Correct Answer - Option 2 : (2, 1, -2)
Concept:
Vector equation of a plane whose perpendicular distance from the origin is d and \(\hat n\) is the unit normal vector to the plane through origin is given by \(\vec{r}.\hat{n}=d\)
Calculation:
Let the coordinate of foot of the perpendicular P from the origin to the plane is (a, b, c).
Then, the direction ratios of the line joining origin to point P are a, b, c.
Now, \(2\hat i + \hat j - 2\hat k\) is the normal vector to the plane 2x + y - 2z = 9
⇒ \(|2\hat{i}+\hat{j}-2\hat{k}| =\sqrt{2^2+1^2+(-2)^2}=3\).
Therefore, Divide both side the plane equation by 3, we get
⇒ \(\frac{2}{3}x+\frac{1}{3}y-\frac{2}{3}z=\frac{9}{3}\)
⇒ \(\frac{2}{3}x+\frac{1}{3}y-\frac{2}{3}z=3\).
This is the equation of plane in the form \(\vec{r}.\hat{n}=d\).
Therefore, direction cosine of line joining origin to point P is:\(\frac{2}{3},\frac{1}{3},\frac{-2}{3}\).
∵ direction cosine and direction ratios are proportional, we have
⇒ \(\frac{a}{\frac{2}{3}} = \frac{b}{\frac{1}{3}} = \frac{c}{\frac{-2}{3}} = k\)
⇒ a = 2k/3, b = k/3 and c = -2k/3.
Now, point (a, b, c) lies on the plane 2x + y - 2z = 9. So, the point (a, b, c) will satisfy the equation of plane
⇒ \(\frac{4k}{3}+\frac{k}{3}+\frac{4k}{3}= 9\) ⇒ 3k = 9 ⇒ k = 3.
So, coordinates of point P = (a, b, c) = (2, 1, -2)
Hence, option 2 is correct.
