Answer on Question #46099-Math-Analytic Geometry
(a) Show that the angle between the two lines in which the plane x − y + 2 z = 0 x - y + 2z = 0 x − y + 2 z = 0 x 2 + y 2 − 4 z 2 + 6 y z = 0 x^2 + y^2 - 4z^2 + 6yz = 0 x 2 + y 2 − 4 z 2 + 6 yz = 0 tan − 1 6 7 \tan^{-1}\frac{\sqrt{6}}{7} tan − 1 7 6
(b) Using projection show that the line passing through ( − 1 , 8 , 8 ) (-1, 8, 8) ( − 1 , 8 , 8 ) ( 6 , 2 , 0 ) (6, 2, 0) ( 6 , 2 , 0 ) ( 4 , 2 , 3 ) (4, 2, 3) ( 4 , 2 , 3 ) ( 2 , 1 , 2 ) (2, 1, 2) ( 2 , 1 , 2 )
(c) At what point the origin must be shifted so that linear terms in the conicoid x 2 + 2 y 2 − z 2 − 2 y z + 2 x z + x − 3 y + z + 4 = 0 x^{2} + 2y^{2} - z^{2} - 2yz + 2xz + x - 3y + z + 4 = 0 x 2 + 2 y 2 − z 2 − 2 yz + 2 x z + x − 3 y + z + 4 = 0
Solution
(a) x − y + 2 z = 0 → x = y − 2 z x - y + 2z = 0 \rightarrow x = y - 2z x − y + 2 z = 0 → x = y − 2 z
( y − 2 z ) 2 + y 2 − 4 z 2 + 6 y z = 0 → y 2 + 4 z 2 − 4 y z + y 2 − 4 z 2 + 6 y z = 0 → 2 y 2 + 2 y z = 0 → y ( y + z ) = 0. \begin{array}{l}
(y - 2z)^{2} + y^{2} - 4z^{2} + 6yz = 0 \rightarrow y^{2} + 4z^{2} - 4yz + y^{2} - 4z^{2} + 6yz = 0 \rightarrow 2y^{2} + 2yz = 0 \\
\rightarrow y(y + z) = 0.
\end{array} ( y − 2 z ) 2 + y 2 − 4 z 2 + 6 yz = 0 → y 2 + 4 z 2 − 4 yz + y 2 − 4 z 2 + 6 yz = 0 → 2 y 2 + 2 yz = 0 → y ( y + z ) = 0.
First case y = − z y = -z y = − z
x = − z − 2 z = − 3 z . x = -z - 2z = -3z. x = − z − 2 z = − 3 z .
The equation of a line:
( x y z ) = t ( − 3 − 1 1 ) . \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = t \left( \begin{array}{c} -3 \\ -1 \\ 1 \end{array} \right). ⎝ ⎛ x y z ⎠ ⎞ = t ⎝ ⎛ − 3 − 1 1 ⎠ ⎞ .
Second case y = 0 y = 0 y = 0
x = − 2 z . x = -2z. x = − 2 z .
The equation of a line:
( x y z ) = t ( − 2 0 1 ) . \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = t \left( \begin{array}{c} -2 \\ 0 \\ 1 \end{array} \right). ⎝ ⎛ x y z ⎠ ⎞ = t ⎝ ⎛ − 2 0 1 ⎠ ⎞ .
The angle between the two lines
cos θ = − 3 ( − 2 ) − 1 ⋅ 0 + 1 ⋅ 1 ( − 3 ) 2 + ( − 1 ) 2 + ( 1 ) 2 ( − 2 ) 2 + ( 0 ) 2 + ( 1 ) 2 = 7 55 ; sin θ = 1 − ( 7 55 ) 2 = 6 55 . \cos \theta = \frac{-3(-2) - 1 \cdot 0 + 1 \cdot 1}{\sqrt{(-3)^{2} + (-1)^{2} + (1)^{2}} \sqrt{(-2)^{2} + (0)^{2} + (1)^{2}}} = \frac{7}{\sqrt{55}}; \sin \theta = \sqrt{1 - \left(\frac{7}{\sqrt{55}}\right)^{2}} = \sqrt{\frac{6}{55}}. cos θ = ( − 3 ) 2 + ( − 1 ) 2 + ( 1 ) 2 ( − 2 ) 2 + ( 0 ) 2 + ( 1 ) 2 − 3 ( − 2 ) − 1 ⋅ 0 + 1 ⋅ 1 = 55 7 ; sin θ = 1 − ( 55 7 ) 2 = 55 6 . θ = tan − 1 6 55 7 55 = tan − 1 6 7 . \theta = \tan^{-1} \frac{\sqrt{\frac{6}{55}}}{\frac{7}{\sqrt{55}}} = \tan^{-1} \frac{\sqrt{6}}{7}. θ = tan − 1 55 7 55 6 = tan − 1 7 6 .
