The question
Electrical transmission wires which are laid down in winters are stretched tightly to accommodate expansion in summers.
Two such wires lie along the following lines :
l 1 : x + 1 3 = y − 3 − 2 = z + 2 − 1 l_1 : \frac{x + 1}{3} = \frac{y - 3}{-2} = \frac{z + 2}{-1} l 1 : 3 x + 1 = − 2 y − 3 = − 1 z + 2 l 2 : x − 1 = y − 7 3 = z + 7 − 2 l_2 : \frac{x}{-1} = \frac{y - 7}{3} = \frac{z + 7}{-2} l 2 : − 1 x = 3 y − 7 = − 2 z + 7 Based on the given information, answer the following questions :
(i) Are the lines l 1 l_1 l 1 l 2 l_2 l 2 (2 marks)
(ii) Find the point of intersection of the lines l 1 l_1 l 1 l 2 l_2 l 2 (2 marks)
Part (i), coplanarity of two lines, is no longer in the CBSE syllabus (2025-26).
Exam answer Show all explanations
(i) l 1 l_1 l 1 ( x 1 , y 1 , z 1 ) = ( − 1 , 3 , − 2 ) (x_1, y_1, z_1) = (-1, 3, -2) ( x 1 , y 1 , z 1 ) = ( − 1 , 3 , − 2 ) 3 , − 2 , − 1 3, -2, -1 3 , − 2 , − 1
l 2 l_2 l 2 ( x 2 , y 2 , z 2 ) = ( 0 , 7 , − 7 ) (x_2, y_2, z_2) = (0, 7, -7) ( x 2 , y 2 , z 2 ) = ( 0 , 7 , − 7 ) − 1 , 3 , − 2 -1, 3, -2 − 1 , 3 , − 2
Why?
A line in the form
x − x 1 a = y − y 1 b = z − z 1 c \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} a x − x 1 = b y − y 1 = c z − z 1 passes through ( x 1 , y 1 , z 1 ) (x_1, y_1, z_1) ( x 1 , y 1 , z 1 ) a , b , c a, b, c a , b , c
Watch the signs: x + 1 = x − ( − 1 ) x + 1 = x - (-1) x + 1 = x − ( − 1 ) x − 1 \frac{x}{-1} − 1 x x − 0 − 1 \frac{x - 0}{-1} − 1 x − 0
Consider
∣ x 2 − x 1 y 2 − y 1 z 2 − z 1 a 1 b 1 c 1 a 2 b 2 c 2 ∣ = ∣ 1 4 − 5 3 − 2 − 1 − 1 3 − 2 ∣ \begin{aligned} &\begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} \\ &= \begin{vmatrix} 1 & 4 & -5 \\ 3 & -2 & -1 \\ -1 & 3 & -2 \end{vmatrix} \end{aligned} x 2 − x 1 a 1 a 2 y 2 − y 1 b 1 b 2 z 2 − z 1 c 1 c 2 = 1 3 − 1 4 − 2 3 − 5 − 1 − 2 Why?
Two lines are coplanar exactly when this determinant is 0. It is the scalar triple product of the vector joining the two points and the two direction vectors; it is 0 when the three vectors lie in one plane.
First row: 0 − ( − 1 ) = 1 0 - (-1) = 1 0 − ( − 1 ) = 1 7 − 3 = 4 7 - 3 = 4 7 − 3 = 4 − 7 − ( − 2 ) = − 5 -7 - (-2) = -5 − 7 − ( − 2 ) = − 5
= 1 ( 4 + 3 ) − 4 ( − 6 − 1 ) − 5 ( 9 − 2 ) = 7 + 28 − 35 = 0 \begin{aligned} &= 1(4 + 3) - 4(-6 - 1) \\ &\quad - 5(9 - 2) \\ &= 7 + 28 - 35 = 0 \end{aligned} = 1 ( 4 + 3 ) − 4 ( − 6 − 1 ) − 5 ( 9 − 2 ) = 7 + 28 − 35 = 0 Why?
