Question

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

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Therefore the equation of the plane in vector form is

$\vec{r}.(\frac{3\hat{i}+4\hat{j}-2\hat{k}}{\sqrt{29} } )=\frac{5}{\sqrt{29} }$

Step-by-step explanation:

Given equation of plane is

3x+4y-2z=5

The direction ratio of the plane is (3,4,-2)    [The coefficient of x,y and z]

The vector form of direction ratio(d)= $(3\hat{i}+4\hat{j}-2\hat{k})$

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

Then, $\vec{r}.d$

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

       $=3x+4y-2z$

The unit vector of $(a\hat{i}+b\hat{j}+c\hat{k})$ is $\frac{(a\hat{i}+b\hat{j}+c\hat{k})}{\sqrt{a^2+b^2+c^2} }$

Since $(3\hat{i}+4\hat{j}-2\hat{k})$ is not a unit vector.

Therefore the unit vector of $(3\hat{i}+4\hat{j}-2\hat{k})$ is $\frac{(3\hat{i}+4\hat{j}-2\hat{k})}{\sqrt{3^2+4^2+(-2)^2} }$ $=\frac{(3\hat{i}+4\hat{j}-2\hat{k})}{\sqrt{29} }$

Therefore the above equation of plane can be rewrite as,

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

$\Rightarrow \vec{r}.(\frac{3\hat{i}+4\hat{j}-2\hat{k}}{\sqrt{29} } )=\frac{5}{\sqrt{29} }$    [dividing $\sqrt{29}$ both sides of the equation for the unit vector]

Therefore the equation of the plane in vector form is

$\vec{r}.(\frac{3\hat{i}+4\hat{j}-2\hat{k}}{\sqrt{29} } )=\frac{5}{\sqrt{29} }$