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The distance of the plane  \vec{r}.\left ( \frac{2}{7}\hat{i}+\frac{3}{7}\hat{j}-\frac{6}{7}\hat{k} \right )=1   from the origin is:

A. 1
B. 7
C. 1/7
D. None of these

Answers (1)

Given plane is

\vec{r}.\left ( \frac{2}{7}\hat{i}+\frac{3}{7}\hat{j}-\frac{6}{7}\hat{k} \right )=1

Let

\vec{n}=\frac{2}{7}\hat{i}+\frac{3}{7}\hat{j}-\frac{6}{7}\hat{k}

\left |\vec{n} \right |=\sqrt{\frac{2}{7}\hat{i}+\frac{3}{7}\hat{j}-\frac{6}{7}\hat{k} }=1

=> n is a unit vector

Thus, the equation of the plane is of the form  \vec{r}.\hat{n} =d, where n is

the unit vector and d is the distance from the origin.

Comparing, we get d =1, hence the distance of the plane from origin is 1        

   (Option A)

Posted by

infoexpert24

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