Expand along the first row with signs + + + − - − + + +
( − 2 ) ( − 2 ) − ( − 1 ) ( 3 ) = 7 ( 3 ) ( − 2 ) − ( − 1 ) ( − 1 ) = − 7 ( 3 ) ( 3 ) − ( − 2 ) ( − 1 ) = 7 \begin{aligned} &(-2)(-2) - (-1)(3) = 7 \\ &(3)(-2) - (-1)(-1) = -7 \\ &(3)(3) - (-2)(-1) = 7 \end{aligned} ( − 2 ) ( − 2 ) − ( − 1 ) ( 3 ) = 7 ( 3 ) ( − 2 ) − ( − 1 ) ( − 1 ) = − 7 ( 3 ) ( 3 ) − ( − 2 ) ( − 1 ) = 7 Each one worked out:
∣ − 2 − 1 3 − 2 ∣ = ( − 2 ) ( − 2 ) − ( − 1 ) ( 3 ) = 4 − ( − 3 ) = 4 + 3 = 7 ∣ 3 − 1 − 1 − 2 ∣ = ( 3 ) ( − 2 ) − ( − 1 ) ( − 1 ) = − 6 − 1 = − 7 ∣ 3 − 2 − 1 3 ∣ = ( 3 ) ( 3 ) − ( − 2 ) ( − 1 ) = 9 − 2 = 7 \begin{aligned} &\begin{vmatrix} -2 & -1 \\ 3 & -2 \end{vmatrix} \\ &= (-2)(-2) - (-1)(3) \\ &= 4 - (-3) = 4 + 3 = 7 \\[4pt] &\begin{vmatrix} 3 & -1 \\ -1 & -2 \end{vmatrix} \\ &= (3)(-2) - (-1)(-1) \\ &= -6 - 1 = -7 \\[4pt] &\begin{vmatrix} 3 & -2 \\ -1 & 3 \end{vmatrix} \\ &= (3)(3) - (-2)(-1) \\ &= 9 - 2 = 7 \end{aligned} − 2 3 − 1 − 2 = ( − 2 ) ( − 2 ) − ( − 1 ) ( 3 ) = 4 − ( − 3 ) = 4 + 3 = 7 3 − 1 − 1 − 2 = ( 3 ) ( − 2 ) − ( − 1 ) ( − 1 ) = − 6 − 1 = − 7 3 − 1 − 2 3 = ( 3 ) ( 3 ) − ( − 2 ) ( − 1 ) = 9 − 2 = 7 Multiply by the first-row entries with their signs, then add:
+ 1 ( 7 ) = 7 − 4 ( − 7 ) = + 28 + ( − 5 ) ( 7 ) = − 35 7 + 28 = 35 35 − 35 = 0 \begin{aligned} +1(7) &= 7 \\ -4(-7) &= +28 \\ +(-5)(7) &= -35 \\[4pt] 7 + 28 &= 35 \\ 35 - 35 &= 0 \end{aligned} + 1 ( 7 ) − 4 ( − 7 ) + ( − 5 ) ( 7 ) 7 + 28 35 − 35 = 7 = + 28 = − 35 = 35 = 0 ∴ \therefore ∴ l 1 l_1 l 1 l 2 l_2 l 2
(ii) Any point on l 1 l_1 l 1
( 3 λ − 1 , − 2 λ + 3 , − λ − 2 ) (3\lambda - 1, -2\lambda + 3, -\lambda - 2) ( 3 λ − 1 , − 2 λ + 3 , − λ − 2 ) Why?
Set each ratio of l 1 l_1 l 1 λ \lambda λ
x + 1 = 3 λ y − 3 = − 2 λ z + 2 = − λ \begin{aligned} x + 1 &= 3\lambda \\ y - 3 &= -2\lambda \\ z + 2 &= -\lambda \end{aligned} x + 1 y − 3 z + 2 = 3 λ = − 2 λ = − λ Each value of λ \lambda λ l 1 l_1 l 1
Substituting in the equation of l 2 l_2 l 2
3 λ − 1 − 1 = − 2 λ + 3 − 7 3 \frac{3\lambda - 1}{-1} = \frac{-2\lambda + 3 - 7}{3} − 1 3 λ − 1 = 3 − 2 λ + 3 − 7 ⇒ 9 λ − 3 = 2 λ + 4 \Rightarrow 9\lambda - 3 = 2\lambda + 4 ⇒ 9 λ − 3 = 2 λ + 4 Why?
