The question
(a) Find the shortest distance between the lines
r ⃗ = 3 i ^ + 2 j ^ − 4 k ^ + λ ( i ^ + 2 j ^ + 2 k ^ ) \begin{aligned} \vec r &= 3\hat i + 2\hat j - 4\hat k \\ &\quad + \lambda(\hat i + 2\hat j + 2\hat k) \end{aligned} r = 3 i ^ + 2 j ^ − 4 k ^ + λ ( i ^ + 2 j ^ + 2 k ^ ) and
r ⃗ = 5 i ^ − 2 j ^ + μ ( 3 i ^ + 2 j ^ + 6 k ^ ) \begin{aligned} \vec r &= 5\hat i - 2\hat j \\ &\quad + \mu(3\hat i + 2\hat j + 6\hat k) \end{aligned} r = 5 i ^ − 2 j ^ + μ ( 3 i ^ + 2 j ^ + 6 k ^ ) If the lines intersect find their point of intersection
OR
(b) Find the foot of the perpendicular drawn from the point (-1, 3, -6) to the plane 2 x + y − 2 z + 5 = 0 2x + y - 2z + 5 = 0 2 x + y − 2 z + 5 = 0
Part (b) uses the equation of a plane. This topic is no longer in the CBSE syllabus (2025-26).
Exam answer Show all explanations
(a)
We have
a ⃗ 1 = 3 i ^ + 2 j ^ − 4 k ^ b ⃗ 1 = i ^ + 2 j ^ + 2 k ^ a ⃗ 2 = 5 i ^ − 2 j ^ b ⃗ 2 = 3 i ^ + 2 j ^ + 6 k ^ \begin{aligned} \vec a_1 &= 3\hat i + 2\hat j - 4\hat k \\ \vec b_1 &= \hat i + 2\hat j + 2\hat k \\ \vec a_2 &= 5\hat i - 2\hat j \\ \vec b_2 &= 3\hat i + 2\hat j + 6\hat k \end{aligned} a 1 b 1 a 2 b 2 = 3 i ^ + 2 j ^ − 4 k ^ = i ^ + 2 j ^ + 2 k ^ = 5 i ^ − 2 j ^ = 3 i ^ + 2 j ^ + 6 k ^ Why?
A line r ⃗ = a ⃗ + λ b ⃗ \vec r = \vec a + \lambda\vec b r = a + λ b a ⃗ \vec a a b ⃗ \vec b b a ⃗ \vec a a λ \lambda λ μ \mu μ b ⃗ \vec b b λ \lambda λ μ \mu μ
a ⃗ 2 − a ⃗ 1 = 2 i ^ − 4 j ^ + 4 k ^ \vec a_2 - \vec a_1 = 2\hat i - 4\hat j + 4\hat k a 2 − a 1 = 2 i ^ − 4 j ^ + 4 k ^ b ⃗ 1 × b ⃗ 2 = ∣ i ^ j ^ k ^ 1 2 2 3 2 6 ∣ = i ^ ( 12 − 4 ) − j ^ ( 6 − 6 ) + k ^ ( 2 − 6 ) = 8 i ^ − 4 k ^ \begin{aligned} \vec b_1 \times \vec b_2 &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 2 & 2 \\ 3 & 2 & 6 \end{vmatrix} \\ &= \hat i(12 - 4) - \hat j(6 - 6) \\ &\quad + \hat k(2 - 6) \\ &= 8\hat i - 4\hat k \end{aligned} b 1 × b 2 = i ^ 1 3 j ^ 2 2 k ^ 2 6 = i ^ ( 12 − 4 ) − j ^ ( 6 − 6 ) + k ^ ( 2 − 6 ) = 8 i ^ − 4 k ^ Why?
