(i) Find the equation of the plane...

MathematicsEquation of a PlaneJEE Advanced 2003Moderate

(i) Find the equation of the plane passing through the points $(2, 1, 0), (5, 0, 1)$ and $(4, 1, 1)$ . (ii) If $P$ is the point $(2, 1, 6)$ then find the point $Q$ such that $P Q$ is perpendicular to the plane in (i) and the midpoint of $P Q$ lies on it.

Answer:(i)

x + y - 2 z = 3

(ii)

Q (6, 5, - 2)

Visualized Solution (Hindi)

Visualizing the Problem

Given points: $A (2, 1, 0)$ , $B (5, 0, 1)$ , and $C (4, 1, 1)$
Objective: Find the equation of the plane passing through $A, B,$ and $C$

The Determinant Formula

The equation of a plane passing through $(x_{1}, y_{1}, z_{1})$ , $(x_{2}, y_{2}, z_{2})$ , and $(x_{3}, y_{3}, z_{3})$ is:
$x - x_{1} x_{2} - x_{1} x_{3} - x_{1} y - y_{1} y_{2} - y_{1} y_{3} - y_{1} z - z_{1} z_{2} - z_{1} z_{3} - z_{1} = 0$

Substituting the Coordinates

Substituting $A (2, 1, 0)$ , $B (5, 0, 1)$ , and $C (4, 1, 1)$ :
$x - 2 5 - 2 4 - 2 y - 1 0 - 1 1 - 1 z - 0 1 - 0 1 - 0 = 0$

Simplifying the Determinant

Simplifying the rows:
$x - 2 32 y - 1 - 1 0 z 11 = 0$

Expanding the Determinant

Expanding along the first row:
$(x - 2) (- 1 - 0) - (y - 1) (3 - 2) + z (0 - (- 2)) = 0$

Final Plane Equation

Simplifying the expression:
$- 1 (x - 2) - 1 (y - 1) + 2 z = 0$
$- x + 2 - y + 1 + 2 z = 0$
Equation of the plane: $x + y - 2 z = 3$

Introducing Point P and Image Q

Given point $P (2, 1, 6)$
Let $Q (α, β, γ)$ be the image of $P$ with respect to the plane $x + y - 2 z = 3$
Line $P Q$ is perpendicular to the plane, and its midpoint $M$ lies on the plane.

Direction of Line PQ

Direction ratios of the normal to the plane: $(1, 1, - 2)$
Equation of line $P Q$ passing through $P (2, 1, 6)$ :
$\frac{x - 2}{1} = \frac{y - 1}{1} = \frac{z - 6}{- 2} = λ$

General Coordinates of Q

Coordinates of any point $Q$ on the line:
$α = λ + 2$
$β = λ + 1$
$γ = - 2 λ + 6$

Finding the Midpoint M

Midpoint $M$ of $P Q$ :
$M = (\frac{( λ + 2 ) + 2}{2}, \frac{( λ + 1 ) + 1}{2}, \frac{( - 2 λ + 6 ) + 6}{2})$
$M = (\frac{λ + 4}{2}, \frac{λ + 2}{2}, \frac{12 - 2 λ}{2})$

M Lies on the Plane

Since $M$ lies on $x + y - 2 z = 3$ :
$(\frac{λ + 4}{2}) + (\frac{λ + 2}{2}) - 2 (\frac{12 - 2 λ}{2}) = 3$

Solving for Lambda

Multiplying by $2$ :
$(λ + 4) + (λ + 2) - 2 (12 - 2 λ) = 6$
$λ + 4 + λ + 2 - 24 + 4 λ = 6$
$6 λ - 18 = 6 \Rightarrow 6 λ = 24 \Rightarrow λ = 4$

Calculating Coordinates of Q

Substituting $λ = 4$ into $Q (α, β, γ)$ :
$α = 4 + 2 = 6$
$β = 4 + 1 = 5$
$γ = - 2 (4) + 6 = - 2$
Point Q: $(6, 5, - 2)$

Summary and Conclusion

Key Takeaways:
Equation of plane: $x + y - 2 z = 3$
Image point $Q$ : $(6, 5, - 2)$
The line $P Q$ is parallel to the normal vector $n = \hat{i} + \hat{j} - 2 \hat{k}$

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The equation of a plane passing through the line of intersection of the planes $x + 2 y + 3 z = 2$ and $x - y + z = 3$ and at a distance $\frac{2}{3}$ from the point $(3, 1, - 1)$ is

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The unit vector perpendicular to the plane determined by $P (1, - 1, 2)$ , $Q (2, 0, - 1)$ and $R (0, 2, 1)$ is .........

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Visualized Solution (Hindi)

Visualizing the Problem

The Determinant Formula

Substituting the Coordinates

Simplifying the Determinant

Expanding the Determinant

Final Plane Equation

Introducing Point P and Image Q

Direction of Line PQ

General Coordinates of Q

Finding the Midpoint M

M Lies on the Plane

Solving for Lambda

Calculating Coordinates of Q

Summary and Conclusion

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