Question

A rectangle whose vertices are

(5, 3, - 3), (5, 9, 9), (0, 5, 11)

and

(0, - 1, - 1)

is rotated about its diagonal (whose direction cosines are

(\frac{1}{3}, \frac{2}{3}, \frac{2}{3})

in such a way that new position of rectangle is perpendicular to its old position find the coordinates of new position of the vertices whose position is changed.

Accepted Answer

Let, $A = (5, 3, - 3), B = (5, 9, 9), C (0, 5, 11) & D = (0, - 1, - 1)$ are the vertices of a rectangle $A B C D$ .

$\Rightarrow \vec{A C} = - 5 \hat{i} + 2 \hat{j} + 12 \hat{k} & \vec{D B} = 5 \hat{i} + 10 \hat{j} + 10 \hat{k}$

The direction cosine of diagonals $A C & D B$ are $(\frac{- 5}{\sqrt{173}}, \frac{2}{\sqrt{173}}, \frac{12}{\sqrt{173}}) & (\frac{1}{3}, \frac{2}{3}, \frac{2}{3})$ respectively.

According to the question the rectangle $A B C D$ is rotated about the diagonal $D B$ .

Let, $A' B C' D$ is the new rectangle such that $A' B C' D & A B C D$ are perpendicular to each other.

Let, $A' = (α, β, γ)$

The equation of planes $A B C D & A' B C' D$ are $|\begin{matrix} x - 5 & y - 3 & z + 3 \\ 5 - 5 & 9 - 3 & 9 + 3 \\ 0 - 5 & - 1 - 3 & - 1 + 3 \end{matrix}| = 0 & |\begin{matrix} x - α & y - β & z - γ \\ 5 - α & 9 - β & 9 - γ \\ 0 - α & - 1 - β & - 1 - λ \end{matrix}| = 0$ .

$\Rightarrow 2 x - 2 y + z - 1 = 0 & 2 (β - γ) x + (1 + γ - 2 α) y - (1 + β - 2 α) z - (β - γ) = 0$

$∵$ The planes are perpendicular to each other.

$∴ 2 \cdot 2 (β - γ) - 2 \cdot (1 + γ - 2 α) - (1 + β - 2 α) = 0$

$\Rightarrow (4 β - 4 γ) - (2 + 2 γ - 4 α) - (1 + β - 2 α) = 0$

$\Rightarrow β = 2 γ - 2 α + 1 . . . . (i)$

Here, $|A' B| = |A B| & |A' D| = |A D|$

$\Rightarrow \sqrt{{(α - 5)}^{2} + {(β - 9)}^{2} + {(γ - 9)}^{2}} = \sqrt{{(5 - 5)}^{2} + {(3 - 9)}^{2} + {(- 3 - 9)}^{2}} & \sqrt{{(α - 0)}^{2} + {(β + 1)}^{2} + {(γ + 1)}^{2}} = \sqrt{{(5 - 0)}^{2} + {(3 + 1)}^{2} + {(- 3 + 1)}^{2}}$

$\Rightarrow {(α - 5)}^{2} + {(2 γ - 2 α + 1 - 9)}^{2} + {(γ - 9)}^{2} = 180 & α^{2} + {(2 γ - 2 α + 1 + 1)}^{2} + {(γ + 1)}^{2} = 45$

$\Rightarrow {(α - 5)}^{2} + 4 {(γ - α - 4)}^{2} + {(γ - 9)}^{2} = 180 & α^{2} + 4 {(γ - α + 1)}^{2} + {(γ + 1)}^{2} = 45$

$\Rightarrow (α^{2} - 10 α + 25) + 4 (γ^{2} + α^{2} + 16 - 2 γ α + 8 α - 8 γ) + (γ^{2} - 18 γ + 81) = 180 & α^{2} + 4 (γ^{2} + α^{2} + 1 - 2 γ α - 2 α + 2 γ) + (γ^{2} + 2 γ + 1) = 45$

$\Rightarrow 5 α^{2} - 8 γ α + 5 γ^{2} = 10 - 22 α + 50 γ . . . . . (i i)$

And $5 α^{2} - 8 γ α + 5 γ^{2} = 40 + 8 α - 10 γ . . . . . (i i i)$

$\Rightarrow 10 - 22 α + 50 γ = 40 + 8 α - 10 γ$

$\Rightarrow α = 2 γ - 1 . . . . . (i v)$

From the equation $(i i) & (i v)$

$\Rightarrow 5 {(2 γ - 1)}^{2} - 8 γ (2 γ - 1) + 5 γ^{2} = 10 - 22 (2 γ - 1) + 50 γ$

$\Rightarrow 9 γ^{2} - 18 γ - 27 = 0$

$\Rightarrow γ^{2} - 3 γ + γ - 3 = 0$

$\Rightarrow (γ - 3) (γ + 1) = 0$

$\Rightarrow γ = - 1, 3$

$\Rightarrow (α, β, γ) = (- 3, 5, - 1) & (5, - 3, 3)$

Let, if the rectangle $A B C D$ is rotated in opposite direction about the diagonal $B D$ such that the rectangle will become $A'' B C'' D$ .

Let, $A'' = (μ, ψ, ω)$

Here, the mid point of $A' A'', C' C'' & B D$ are same.

$\Rightarrow (\frac{μ + α}{2}, \frac{ψ + β}{2}, \frac{ω + γ}{2}) = (\frac{5 + 0}{2}, \frac{9 - 1}{2}, \frac{9 - 1}{2})$

$\Rightarrow (μ, ψ, ω) = (5 - α, 8 - β, 8 - γ)$

$\Rightarrow (μ, ψ, ω) = (8, 3, 9) & (0, 11, 5)$

Hence, the coordinates of new position of the vertices is either $(- 3, 5, - 1) & (5, - 3, 3)$ or $(8, 3, 9) & (0, 11, 5)$ .

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