Find the distance of the point (–1,–5,–10) from the point of intersection of the line and plane .

Engineering

Mathematics

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

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Eqaution of line is $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{2}$

(–1,–5,–10)

Any point on the line is

(2 + 3λ, –1 + 4λ, 2 + 2λ).

But this point lies on the plane,

(2 + 3λ) – (–1 + 4λ) + (2 + 2λ) = 5

point of interesction (2,–1,2)

distance of point

P(–1,–5,–10) from (2,–1,2)

$= \sqrt{9 + 16 + 144} = \sqrt{160} = 13$