Question

Write the vector equation of the plane passing through the point (a, b, c) and parallel to the plane \[\vec{r}\cdot \left( \hat{i}+\hat{j}+\hat{k} \right)=2\]

Answer 1

Hint: To solve this question, we will use general equation of a plane where normal vector is $\vec{n}$ and distance $\vec{d}$ the general equation of a plane with normal vector $\vec{n}$ and distance $\vec{d}$ from point of a position vector $\vec{r}$ is given by: \[\vec{r}\cdot \vec{n}=\vec{d}\]
We will compare this to \[\vec{r}\cdot \left( \hat{i}+\hat{j}+\hat{k} \right)=2\] and try to get the value of 2 in term of a, b and c. In between while calculation, we will use dot product of two vector.

Complete step-by-step solution:
Dot product of two vector $\vec{p}\text{ and }\vec{q}$ where \[p=x\hat{i}+y\hat{j}+z\hat{k}\text{ and q=}x'\hat{i}+y'\hat{j}+z'\hat{k}\] given by
\[\vec{p}\cdot \vec{q}=xx'+yy'+zz\text{ }\!\!'\!\!\text{ }\]
The general equation of a plane with normal vector $\vec{n}$ and distance $\vec{d}$ from point of a position vector $\vec{r}$ is given by:
\[\vec{r}\cdot \vec{n}=\vec{d}\]

Comparison our given equation from given equation \[\vec{r}\cdot \left( \hat{i}+\hat{j}+\hat{k} \right)=2\] we get that the normal vector \[\vec{n}=\hat{i}+\hat{j}+\hat{k}\] and distance d.
So, we have \[\vec{r}\cdot \left( \hat{i}+\hat{j}+\hat{k} \right)=d\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}\]
Where \[\hat{i}+\hat{j}+\hat{k}\] is the normal vector.
We are given that, the point is (a, b, c) then, writing this \[\left( a,b,c \right)=a\hat{i}+b\hat{j}+c\hat{k}\] is the position vector of the point (a, b, c).
Position vector of a point (p, q, r) is given as \[p\hat{i}+q\hat{j}+r\hat{k}\] because \[\vec{r}=a\hat{i}+b\hat{j}+c\hat{k}\] is the position vector of point (a, b, c).
Substituting this \[\vec{r}=a\hat{i}+b\hat{j}+c\hat{k}\] in equation (i) we get
\[\Rightarrow \left( a\hat{i}+b\hat{j}+c\hat{k} \right)\cdot \left( \hat{i}+\hat{j}+\hat{k} \right)=d\]
Dot product of two vector $\vec{p}\text{ and }\vec{q}$ where \[p=x\hat{i}+y\hat{j}+z\hat{k}\text{ and q=}x'\hat{i}+y'\hat{j}+z'\hat{k}\] given by
\[\vec{p}\cdot \vec{q}=xx'+yy'+zz\text{ }\!\!'\!\!\text{ }\]
Using this above to get dot product of above vector, we get:
\[\Rightarrow a+b+c=d\]
So, we have value of \[d=a+b+c\]
Substituting value of \[d=a+b+c\] in equation (i) we get
\[\vec{r}\cdot \left( \hat{i}+\hat{j}+\hat{k} \right)=a+b+c\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)}\]
This is the vector equation of required plane.
Substituting $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\text{ }$ in equation (ii) we get:
\[\left( x\hat{i}+y\hat{j}+z\hat{k}\text{ } \right)\cdot \left( \hat{i}+\hat{j}+\hat{k} \right)=a+b+c\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (iii)}\]
Both equation (ii) and equation (iii) are our required vector equation of a plane passing through the point (a, b, c) and parallel to the plane \[\vec{r}\cdot \left( \hat{i}+\hat{j}+\hat{k} \right)=2\]

Note: We can also use \[\vec{r}\cdot \left( \hat{i}+\hat{j}+\hat{k} \right)=2\] as given in the question and substitute d = 2 then, the value of \[a+b+c=2\] then, we can directly substitute \[a+b+c=2\] in equation (ii) that is \[\vec{r}\cdot \left( \hat{i}+\hat{j}+\hat{k} \right)=a+b+c=2\Rightarrow \vec{r}\cdot \left( \hat{i}+\hat{j}+\hat{k} \right)=2\] then, the final answer would be \[\left( x\hat{i}+y\hat{j}+z\hat{k}\text{ } \right)\cdot \left( \hat{i}+\hat{j}+\hat{k} \right)=2\] which is required result.