(b) The displacement vector of the first line is
v ~ = ( 6 − ( − 1 ) 2 − 8 0 − 8 ) = ( 7 − 6 − 8 ) . \tilde{v} = \left( \begin{array}{c} 6 - (-1) \\ 2 - 8 \\ 0 - 8 \end{array} \right) = \left( \begin{array}{c} 7 \\ -6 \\ -8 \end{array} \right). v ~ = ⎝ ⎛ 6 − ( − 1 ) 2 − 8 0 − 8 ⎠ ⎞ = ⎝ ⎛ 7 − 6 − 8 ⎠ ⎞ .
The displacement vector of the second line is
s ~ = ( 4 − 2 2 − 1 3 − 2 ) = ( 2 1 1 ) . \tilde{s} = \left( \begin{array}{c} 4 - 2 \\ 2 - 1 \\ 3 - 2 \end{array} \right) = \left( \begin{array}{c} 2 \\ 1 \\ 1 \end{array} \right). s ~ = ⎝ ⎛ 4 − 2 2 − 1 3 − 2 ⎠ ⎞ = ⎝ ⎛ 2 1 1 ⎠ ⎞ .
The orthogonal projection of v ~ \tilde{v} v ~
( 7 − 6 ) ⋅ ( 2 1 ) ( 2 1 ) ⋅ ( 2 1 ) ( 2 1 ) = 7 ⋅ 2 − 6 ⋅ 1 − 8 ⋅ 1 2 2 + 1 2 + 1 2 ( 2 1 ) = 0 6 ( 2 1 ) = ( 0 0 ) . \frac {\binom {7} {- 6} \cdot \binom {2} {1}}{\binom {2} {1} \cdot \binom {2} {1}} \binom {2} {1} = \frac {7 \cdot 2 - 6 \cdot 1 - 8 \cdot 1}{2 ^ {2} + 1 ^ {2} + 1 ^ {2}} \binom {2} {1} = \frac {0}{6} \binom {2} {1} = \binom {0} {0}. ( 1 2 ) ⋅ ( 1 2 ) ( − 6 7 ) ⋅ ( 1 2 ) ( 1 2 ) = 2 2 + 1 2 + 1 2 7 ⋅ 2 − 6 ⋅ 1 − 8 ⋅ 1 ( 1 2 ) = 6 0 ( 1 2 ) = ( 0 0 ) .
That's why these lines are perpendicular to each other.
(c) Suppose that the origin is shifted to point ( a , b , c ) (a, b, c) ( a , b , c ) ( x , y , z ) (x, y, z) ( x , y , z ) ( x ′ , y ′ , z ′ ) (x', y', z') ( x ′ , y ′ , z ′ ) x = x ′ + a x = x' + a x = x ′ + a y = y ′ + b y = y' + b y = y ′ + b z = z ′ + c z = z' + c z = z ′ + c x 2 + 2 y 2 − z 2 − 2 y z + 2 x z + x − 3 y + z + 4 = 0 x^2 + 2y^2 - z^2 - 2yz + 2xz + x - 3y + z + 4 = 0 x 2 + 2 y 2 − z 2 − 2 yz + 2 x z + x − 3 y + z + 4 = 0
( x ′ ) 2 + 2 x ′ a + a 2 + 2 ( y ′ ) 2 + 4 y ′ b + 2 b 2 − ( z ′ ) 2 − 2 z ′ c − c 2 − 2 y ′ z ′ − 2 y ′ c − 2 z ′ b − 2 b c + 2 x ′ z ′ + 2 x ′ c + 2 z ′ a + 2 a c + x ′ + a − 3 y ′ − 3 b + z ′ + c + 4 = 0 \begin{array}{l} (x ^ {\prime}) ^ {2} + 2 x ^ {\prime} a + a ^ {2} + 2 (y ^ {\prime}) ^ {2} + 4 y ^ {\prime} b + 2 b ^ {2} - (z ^ {\prime}) ^ {2} - 2 z ^ {\prime} c - c ^ {2} - 2 y ^ {\prime} z ^ {\prime} - 2 y ^ {\prime} c - 2 z ^ {\prime} b - 2 b c + 2 x ^ {\prime} z ^ {\prime} \\ + 2 x ^ {\prime} c + 2 z ^ {\prime} a + 2 a c + x ^ {\prime} + a - 3 y ^ {\prime} - 3 b + z ^ {\prime} + c + 4 = 0 \\ \end{array} ( x ′ ) 2 + 2 x ′ a + a 2 + 2 ( y ′ ) 2 + 4 y ′ b + 2 b 2 − ( z ′ ) 2 − 2 z ′ c − c 2 − 2 y ′ z ′ − 2 y ′ c − 2 z ′ b − 2 b c + 2 x ′ z ′ + 2 x ′ c + 2 z ′ a + 2 a c + x ′ + a − 3 y ′ − 3 b + z ′ + c + 4 = 0
