The coordinates of the point where the line through (3,−4,−5) and (2,−3,1) crosses the plane passing through three points (2,2,1),(3,0,1) and (4,−1,0) is

Engineering

Mathematics

Plane and Its Different Forms

Distance from a Plane

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

Question

The coordinates of the point where the line through (3,−4,−5) and (2,−3,1) crosses the plane passing through three points (2,2,1),(3,0,1) and (4,−1,0) is

(−1,2,−7)

(1,−2,7)

None of these

(1,2,7)

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The equation of line AB is given by

$\vec{r} = (3 \hat{i} - 4 \hat{j} - 5 \hat{k}) + λ (\hat{i} - \hat{j} - 6 \hat{k})$

The normal to the plane will be

$\vec{LM} \times \vec{MN}$

$= (\hat{i} - 2 \hat{j}) \times (- \hat{i} + \hat{j} + \hat{k}) = 0 + \hat{k} - \hat{j} - 2 \hat{k} - 2 \hat{i} = - 2 \hat{i} - \hat{j} - \hat{k}$

The equation of plane is given by 2x + y + z + d = 0

Since (2,2,1) lies on the plane, we get 4 + 2 + 1 + d = 0

⇒ d = –7

The palne becomes 2x + y + z – 7 = 0

let the point P on the line be (3 + t, –4–t, –5–6t)

Since this lies on the plane 6 + 2t – 4–t–5–6t–7= 0

⇒ –5t – 10 = 0 or t = –2

P = (1, –2,7)

∴ Correct answer is (c).