If the lines meet, the meeting point is on l 2 l_2 l 2 l 2 l_2 l 2 λ \lambda λ
On the right, − 2 λ + 3 − 7 = − 2 λ − 4 -2\lambda + 3 - 7 = -2\lambda - 4 − 2 λ + 3 − 7 = − 2 λ − 4
Cross-multiplying: 3 ( 3 λ − 1 ) = ( − 1 ) ( − 2 λ − 4 ) 3(3\lambda - 1) = (-1)(-2\lambda - 4) 3 ( 3 λ − 1 ) = ( − 1 ) ( − 2 λ − 4 ) 9 λ − 3 = 2 λ + 4 9\lambda - 3 = 2\lambda + 4 9 λ − 3 = 2 λ + 4
3 ( 3 λ − 1 ) = 9 λ − 3 ( − 1 ) ( − 2 λ − 4 ) = 2 λ + 4 \begin{aligned} 3(3\lambda - 1) &= 9\lambda - 3 \\ (-1)(-2\lambda - 4) &= 2\lambda + 4 \end{aligned} 3 ( 3 λ − 1 ) ( − 1 ) ( − 2 λ − 4 ) = 9 λ − 3 = 2 λ + 4 ⇒ 7 λ = 7 ⇒ λ = 1 \Rightarrow 7\lambda = 7 \Rightarrow \lambda = 1 ⇒ 7 λ = 7 ⇒ λ = 1 Why?
Move the λ \lambda λ
9 λ − 3 = 2 λ + 4 9 λ − 2 λ = 4 + 3 7 λ = 7 λ = 7 7 = 1 \begin{aligned} 9\lambda - 3 &= 2\lambda + 4 \\ 9\lambda - 2\lambda &= 4 + 3 \\ 7\lambda &= 7 \\ \lambda &= \frac77 = 1 \end{aligned} 9 λ − 3 9 λ − 2 λ 7 λ λ = 2 λ + 4 = 4 + 3 = 7 = 7 7 = 1 At λ = 1 \lambda = 1 λ = 1
( 3 − 1 , − 2 + 3 , − 1 − 2 ) = ( 2 , 1 , − 3 ) \begin{aligned} &(3 - 1, -2 + 3, -1 - 2) \\ &= (2, 1, -3) \end{aligned} ( 3 − 1 , − 2 + 3 , − 1 − 2 ) = ( 2 , 1 , − 3 ) Why?
Put λ = 1 \lambda = 1 λ = 1
x = 3 ( 1 ) − 1 = 3 − 1 = 2 y = − 2 ( 1 ) + 3 = − 2 + 3 = 1 z = − ( 1 ) − 2 = − 1 − 2 = − 3 \begin{aligned} x &= 3(1) - 1 = 3 - 1 = 2 \\ y &= -2(1) + 3 = -2 + 3 = 1 \\ z &= -(1) - 2 = -1 - 2 = -3 \end{aligned} x y z = 3 ( 1 ) − 1 = 3 − 1 = 2 = − 2 ( 1 ) + 3 = − 2 + 3 = 1 = − ( 1 ) − 2 = − 1 − 2 = − 3 Check with the third ratio:
− 1 − 2 + 7 − 2 = − 2 = 3 − 1 − 1 \frac{-1 - 2 + 7}{-2} = -2 = \frac{3 - 1}{-1} − 2 − 1 − 2 + 7 = − 2 = − 1 3 − 1 Why?
λ \lambda λ λ = 1 \lambda = 1 λ = 1 z = − λ − 2 = − 3 z = -\lambda - 2 = -3 z = − λ − 2 = − 3 z + 7 − 2 = 4 − 2 = − 2 \frac{z + 7}{-2} = \frac{4}{-2} = -2 − 2 z + 7 = − 2 4 = − 2
third: z + 7 − 2 = − 3 + 7 − 2 = 4 − 2 = − 2 first: x − 1 = 3 ( 1 ) − 1 − 1 = 2 − 1 = − 2 \begin{aligned} \text{third: } \frac{z + 7}{-2} &= \frac{-3 + 7}{-2} \\ &= \frac{4}{-2} = -2 \\[4pt] \text{first: } \frac{x}{-1} &= \frac{3(1) - 1}{-1} \\ &= \frac{2}{-1} = -2 \end{aligned} third: − 2 z + 7 first: − 1 x = − 2 − 3 + 7 = − 2 4 = − 2 = − 1 3 ( 1 ) − 1 = − 1 2 = − 2 If it did not match, the lines would not meet.
∴ \therefore ∴ ( 2 , 1 , − 3 ) (2, 1, -3) ( 2 , 1 , − 3 )