or
( x ′ ) 2 + 2 ( y ′ ) 2 − ( z ′ ) 2 − 2 y ′ z ′ + 2 x ′ z ′ + x ′ ( 2 a + 2 c + 1 ) + y ′ ( 4 b − 2 c − 3 ) − z ′ ( 2 b + 2 c − 2 a − 1 ) + a 2 + 2 b 2 − c 2 − 2 b c + 2 a c + a − 3 b + c + 4 = 0. \begin{array}{l} (x ^ {\prime}) ^ {2} + 2 (y ^ {\prime}) ^ {2} - (z ^ {\prime}) ^ {2} - 2 y ^ {\prime} z ^ {\prime} + 2 x ^ {\prime} z ^ {\prime} + x ^ {\prime} (2 a + 2 c + 1) + y ^ {\prime} (4 b - 2 c - 3) - z ^ {\prime} (2 b + 2 c - 2 a - 1) \\ + a ^ {2} + 2 b ^ {2} - c ^ {2} - 2 b c + 2 a c + a - 3 b + c + 4 = 0. \\ \end{array} ( x ′ ) 2 + 2 ( y ′ ) 2 − ( z ′ ) 2 − 2 y ′ z ′ + 2 x ′ z ′ + x ′ ( 2 a + 2 c + 1 ) + y ′ ( 4 b − 2 c − 3 ) − z ′ ( 2 b + 2 c − 2 a − 1 ) + a 2 + 2 b 2 − c 2 − 2 b c + 2 a c + a − 3 b + c + 4 = 0.
The linear terms should vanish, so
{ 2 a + 2 c + 1 = 0 4 b − 2 c − 3 = 0 2 c + 2 b − 2 a − 1 = 0 → { 2 a + 2 c + 1 = 0 4 b − 2 c − 3 = 0 ( 2 c + 2 b − 2 a − 1 ) + ( 2 a + 2 c + 1 ) = 2 b + 4 c = 0 \left\{ \begin{array}{c} 2 a + 2 c + 1 = 0 \\ 4 b - 2 c - 3 = 0 \\ 2 c + 2 b - 2 a - 1 = 0 \end{array} \right. \to \left\{ \begin{array}{c} 2 a + 2 c + 1 = 0 \\ 4 b - 2 c - 3 = 0 \\ (2 c + 2 b - 2 a - 1) + (2 a + 2 c + 1) = 2 b + 4 c = 0 \end{array} \right. ⎩ ⎨ ⎧ 2 a + 2 c + 1 = 0 4 b − 2 c − 3 = 0 2 c + 2 b − 2 a − 1 = 0 → ⎩ ⎨ ⎧ 2 a + 2 c + 1 = 0 4 b − 2 c − 3 = 0 ( 2 c + 2 b − 2 a − 1 ) + ( 2 a + 2 c + 1 ) = 2 b + 4 c = 0 { 2 a + 2 c + 1 = 0 4 b − 2 c − 3 = 0 → { a = − 0.2 b = 0.6 c = − 0.3 b = − 2 c \left\{ \begin{array}{l} 2 a + 2 c + 1 = 0 \\ 4 b - 2 c - 3 = 0 \to \left\{ \begin{array}{c} a = - 0. 2 \\ b = 0. 6 \\ c = - 0. 3 \end{array} \right. \\ b = - 2 c \end{array} \right. ⎩ ⎨ ⎧ 2 a + 2 c + 1 = 0 4 b − 2 c − 3 = 0 → ⎩ ⎨ ⎧ a = − 0.2 b = 0.6 c = − 0.3 b = − 2 c
Answer: ( − 0.2 ; 0.6 ; − 0.3 ) (-0.2; 0.6; -0.3) ( − 0.2 ; 0.6 ; − 0.3 )
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