Here, A(a, b, c) is the given point in the plane.

Substituting the respective vectors, we get

⇒ a(x – a) + b(y – b) + c(z – c) = 0

We can either further simplify or leave it as be.

From simplifying, we get

⇒ ax – a² + by – b² + cz – c² = 0

⇒ ax + by + cz – a² – b² – c² = 0

⇒ a² + b² + c² = ax + by + cz

Thus, the required equation of the plane is a² + b² + c² = ax + by + cz.