To find the Cartesian equation of the given plane, we can rewrite the parametric equation of the plane as:
r ⃗ = k ^ + t ( − i ^ + j ^ ) + p ( − i ^ + k ^ ) \vec{r} = \hat{k} + t(-\hat{i} + \hat{j}) + p(-\hat{i} + \hat{k}) r = k ^ + t ( − i ^ + j ^ ) + p ( − i ^ + k ^ )
The plane passes through the point ( 0 , 0 , 1 ) (0, 0, 1) ( 0 , 0 , 1 ) u ⃗ = − i ^ + j ^ and v ⃗ = − i ^ + k ^ \vec{u} = -\hat{i} + \hat{j} \quad \text{and} \quad \vec{v} = -\hat{i} + \hat{k} u = − i ^ + j ^ and v = − i ^ + k ^
The normal vector n ⃗ \vec{n} n u ⃗ \vec{u} u v ⃗ \vec{v} v n ⃗ = u ⃗ × v ⃗ = ∣ i ^ j ^ k ^ − 1 1 0 − 1 0 1 ∣ = i ^ ( 1 − 0 ) − j ^ ( − 1 − 0 ) + k ^ ( 0 − ( − 1 ) ) = i ^ + j ^ + k ^ \vec{n} = \vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{vmatrix} = \hat{i}(1 - 0) - \hat{j}(-1 - 0) + \hat{k}(0 - (-1)) = \hat{i} + \hat{j} + \hat{k} n = u × v = i ^ − 1 − 1 j ^ 1 0 k ^ 0 1 = i ^ ( 1 − 0 ) − j ^ ( − 1 − 0 ) + k ^ ( 0 − ( − 1 )) = i ^ + j ^ + k ^
Thus, the Cartesian equation of the plane is:
1 ( x − 0 ) + 1 ( y − 0 ) + 1 ( z − 1 ) = 0 ⟹ x + y + z − 1 = 0 1(x - 0) + 1(y - 0) + 1(z - 1) = 0 \implies x + y + z - 1 = 0 1 ( x − 0 ) + 1 ( y − 0 ) + 1 ( z − 1 ) = 0 ⟹ x + y + z − 1 = 0
The given point is Q ( 10 , 15 , 20 ) Q(10, 15, 20) Q ( 10 , 15 , 20 ) S ( α , β , γ ) S(\alpha, \beta, \gamma) S ( α , β , γ )
Using the formula for the reflection of a point ( x 1 , y 1 , z 1 ) (x_1, y_1, z_1) ( x 1 , y 1 , z 1 ) a x + b y + c z + d = 0 ax + by + cz + d = 0 a x + b y + cz + d = 0 α − x 1 a = β − y 1 b = γ − z 1 c = − 2 a x 1 + b y 1 + c z 1 + d a 2 + b 2 + c 2 \frac{\alpha - x_1}{a} = \frac{\beta - y_1}{b} = \frac{\gamma - z_1}{c} = -2\frac{a x_1 + b y_1 + c z_1 + d}{a^2 + b^2 + c^2} a α − x 1 = b β − y 1 = c γ − z 1 = − 2 a 2 + b 2 + c 2 a x 1 + b y 1 + c z 1 + d
Substituting a = 1 , b = 1 , c = 1 , d = − 1 a = 1, b = 1, c = 1, d = -1 a = 1 , b = 1 , c = 1 , d = − 1 ( x 1 , y 1 , z 1 ) = ( 10 , 15 , 20 ) (x_1, y_1, z_1) = (10, 15, 20) ( x 1 , y 1 , z 1 ) = ( 10 , 15 , 20 ) α − 10 1 = β − 15 1 = γ − 20 1 = − 2 10 + 15 + 20 − 1 1 2 + 1 2 + 1 2 = − 2 ⋅ 44 3 = − 88 3 \frac{\alpha - 10}{1} = \frac{\beta - 15}{1} = \frac{\gamma - 20}{1} = -2 \frac{10 + 15 + 20 - 1}{1^2 + 1^2 + 1^2} = -2 \cdot \frac{44}{3} = -\frac{88}{3} 1 α − 10 = 1 β − 15 = 1 γ − 20 = − 2 1 2 + 1 2 + 1 2 10 + 15 + 20 − 1 = − 2 ⋅ 3 44 = − 3 88
From this, we solve for α , β , \alpha, \beta, α , β , γ \gamma γ α = 10 − 88 3 = − 58 3 \alpha = 10 - \frac{88}{3} = -\frac{58}{3} α = 10 − 3 88 = − 3 58 β = 15 − 88 3 = − 43 3 \beta = 15 - \frac{88}{3} = -\frac{43}{3} β = 15 − 3 88 = − 3 43 γ = 20 − 88 3 = − 28 3 \gamma = 20 - \frac{88}{3} = -\frac{28}{3} γ = 20 − 3 88 = − 3 28
Now, checking the options:
Option (A):
3 ( α + β ) = 3 ( − 58 3 − 43 3 ) = − 58 − 43 = − 101 (TRUE) 3(\alpha + \beta) = 3\left(-\frac{58}{3} - \frac{43}{3}\right) = -58 - 43 = -101 \quad \text{(TRUE)} 3 ( α + β ) = 3 ( − 3 58 − 3 43 ) = − 58 − 43 = − 101 (TRUE)
Option (B):
3 ( β + γ ) = 3 ( − 43 3 − 28 3 ) = − 43 − 28 = − 71 (TRUE) 3(\beta + \gamma) = 3\left(-\frac{43}{3} - \frac{28}{3}\right) = -43 - 28 = -71 \quad \text{(TRUE)} 3 ( β + γ ) = 3 ( − 3 43 − 3 28 ) = − 43 − 28 = − 71 (TRUE)
Option (C):
3 ( γ + α ) = 3 ( − 28 3 − 58 3 ) = − 28 − 58 = − 86 (TRUE) 3(\gamma + \alpha) = 3\left(-\frac{28}{3} - \frac{58}{3}\right) = -28 - 58 = -86 \quad \text{(TRUE)} 3 ( γ + α ) = 3 ( − 3 28 − 3 58 ) = − 28 − 58 = − 86 (TRUE)
Option (D):
3 ( α + β + γ ) = 3 ( − 58 + 43 + 28 3 ) = − 129 ≠ − 121 (FALSE) 3(\alpha + \beta + \gamma) = 3\left(-\frac{58 + 43 + 28}{3}\right) = -129 \neq -121 \quad \text{(FALSE)} 3 ( α + β + γ ) = 3 ( − 3 58 + 43 + 28 ) = − 129 = − 121 (FALSE)
Therefore, the correct options are A, B, and C .