The question
The image of point P ( x , y , z ) P(x, y, z) P ( x , y , z ) x 1 = y − 1 2 = z − 2 3 \dfrac{x}{1} = \dfrac{y - 1}{2} = \dfrac{z - 2}{3} 1 x = 2 y − 1 = 3 z − 2 P ′ ( 1 , 0 , 7 ) P'(1, 0, 7) P ′ ( 1 , 0 , 7 )
Exam answer Show all explanations
Any point of the line x 1 = y − 1 2 = z − 2 3 = λ \frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} = \lambda 1 x = 2 y − 1 = 3 z − 2 = λ ( λ , 2 λ + 1 , 3 λ + 2 ) (\lambda, 2\lambda + 1, 3\lambda + 2) ( λ , 2 λ + 1 , 3 λ + 2 )
Why?
Set each of the equal fractions equal to a parameter λ \lambda λ x = λ x = \lambda x = λ y − 1 = 2 λ y - 1 = 2\lambda y − 1 = 2 λ z − 2 = 3 λ z - 2 = 3\lambda z − 2 = 3 λ λ \lambda λ
x 1 = λ ⇒ x = λ y − 1 2 = λ ⇒ y − 1 = 2 λ ⇒ y = 2 λ + 1 z − 2 3 = λ ⇒ z − 2 = 3 λ ⇒ z = 3 λ + 2 \begin{aligned} \frac{x}{1} = \lambda &\Rightarrow x = \lambda \\ \frac{y - 1}{2} = \lambda &\Rightarrow y - 1 = 2\lambda \\ &\Rightarrow y = 2\lambda + 1 \\ \frac{z - 2}{3} = \lambda &\Rightarrow z - 2 = 3\lambda \\ &\Rightarrow z = 3\lambda + 2 \end{aligned} 1 x = λ 2 y − 1 = λ 3 z − 2 = λ ⇒ x = λ ⇒ y − 1 = 2 λ ⇒ y = 2 λ + 1 ⇒ z − 2 = 3 λ ⇒ z = 3 λ + 2 Let M ( λ , 2 λ + 1 , 3 λ + 2 ) M(\lambda, 2\lambda + 1, 3\lambda + 2) M ( λ , 2 λ + 1 , 3 λ + 2 ) P ′ ( 1 , 0 , 7 ) P'(1, 0, 7) P ′ ( 1 , 0 , 7 ) P P ′ PP' P P ′
Why?
P′ is the image (mirror point) of P in the line. That means the segment P P ′ PP' P P ′
So the foot M of the perpendicular from P′ is the mid-point of P P ′ PP' P P ′
Direction ratios of P ′ M P'M P ′ M P P ′ PP' P P ′ λ − 1 , 2 λ + 1 , 3 λ − 5 \lambda - 1,\ 2\lambda + 1,\ 3\lambda - 5 λ − 1 , 2 λ + 1 , 3 λ − 5
Why?
The direction ratios of the segment joining two points are the differences of their coordinates, M minus P′:
λ − 1 , ( 2 λ + 1 ) − 0 , ( 3 λ + 2 ) − 7 \begin{aligned} &\lambda - 1,\ (2\lambda + 1) - 0, \\ &(3\lambda + 2) - 7 \end{aligned} λ − 1 , ( 2 λ + 1 ) − 0 , ( 3 λ + 2 ) − 7 Simplify each one:
( 2 λ + 1 ) − 0 = 2 λ + 1 ( 3 λ + 2 ) − 7 = 3 λ − 5 \begin{aligned} (2\lambda + 1) - 0 &= 2\lambda + 1 \\ (3\lambda + 2) - 7 &= 3\lambda - 5 \end{aligned} ( 2 λ + 1 ) − 0 ( 3 λ + 2 ) − 7 = 2 λ + 1 = 3 λ − 5 M lies on P P ′ PP' P P ′ P P ′ PP' P P ′
P P ′ PP' P P ′
1 ( λ − 1 ) + 2 ( 2 λ + 1 ) + 3 ( 3 λ − 5 ) = 0 ⇒ λ − 1 + 4 λ + 2 + 9 λ − 15 = 0 ⇒ 14 λ − 14 = 0 ⇒ λ = 1 \begin{aligned} &1(\lambda - 1) + 2(2\lambda + 1) \\ &\quad + 3(3\lambda - 5) = 0 \\ \Rightarrow\ &\lambda - 1 + 4\lambda + 2 \\ &\quad + 9\lambda - 15 = 0 \\ \Rightarrow\ &14\lambda - 14 = 0 \\ \Rightarrow\ &\lambda = 1 \end{aligned} ⇒ ⇒ ⇒ 1 ( λ − 1 ) + 2 ( 2 λ + 1 ) + 3 ( 3 λ − 5 ) = 0 λ − 1 + 4 λ + 2 + 9 λ − 15 = 0 14 λ − 14 = 0 λ = 1 Why?
