Given: x ˉ = i ^ + 2 j ^ + 3 k ^ \text{Given: } \bar{x}=\hat{i}+2\hat{j}+3\hat{k} Given: x ˉ = i ^ + 2 j ^ + 3 k ^
y ˉ = 2 i ^ + 3 j ^ + k ^ \bar{y}=2\hat{i}+3\hat{j}+\hat{k} y ˉ = 2 i ^ + 3 j ^ + k ^
z ˉ = 3 i ^ + j ^ + 2 k ^ \bar{z}=3\hat{i}+\hat{j}+2\hat{k} z ˉ = 3 i ^ + j ^ + 2 k ^
X ‾ = α x ˉ + β y ˉ − z ˉ , Y ‾ = α y ˉ + β z ˉ − x ˉ \overline{X}=\alpha\bar{x}+\beta\bar{y}-\bar{z},\ \ \overline{Y}=\alpha\bar{y}+\beta\bar{z}-\bar{x} X = α x ˉ + β y ˉ − z ˉ , Y = α y ˉ + β z ˉ − x ˉ
Z ‾ = α z ˉ + β x ˉ − y ˉ \overline{Z}=\alpha\bar{z}+\beta\bar{x}-\bar{y} Z = α z ˉ + β x ˉ − y ˉ
X ‾ , Y ‾ , Z ‾ lie in a plane ⇒ find α + β − 3 \overline{X},\overline{Y},\overline{Z} \text{ lie in a plane} \ \Rightarrow\ \text{find } \alpha+\beta-3 X , Y , Z lie in a plane ⇒ find α + β − 3
[ x ˉ y ˉ z ˉ ] ≠ 0 ⇒ x ˉ , y ˉ , z ˉ are non-coplanar [\bar{x}\ \bar{y}\ \bar{z}] \neq 0 \ \Rightarrow \ \bar{x},\bar{y},\bar{z} \text{ are non-coplanar} [ x ˉ y ˉ z ˉ ] = 0 ⇒ x ˉ , y ˉ , z ˉ are non-coplanar Plan: 1. M from the coefficients of X ‾ , Y ‾ , Z ‾ \text{Plan: } 1.\ M \text{ from the coefficients of } \overline{X},\overline{Y},\overline{Z} Plan: 1. M from the coefficients of X , Y , Z 2. X ‾ , Y ‾ , Z ‾ coplanar ⇒ [ X ‾ Y ‾ Z ‾ ] = 0 2.\ \overline{X},\overline{Y},\overline{Z} \text{ coplanar} \Rightarrow [\overline{X}\ \overline{Y}\ \overline{Z}] = 0 2. X , Y , Z coplanar ⇒ [ X Y Z ] = 0 3. [ X ‾ Y ‾ Z ‾ ] = det ( M ) [ x ˉ y ˉ z ˉ ] 3.\ [\overline{X}\ \overline{Y}\ \overline{Z}] = \det(M)\,[\bar{x}\ \bar{y}\ \bar{z}] 3. [ X Y Z ] = det ( M ) [ x ˉ y ˉ z ˉ ] [ x ˉ y ˉ z ˉ ] = ∣ 1 2 3 2 3 1 3 1 2 ∣ [\bar{x}\ \bar{y}\ \bar{z}] = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{vmatrix} [ x ˉ y ˉ z ˉ ] = 1 2 3 2 3 1 3 1 2 = 1 ( 6 − 1 ) − 2 ( 4 − 3 ) + 3 ( 2 − 9 ) = 1(6-1) - 2(4-3) + 3(2-9) = 1 ( 6 − 1 ) − 2 ( 4 − 3 ) + 3 ( 2 − 9 ) = 5 − 2 − 21 = − 18 = 5 - 2 - 21 = -18 = 5 − 2 − 21 = − 18 [ x ˉ y ˉ z ˉ ] = − 18 ≠ 0 [\bar{x}\ \bar{y}\ \bar{z}] = -18 \neq 0 [ x ˉ y ˉ z ˉ ] = − 18 = 0 ⇒ x ˉ , y ˉ , z ˉ form a basis of R 3 \Rightarrow \bar{x},\bar{y},\bar{z} \text{ form a basis of } \mathbb{R}^3 ⇒ x ˉ , y ˉ , z ˉ form a basis of R 3 X ‾ = α x ˉ + β y ˉ − z ˉ \overline{X} = \alpha\bar{x} + \beta\bar{y} - \bar{z} X = α x ˉ + β y ˉ − z ˉ Y ‾ = − x ˉ + α y ˉ + β z ˉ \overline{Y} = -\bar{x} + \alpha\bar{y} + \beta\bar{z} Y = − x ˉ + α y ˉ + β z ˉ Z ‾ = β x ˉ − y ˉ + α z ˉ \overline{Z} = \beta\bar{x} - \bar{y} + \alpha\bar{z} Z = β x ˉ − y ˉ + α z ˉ M = ( α β − 1 − 1 α β β − 1 α ) M = \begin{pmatrix} \alpha & \beta & -1 \\ -1 & \alpha & \beta \\ \beta & -1 & \alpha \end{pmatrix} M = α − 1 β β α − 1 − 1 β α ( X ‾ Y ‾ Z ‾ ) = M ( x ˉ y ˉ z ˉ ) \begin{pmatrix}\overline{X}\\ \overline{Y}\\ \overline{Z}\end{pmatrix} = M \begin{pmatrix}\bar{x}\\ \bar{y}\\ \bar{z}\end{pmatrix} X Y Z = M x ˉ y ˉ z ˉ [ X ‾ Y ‾ Z ‾ ] = det ( M ) [ x ˉ y ˉ z ˉ ] [\overline{X}\ \overline{Y}\ \overline{Z}] = \det(M)\,[\bar{x}\ \bar{y}\ \bar{z}] [ X Y Z ] = det ( M ) [ x ˉ y ˉ z ˉ ] [ X ‾ Y ‾ Z ‾ ] = 0 [\overline{X}\ \overline{Y}\ \overline{Z}] = 0 [ X Y Z ] = 0 det ( M ) ⋅ ( − 18 ) = 0 ⇒ det ( M ) = 0 \det(M)\cdot(-18) = 0 \ \Rightarrow \ \det(M) = 0 det ( M ) ⋅ ( − 18 ) = 0 ⇒ det ( M ) = 0 det ( M ) = α ( α 2 + β ) − β ( − α − β 2 ) − ( 1 − α β ) \det(M) = \alpha(\alpha^2+\beta) - \beta(-\alpha-\beta^2) - (1-\alpha\beta) det ( M ) = α ( α 2 + β ) − β ( − α − β 2 ) − ( 1 − α β ) = α 3 + α β + α β + β 3 − 1 + α β = \alpha^3 + \alpha\beta + \alpha\beta + \beta^3 - 1 + \alpha\beta = α 3 + α β + α β + β 3 − 1 + α β det ( M ) = α 3 + β 3 + 3 α β − 1 \det(M) = \alpha^3 + \beta^3 + 3\alpha\beta - 1 det ( M ) = α 3 + β 3 + 3 α β − 1 a 3 + b 3 + c 3 − 3 a b c = ( a + b + c ) ( a 2 + b 2 + c 2 − a b − b c − c a ) a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) a 3 + b 3 + c 3 − 3 ab c = ( a + b + c ) ( a 2 + b 2 + c 2 − ab − b c − c a ) det ( M ) = ( α + β − 1 ) ( α 2 + β 2 − α β + α + β + 1 ) \det(M) = (\alpha+\beta-1)(\alpha^2+\beta^2-\alpha\beta+\alpha+\beta+1) det ( M ) = ( α + β − 1 ) ( α 2 + β 2 − α β + α + β + 1 ) α 2 + β 2 − α β = 1 2 [ ( α − β ) 2 + α 2 + β 2 ] ≥ 0 \alpha^2+\beta^2-\alpha\beta = \tfrac{1}{2}\left[(\alpha-\beta)^2+\alpha^2+\beta^2\right] \geq 0 α 2 + β 2 − α β = 2 1 [ ( α − β ) 2 + α 2 + β 2 ] ≥ 0 α 2 + β 2 − α β + α + β + 1 > 0 \alpha^2+\beta^2-\alpha\beta+\alpha+\beta+1 > 0 α 2 + β 2 − α β + α + β + 1 > 0 α + β − 1 = 0 ⇒ α + β = 1 \alpha+\beta-1 = 0 \ \Rightarrow \ \alpha+\beta = 1 α + β − 1 = 0 ⇒ α + β = 1 α + β − 3 = 1 − 3 = − 2 \alpha+\beta-3 = 1-3 = -2 α + β − 3 = 1 − 3 = − 2 α , β > 0 , α ≠ β , α + β = 1 is possible \alpha,\beta > 0,\ \alpha \neq \beta,\ \alpha+\beta = 1 \ \text{ is possible} α , β > 0 , α = β , α + β = 1 is possible ∴ α + β − 3 = − 2 ( independent of α , β ) \therefore \alpha+\beta-3 = -2\ \ (\text{independent of } \alpha,\beta) ∴ α + β − 3 = − 2 ( independent of α , β )
