Question

What is the distance of the plane $2x-3y+4z=6$ from the origin?

• A. $\frac{2}{\sqrt{29}}$
• B. $\frac{3}{\sqrt{29}}$
• C. $\frac{5}{\sqrt{29}}$
• D. $\frac{6}{\sqrt{29}}$

Answer 1

The correct option is D

$\frac{6}{\sqrt{29}}$

We know that the equation of plane in normal vector form is $\overrightarrow{r}\cdot \hat{n}=d$

Where $\hat{n}$ is the unit vector in the direction of normal.

Let $\overrightarrow{r}$ vector be equal to $x\hat{i}+y\hat{j}+z\hat{k}$

And $\hat{n}$ be equal to $l\hat{i}+m\hat{j}+n\hat{k}$

Where $l^2+m^2+n^2=1$

So the equation of plane will be -

$(x\hat{i}+y\hat{j}+z\hat{k}).(l\hat{i}+m\hat{j}+n\hat{k})=d$

Or $lx+my+nz=d$

Here '$d$' is the distance from origin.

The equation we are given in the question doesn’t have $l,m\text{and}n$ as the coefficients of $x,y\text{and}z$, Since $2^2+(-3{)}^2+(4{)}^2\ne 1$

So, we’ll find appropriate $l,m,n$ first.

Here, We can consider $2\hat{i}-3\hat{j}+4\hat{k}$ as a vector. And now we have to convert it to a unit vector.

We know that $\overrightarrow{n}=|n|.\hat{n}$

Thus, the unit vector here will be $=\frac{2i-3j+4k}{|2i-3j+4k|}$

$=\frac{2i-3j+4k}{\sqrt{29}}$

$So,l=\frac{2}{\sqrt{29}},m=\frac{-3}{\sqrt{29}},n=\frac{4}{\sqrt{29}}$

To have $l,m,n$ in the equation given we’ll divide both L.H.S and R.H.S by $\sqrt{29}$

So, we’ll have $\frac{2x-3y+4z}{\sqrt{29}}=\frac{6}{\sqrt{29}}$

$Or\frac{2}{\sqrt{29}}x-\frac{3}{\sqrt{29}}y+\frac{4}{\sqrt{29}}z=\frac{6}{\sqrt{29}}$

Since the coefficients of $x,y\text{and}z$ are such that they represent the direction cosines. The constant term on the R.H.S will be equal to the distance of the plane from origin.