The direction ratios of the line are the denominators 1, 2, 3. Two lines with direction ratios a 1 , b 1 , c 1 a_1, b_1, c_1 a 1 , b 1 , c 1 a 2 , b 2 , c 2 a_2, b_2, c_2 a 2 , b 2 , c 2
a 1 a 2 + b 1 b 2 + c 1 c 2 = 0 a_1a_2 + b_1b_2 + c_1c_2 = 0 a 1 a 2 + b 1 b 2 + c 1 c 2 = 0 Opening each bracket:
1 ( λ − 1 ) = λ − 1 2 ( 2 λ + 1 ) = 4 λ + 2 3 ( 3 λ − 5 ) = 9 λ − 15 \begin{aligned} 1(\lambda - 1) &= \lambda - 1 \\ 2(2\lambda + 1) &= 4\lambda + 2 \\ 3(3\lambda - 5) &= 9\lambda - 15 \end{aligned} 1 ( λ − 1 ) 2 ( 2 λ + 1 ) 3 ( 3 λ − 5 ) = λ − 1 = 4 λ + 2 = 9 λ − 15 Collecting terms: λ + 4 λ + 9 λ = 14 λ \lambda + 4\lambda + 9\lambda = 14\lambda λ + 4 λ + 9 λ = 14 λ − 1 + 2 − 15 = − 14 -1 + 2 - 15 = -14 − 1 + 2 − 15 = − 14
− 1 + 2 = 1 1 − 15 = − 14 14 λ − 14 = 0 14 λ = 14 ( add 14 ) λ = 1 ( divide by 14 ) \begin{aligned} -1 + 2 &= 1 \\ 1 - 15 &= -14 \\[4pt] 14\lambda - 14 &= 0 \\ 14\lambda &= 14 \quad (\text{add } 14) \\ \lambda &= 1 \quad (\text{divide by } 14) \end{aligned} − 1 + 2 1 − 15 14 λ − 14 14 λ λ = 1 = − 14 = 0 = 14 ( add 14 ) = 1 ( divide by 14 ) So the coordinates of M are ( 1 , 3 , 5 ) (1, 3, 5) ( 1 , 3 , 5 )
Why?
Put λ = 1 \lambda = 1 λ = 1 ( λ , 2 λ + 1 , 3 λ + 2 ) (\lambda, 2\lambda + 1, 3\lambda + 2) ( λ , 2 λ + 1 , 3 λ + 2 ) ( 1 , 2 + 1 , 3 + 2 ) (1, 2 + 1, 3 + 2) ( 1 , 2 + 1 , 3 + 2 )
x = 1 y = 2 ( 1 ) + 1 = 3 z = 3 ( 1 ) + 2 = 5 \begin{aligned} x &= 1 \\ y &= 2(1) + 1 = 3 \\ z &= 3(1) + 2 = 5 \end{aligned} x y z = 1 = 2 ( 1 ) + 1 = 3 = 3 ( 1 ) + 2 = 5 M is the mid-point of P ( x , y , z ) P(x, y, z) P ( x , y , z ) P ′ ( 1 , 0 , 7 ) P'(1, 0, 7) P ′ ( 1 , 0 , 7 )
x + 1 2 = 1 , y + 0 2 = 3 , z + 7 2 = 5 \frac{x + 1}{2} = 1,\ \frac{y + 0}{2} = 3,\ \frac{z + 7}{2} = 5 2 x + 1 = 1 , 2 y + 0 = 3 , 2 z + 7 = 5 Why?
The mid-point of two points is the average of their coordinates:
( x 1 + x 2 2 , y 1 + y 2 2 , z 1 + z 2 2 ) \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right) ( 2 x 1 + x 2 , 2 y 1 + y 2 , 2 z 1 + z 2 ) Each coordinate is set equal to the matching coordinate of M(1, 3, 5).
⇒ x = 1 , y = 6 , z = 3 \Rightarrow x = 1,\ y = 6,\ z = 3 ⇒ x = 1 , y = 6 , z = 3 Why?
x + 1 = 2 x + 1 = 2 x + 1 = 2 y = 6 y = 6 y = 6 z + 7 = 10 z + 7 = 10 z + 7 = 10
Multiply each equation by 2, then move the number across:
x + 1 2 = 1 ⇒ x + 1 = 2 ⇒ x = 2 − 1 = 1 y + 0 2 = 3 ⇒ y = 2 × 3 = 6 z + 7 2 = 5 ⇒ z + 7 = 10 ⇒ z = 10 − 7 = 3 \begin{aligned} \frac{x + 1}{2} = 1 &\Rightarrow x + 1 = 2 \\ &\Rightarrow x = 2 - 1 = 1 \\[4pt] \frac{y + 0}{2} = 3 &\Rightarrow y = 2 \times 3 = 6 \\[4pt] \frac{z + 7}{2} = 5 &\Rightarrow z + 7 = 10 \\ &\Rightarrow z = 10 - 7 = 3 \end{aligned} 2 x + 1 = 1 2 y + 0 = 3 2 z + 7 = 5 ⇒ x + 1 = 2 ⇒ x = 2 − 1 = 1 ⇒ y = 2 × 3 = 6 ⇒ z + 7 = 10 ⇒ z = 10 − 7 = 3 A quick check: P ′ P ⃗ \vec{P'P} P ′ P 0 , 6 , − 4 0, 6, -4 0 , 6 , − 4 1 ( 0 ) + 2 ( 6 ) + 3 ( − 4 ) = 0 1(0) + 2(6) + 3(-4) = 0 1 ( 0 ) + 2 ( 6 ) + 3 ( − 4 ) = 0
1 − 1 = 0 , 6 − 0 = 6 , 3 − 7 = − 4 1 ( 0 ) + 2 ( 6 ) + 3 ( − 4 ) = 0 + 12 − 12 = 0 \begin{aligned} &1 - 1 = 0,\ 6 - 0 = 6,\ 3 - 7 = -4 \\[4pt] &1(0) + 2(6) + 3(-4) \\ &= 0 + 12 - 12 = 0 \end{aligned} 1 − 1 = 0 , 6 − 0 = 6 , 3 − 7 = − 4 1 ( 0 ) + 2 ( 6 ) + 3 ( − 4 ) = 0 + 12 − 12 = 0 So P P ′ PP' P P ′
∴ \therefore ∴ ( 1 , 6 , 3 ) (1, 6, 3) ( 1 , 6 , 3 )