Question

Write the equation of the plane 3x+4y-2z=5 in the vector form

Answer 1

Answer:

$\textbf{Vector form of the given plane is}$

$\bf\vec{r}.(3\vec{i}+4\vec{i}-2\vec{k})=5}$

Step-by-step explanation:

$\textbf{Concept:}$

$\text{The equation of any plane ax+by+cz+d=0 can be}$

$\text{written as in vector form as \vec{r}.(a\vec{i}+b\vec{j}+c\vec{k})=q }$

$\text{That is, \vec{r}.\vec{n}=q }$

$\text{Given plane is 3x+4y-2z=5}$

$\text{It can be written as}$

$\bf\vec{r}.(3\vec{i}+4\vec{i}-2\vec{k})=5}$

Answer 2

The equation of plane, $\vec{r}{\cdot} (3i+4j-2k)=5$

Step-by-step explanation:

We need to find the equation of the plane 3x+4y-2z=5 in the vector form.

Let $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$

Direction ratio of the equation of plane is, $d=3\vec{i}+4\vec{j}-2\vec{k}$

The equation of plane in vector form is given by :

$\vec{r}{\cdot} \hat{n}=d$

Then,

$\vec{r}.d\\\\ =(x\hat{i}+y\hat{j}+z\hat{k}).(3\hat{i}+4\hat{j}-2\hat{k})\\\\=3x+4y-2z$

So, the equation of plane is given by :

$\vec{r}{\cdot} (3i+4j-2k)=5$

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Equation of plane

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