Question

If \[O\] is the origin and is the point $(a, b, c)$, then the equation of the plane through A and at right angles to \[OA\] is

Answer 1

Hint: In the following question, we have to find the equation of plane passing through the point A $(a, b, c)$ whose normal is \[OA\]
Normal is the line parallel to the plane. The general equation of the line in normal form is:
\[({{n}_{1}}\widehat{i}+{{n}_{2}}\widehat{j}+{{n}_{3}}\widehat{k})\centerdot (x\widehat{i}+y\widehat{j}+z\widehat{k})=P\]
Here, \[({{n}_{1}}\widehat{i}+{{n}_{2}}\widehat{j}+{{n}_{3}}\widehat{k})\] is the vector equation of the normal and P is the perpendicular distance.

Complete step by step solution:
As we know that \[OA\] is the normal and point A $(a, b, c)$ is the foot of the perpendicular distance then the direction ratios of the normal will be: \[a-0\], \[b-0\] and \[c-0\].
Therefore, f=direction ratios of \[OA\] are: a, b, c.
So, the vector equations of normal will be \[a\widehat{i}+b\widehat{j}+c\widehat{k}\].
Here,\[P=(a\widehat{i}+b\widehat{j}+c\widehat{k})\centerdot (a\widehat{i}+b\widehat{j}+c\widehat{k})\]
 \[P={{a}^{2}}+{{b}^{2}}+{{c}^{2}}\]
As (a, b, c) are the direction ratios of the normal but $(a, b, c)$ is also the point that lies on the plane.
So, equation of the plane will be:
\[(a\widehat{i}+b\widehat{j}+c\widehat{k})\centerdot (x\widehat{i}+y\widehat{j}+z\widehat{k})={{a}^{2}}+{{b}^{2}}+{{c}^{2}}\]
\[ax+by+cz={{a}^{2}}+{{b}^{2}}+{{c}^{2}}\]
\[ax+by+cz-{{a}^{2}}-{{b}^{2}}-{{c}^{2}}=0\]
The equation can be rearranged as follows:
\[ax-{{a}^{2}}+by-{{b}^{2}}+cz-{{c}^{2}}=0\]
Now, take the common terms:
\[a(x-a)+b(y-b)+c(z-c)=0\]

Equation of a plane is \[a(x-a)+b(y-b)+c(z-c)=0\].

Note: Always remember when a plane is passing through the point say \[({{a}_{1}},{{a}_{2}},{{a}_{3}})\] consider it as \[\overrightarrow{a}\] and has a perpendicular vector (normal vector) say \[\overrightarrow{b}\].
Then equation of plane will be
\[(\overrightarrow{r}-\overrightarrow{a})\centerdot \overrightarrow{b}=0\]
\[\overrightarrow{r}\centerdot \overrightarrow{b}=\overrightarrow{a}\centerdot \overrightarrow{b}\]
Here, \[\overrightarrow{r}=(x\widehat{i}+y\widehat{j}+z\widehat{k})\]
\[\overrightarrow{a}=({{a}_{1}}\widehat{i}+{{a}_{2}}\widehat{j}+{{a}_{3}}\widehat{k})\]
And \[\overrightarrow{b}=({{b}_{1}}\widehat{i}+{{b}_{2}}\widehat{j}+{{b}_{3}}\widehat{k})\] (say)
So, equation of plane in Cartesian form will be: \[{{b}_{1}}x+{{b}_{2}}y+{{b}_{3}}z={{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}}\]
Don’t confuse between the dot and cross product. In this question we used dot product as follow:
\[\widehat{i}\widehat{i}=\widehat{j}\widehat{j}=\widehat{k}\widehat{k}=1\]
And \[\widehat{i}\widehat{j}=\widehat{j}\widehat{i}=\widehat{i}\widehat{k}=\widehat{k}\widehat{i}=\widehat{j}\widehat{k}=\widehat{k}\widehat{j}=0\]
Dot product always gives the scalar quantity while cross product gives the vector quantities which have both magnitude as well as direction. But in the scalar product we get only magnitude not direction.