Question

Find the equation of the plane which is at a distance of $\frac{6}{\sqrt{29}}$ from the origin and its normal vector from the origin is $2\hat{i}-3\hat{j}+4\hat{k}$

• A. $2x+3y+4z=6$
• B. $3x-4y+2z=6$
• C. $2x-3y+4z=6$
• D. $2x-3y-4z=6$

Answer 1

The correct option is C $2x-3y+4z=6$

Let
$\overrightarrow{n}=2\hat{i}-3\hat{j}+4\hat{k}\hat{n}=\frac{2\hat{i}-3\hat{j}+4\hat{k}}{\sqrt{4+9+16}}=\frac{2\hat{i}-3\hat{j}+4\hat{k}}{\sqrt{29}}$
And let r be a position vector$\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$
Hence, the required equation of the plane is
$\overrightarrow{r}.\hat{n}=\frac{6}{\sqrt{29}}\left(x\hat{i}+y\hat{j}+z\hat{k}\right).\left(\frac{2\hat{i}-3\hat{j}+4\hat{k}}{\sqrt{29}}\right)=\frac{6}{\sqrt{29}}\frac{2x-3y+4z}{\sqrt{29}}=\frac{6}{\sqrt{29}}2x-3y+4z=6$
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