Answer on Question #46185 - Math - Vector Calculus
Find co-ordinates of the points on the line ( x + 2 ) / 3 = ( y + 1 ) / 2 = ( z − 3 ) / 2 (x+2)/3 = (y+1)/2 = (z-3)/2 ( x + 2 ) /3 = ( y + 1 ) /2 = ( z − 3 ) /2 3 2 3\sqrt{2} 3 2 ( 1 , 2 , 3 ) (1, 2, 3) ( 1 , 2 , 3 )
Solution.
( x ′ , y ′ , z ′ ) (x', y', z') ( x ′ , y ′ , z ′ )
Distance between points ( x ′ , y ′ , z ′ ) (x', y', z') ( x ′ , y ′ , z ′ ) ( 1 , 2 , 3 ) (1,2,3) ( 1 , 2 , 3 )
( x ′ − 1 ) 2 + ( y ′ − 2 ) 2 + ( z ′ − 3 ) 2 = 3 2 \sqrt{(x' - 1)^2 + (y' - 2)^2 + (z' - 3)^2} = 3\sqrt{2} ( x ′ − 1 ) 2 + ( y ′ − 2 ) 2 + ( z ′ − 3 ) 2 = 3 2
Then
( x ′ − 1 ) 2 + ( y ′ − 2 ) 2 + ( z ′ − 3 ) 2 = 18 (x' - 1)^2 + (y' - 2)^2 + (z' - 3)^2 = 18 ( x ′ − 1 ) 2 + ( y ′ − 2 ) 2 + ( z ′ − 3 ) 2 = 18
Point ( x ′ , y ′ , z ′ ) (x', y', z') ( x ′ , y ′ , z ′ )
x ′ + 2 3 = y ′ + 1 2 = z ′ − 3 2 \frac{x' + 2}{3} = \frac{y' + 1}{2} = \frac{z' - 3}{2} 3 x ′ + 2 = 2 y ′ + 1 = 2 z ′ − 3
So we have system of equations
{ ( x ′ − 1 ) 2 + ( y ′ − 2 ) 2 + ( z ′ − 3 ) 2 = 18 x ′ + 2 3 = y ′ + 1 2 = z ′ − 3 2 \left\{
\begin{array}{c}
(x' - 1)^2 + (y' - 2)^2 + (z' - 3)^2 = 18 \\
\frac{x' + 2}{3} = \frac{y' + 1}{2} = \frac{z' - 3}{2}
\end{array}
\right. { ( x ′ − 1 ) 2 + ( y ′ − 2 ) 2 + ( z ′ − 3 ) 2 = 18 3 x ′ + 2 = 2 y ′ + 1 = 2 z ′ − 3 y ′ + 1 2 = z ′ − 3 2 z ′ − 3 = y ′ + 1 \frac{y' + 1}{2} = \frac{z' - 3}{2} \quad z' - 3 = y' + 1 2 y ′ + 1 = 2 z ′ − 3 z ′ − 3 = y ′ + 1 x ′ + 2 3 = y ′ + 1 2 x ′ + 2 = 3 ( y ′ + 1 ) 2 x ′ − 1 = 3 ( y ′ + 1 ) 2 − 3 = 3 2 ( y ′ − 1 ) \frac{x' + 2}{3} = \frac{y' + 1}{2} \quad x' + 2 = \frac{3(y' + 1)}{2} \quad x' - 1 = \frac{3(y' + 1)}{2} - 3 = \frac{3}{2}(y' - 1) 3 x ′ + 2 = 2 y ′ + 1 x ′ + 2 = 2 3 ( y ′ + 1 ) x ′ − 1 = 2 3 ( y ′ + 1 ) − 3 = 2 3 ( y ′ − 1 ) { ( 3 2 ( y ′ − 1 ) ) 2 + ( y ′ − 2 ) 2 + ( y ′ + 1 ) 2 = 18 x ′ + 2 3 = y ′ + 1 2 = z ′ − 3 2 \left\{
\begin{array}{c}
\left(\frac{3}{2}(y' - 1)\right)^2 + (y' - 2)^2 + (y' + 1)^2 = 18 \\
