Figure-out the distance of a point (2, 5, -3) from the plane \(\vec{r}\). \((6\hat{i}-3\hat{j}+2\hat{k})\) = 6

Question

Question

This question was previously asked in

AAI ATC Junior Executive 25 March 2021 Official Paper (Shift 3)

Answer 1

Given:

Point = (2, 5, -3)

Equation of the plane : \(\vec{r}\).\((6̂{i}-3̂{j}+2̂{k})\)= 6

Concept:

Equation of a plane in cartesian form can be written by putting \(\vec{r} = x̂{i} + ŷ{j} + ẑ{k}\) in the vector equation of the plane.

Formula:

The distance of a point (p, q, r) from a plane ax + by + cz = d is given by,

\(D = |\frac{ap + bq + rc -d}{\sqrt{a^2 + b^2 + c^2 }}|\)

Solution:

Writing the equation of a plane in the cartesian form :

\(\vec{r}\).\((6̂{i}-3̂{j}+2̂{k})\)= 6

⇒ (xî + yĵ + zk̂).(6î - 3ĵ + 2k̂) = 6

⇒ 6x -3y + 2z = 6

Now, using distance formula -

D = \(|\frac{6(2) - 3(5) + 2(-3) -6}{\sqrt{6^2 + (-3)^2 + 2^2 }}|\)

⇒ D = 15/7