Find the distance of the point from the point of intersection of the line and the plane .

Engineering

Mathematics

Distance Formulae and Its Application

Distance from a Plane

Equation of Straight Line in 3D and Distance Of a Point From Line

Question

Find the distance of the point $(- 1, - 5, - 10)$ from the point of intersection of the line $\vec{r} = (2 \hat{i} - \hat{j} + 2 \hat{k}) + λ (3 \hat{i} + 4 \hat{j} + 2 \hat{k})$ and the plane $\vec{r} \cdot (\hat{i} - \hat{j} + \hat{k}) = 5$ .

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Hint : To find the intersection of line and plane put the value of r from equation of line into the equation of plane

Step 1:

The equation of the given line is

\vec{r} = (2 \hat{i} - \hat{j} + 2 \hat{k}) + λ (3 \hat{i} + 4 \hat{j} + 2 \hat{k}) .. .. . (1)

The equation of the given plane is

\vec{r} . (\hat{i} - \hat{j} + \hat{k}) = 5 ........ (2)

Step 2 :Substituting the value of

\vec{r}

in equation (2), we get,

[(2 \hat{i} - \hat{j} + 2 \hat{k}) + λ (3 \hat{i} + 4 \hat{j} + 2 \hat{k})] [\hat{i} - \hat{j} + \hat{k}] = 5

⟹ [(2 + 3 λ) \hat{i} + (4 λ - 1) \hat{j} + (2 + 2 λ) \hat{k}] [\hat{i} - \hat{j} + \hat{k}] = 5

⇒ (2 + 3λ) – (4λ – 1) + (2λ + 2) = 5
⇒ λ = 0
Thus the point of intersection of the given line and the plane is,

\vec{r} = 2 \hat{i} - \hat{j} + 2 \hat{k} = (2, - 1, 2)

This means that the position vector of the point of intersection of the line and plane is given by the coordinates (2,–1,2). The point is (–1,–5,–10).

Hence, distance d between the points (2,–1,2) and (–1,–5,–10) is

d = \sqrt{{(- 1 - 2)}^{2}} + {(- 5 + 1)}^{2} + {(- 10 - 2)}^{2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13