\frac{x' + 2}{3} = \frac{y' + 1}{2} = \frac{z' - 3}{2}
\end{array}
\right. { ( 2 3 ( y ′ − 1 ) ) 2 + ( y ′ − 2 ) 2 + ( y ′ + 1 ) 2 = 18 3 x ′ + 2 = 2 y ′ + 1 = 2 z ′ − 3 { 17 4 y ′ 2 − 13 2 y ′ − 43 4 = 0 x ′ + 2 3 = y ′ + 1 2 = z ′ − 3 2 \left\{
\begin{array}{c}
\frac{17}{4}y'^2 - \frac{13}{2}y' - \frac{43}{4} = 0 \\
\frac{x' + 2}{3} = \frac{y' + 1}{2} = \frac{z' - 3}{2}
\end{array}
\right. { 4 17 y ′2 − 2 13 y ′ − 4 43 = 0 3 x ′ + 2 = 2 y ′ + 1 = 2 z ′ − 3 17 4 y ′ 2 − 13 2 y ′ − 43 4 = 0 \frac{17}{4}y'^2 - \frac{13}{2}y' - \frac{43}{4} = 0 4 17 y ′2 − 2 13 y ′ − 4 43 = 0 17 y ′ 2 − 26 y ′ − 43 = 0 D = 2 6 2 + 4 ⋅ 17 ⋅ 43 = 3600 17y'^2 - 26y' - 43 = 0 \quad D = 26^2 + 4 \cdot 17 \cdot 43 = 3600 17 y ′2 − 26 y ′ − 43 = 0 D = 2 6 2 + 4 ⋅ 17 ⋅ 43 = 3600 y 1 ′ = 26 + D 2 ⋅ 17 = 43 17 y 2 ′ = 26 − D 2 ⋅ 17 = − 1 y_1' = \frac{26 + \sqrt{D}}{2 \cdot 17} = \frac{43}{17} \quad y_2' = \frac{26 - \sqrt{D}}{2 \cdot 17} = -1 y 1 ′ = 2 ⋅ 17 26 + D = 17 43 y 2 ′ = 2 ⋅ 17 26 − D = − 1 x 1 ′ = 3 2 ( y 1 ′ − 1 ) + 1 = 56 17 x 2 ′ = 3 2 ( y 2 ′ − 1 ) + 1 = − 2 x _ {1} ^ {\prime} = \frac {3}{2} \left(y _ {1} ^ {\prime} - 1\right) + 1 = \frac {5 6}{1 7} \quad x _ {2} ^ {\prime} = \frac {3}{2} \left(y _ {2} ^ {\prime} - 1\right) + 1 = - 2 x 1 ′ = 2 3 ( y 1 ′ − 1 ) + 1 = 17 56 x 2 ′ = 2 3 ( y 2 ′ − 1 ) + 1 = − 2 z 1 ′ = y 1 ′ + 4 = 111 17 z 2 ′ = y 2 ′ + 4 = 3 z _ {1} ^ {\prime} = y _ {1} ^ {\prime} + 4 = \frac {1 1 1}{1 7} \quad z _ {2} ^ {\prime} = y _ {2} ^ {\prime} + 4 = 3 z 1 ′ = y 1 ′ + 4 = 17 111 z 2 ′ = y 2 ′ + 4 = 3
Our points: ( 56 17 , 43 17 , 111 17 ) \left(\frac{56}{17},\frac{43}{17},\frac{111}{17}\right) ( 17 56 , 17 43 , 17 111 ) ( − 2 , − 1 , 3 ) (-2, - 1,3) ( − 2 , − 1 , 3 )
Answer: ( 56 17 , 43 17 , 111 17 ) \left(\frac{56}{17},\frac{43}{17},\frac{111}{17}\right) ( 17 56 , 17 43 , 17 111 ) ( − 2 , − 1 , 3 ) (-2, - 1,3) ( − 2 , − 1 , 3 )
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#339914 on Dec 2023