Determine the direction cosines of the unit vector perpendicular to the plane \(\vec{r} \cdot(2 \hat{\imath}-6 \hat{\jmath}-3 \hat{k})+1=0\) passing through the origin?

Given:

Equation of plane is \({\vec r} \cdot(2 \hat{\imath}-6 \hat {\jmath}-3 {\hat k})+1=0\)

Concept:

Vector equation of a plane at a distance 'd' from the origin and unit vector to normal form origin \(̂ n\) is \(\vec r. \hat n = d\)

Formula Used:

\(\vec n = \hat n = \frac{1}{|\vec n|}(\vec n)\)

Calculation:

We have,

⇒ \(⃗{r} \cdot(2 \hat{\imath}-6 \hat{\jmath}-3 \hat{k})+1=0\)

⇒ \(⃗{r} \cdot(2 \hat{\imath}-6 \hat{\jmath}-3 \hat{k}) = -1\)

Multiplying with -1 on both sides,

⇒ \(-\vec {r} \cdot(2 \hat{\imath}-6 \hat{\jmath}-3 \hat{k}) = -1 \times -1 \)

⇒ \({\vec r} \cdot(-2 \hat{\imath}+6 \hat{\jmath}+3 \hat{k}) = 1\)     ----- equation (1)

So, \(\vec n = - 2 \hat i + 6 \hat j + 3 \hat k\)

Magnitude of \(\vec n\) = \(\sqrt{ (-2)^2 + 6^2 + 3^2}\)

⇒ \(|\vec n|\) = \(\sqrt{4 + 36 + 9}\) = \(\sqrt{49}\) = 7

Now, \(\hat n = \frac{1}{|\vec n|}(\vec n)\)

⇒ \(\hat n = \frac{1}{7}(- 2 \hat i + 6 \hat j + 3 \hat k)\)

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

∴ Direction cosines of the unit vector perpendicular to the given plane are \(\frac{-2}{7}, \frac{6}{7}, \frac{3}{7}\)