Subtract component by component:
a ⃗ 2 − a ⃗ 1 = ( 5 − 3 ) i ^ + ( − 2 − 2 ) j ^ + ( 0 + 4 ) k ^ \begin{aligned} &\vec a_2 - \vec a_1 \\ &= (5 - 3)\hat i + (-2 - 2)\hat j \\ &\quad + (0 + 4)\hat k \end{aligned} a 2 − a 1 = ( 5 − 3 ) i ^ + ( − 2 − 2 ) j ^ + ( 0 + 4 ) k ^ 5 − 3 = 2 − 2 − 2 = − 4 0 − ( − 4 ) = 0 + 4 = 4 \begin{aligned} 5 - 3 &= 2 \\ -2 - 2 &= -4 \\ 0 - (-4) &= 0 + 4 = 4 \end{aligned} 5 − 3 − 2 − 2 0 − ( − 4 ) = 2 = − 4 = 0 + 4 = 4 The cross product, expanded along the first row: the i ^ \hat i i ^ 2 × 6 − 2 × 2 2 \times 6 - 2 \times 2 2 × 6 − 2 × 2 j ^ \hat j j ^ 1 × 6 − 2 × 3 1 \times 6 - 2 \times 3 1 × 6 − 2 × 3 k ^ \hat k k ^ 1 × 2 − 2 × 3 1 \times 2 - 2 \times 3 1 × 2 − 2 × 3
Cover the i ^ \hat i i ^ j ^ \hat j j ^ k ^ \hat k k ^
i ^ : ∣ 2 2 2 6 ∣ = ( 2 ) ( 6 ) − ( 2 ) ( 2 ) = 12 − 4 = 8 j ^ : ∣ 1 2 3 6 ∣ = ( 1 ) ( 6 ) − ( 2 ) ( 3 ) = 6 − 6 = 0 with the minus: − 0 = 0 k ^ : ∣ 1 2 3 2 ∣ = ( 1 ) ( 2 ) − ( 2 ) ( 3 ) = 2 − 6 = − 4 \begin{aligned} \hat i&: \begin{vmatrix} 2 & 2 \\ 2 & 6 \end{vmatrix} \\ &= (2)(6) - (2)(2) \\ &= 12 - 4 = 8 \\[4pt] \hat j&: \begin{vmatrix} 1 & 2 \\ 3 & 6 \end{vmatrix} \\ &= (1)(6) - (2)(3) \\ &= 6 - 6 = 0 \\ &\text{with the minus: } -0 = 0 \\[4pt] \hat k&: \begin{vmatrix} 1 & 2 \\ 3 & 2 \end{vmatrix} \\ &= (1)(2) - (2)(3) \\ &= 2 - 6 = -4 \end{aligned} i ^ j ^ k ^ : 2 2 2 6 = ( 2 ) ( 6 ) − ( 2 ) ( 2 ) = 12 − 4 = 8 : 1 3 2 6 = ( 1 ) ( 6 ) − ( 2 ) ( 3 ) = 6 − 6 = 0 with the minus: − 0 = 0 : 1 3 2 2 = ( 1 ) ( 2 ) − ( 2 ) ( 3 ) = 2 − 6 = − 4 So b ⃗ 1 × b ⃗ 2 = 8 i ^ + 0 j ^ − 4 k ^ = 8 i ^ − 4 k ^ \vec b_1 \times \vec b_2 = 8\hat i + 0\hat j - 4\hat k = 8\hat i - 4\hat k b 1 × b 2 = 8 i ^ + 0 j ^ − 4 k ^ = 8 i ^ − 4 k ^
( b ⃗ 1 × b ⃗ 2 ) ⋅ ( a ⃗ 2 − a ⃗ 1 ) = ( 8 ) ( 2 ) + ( 0 ) ( − 4 ) + ( − 4 ) ( 4 ) = 16 + 0 − 16 = 0 \begin{aligned} &(\vec b_1 \times \vec b_2)\cdot(\vec a_2 - \vec a_1) \\ &= (8)(2) + (0)(-4) + (-4)(4) \\ &= 16 + 0 - 16 = 0 \end{aligned} ( b 1 × b 2 ) ⋅ ( a 2 − a 1 ) = ( 8 ) ( 2 ) + ( 0 ) ( − 4 ) + ( − 4 ) ( 4 ) = 16 + 0 − 16 = 0 Why?
The dot product multiplies matching components and adds. 8 i ^ − 4 k ^ 8\hat i - 4\hat k 8 i ^ − 4 k ^ j ^ \hat j j ^
So the shortest distance = ∣ ( b ⃗ 1 × b ⃗ 2 ) ⋅ ( a ⃗ 2 − a ⃗ 1 ) ∣ ∣ b ⃗ 1 × b ⃗ 2 ∣ = 0 = \frac{|(\vec b_1 \times \vec b_2)\cdot(\vec a_2 - \vec a_1)|}{|\vec b_1 \times \vec b_2|} = 0 = ∣ b 1 × b 2 ∣ ∣ ( b 1 × b 2 ) ⋅ ( a 2 − a 1 ) ∣ = 0
Why?
This formula gives the shortest distance between two lines that are not parallel. Here b ⃗ 1 × b ⃗ 2 ≠ 0 ⃗ \vec b_1 \times \vec b_2 \ne \vec 0 b 1 × b 2 = 0 ∣ b ⃗ 1 × b ⃗ 2 ∣ = 64 + 16 |\vec b_1 \times \vec b_2| = \sqrt{64 + 16} ∣ b 1 × b 2 ∣ = 64 + 16
∣ b ⃗ 1 × b ⃗ 2 ∣ = 8 2 + 0 2 + ( − 4 ) 2 = 64 + 0 + 16 = 80 distance = ∣ 0 ∣ 80 = 0 \begin{aligned} |\vec b_1 \times \vec b_2| &= \sqrt{8^2 + 0^2 + (-4)^2} \\ &= \sqrt{64 + 0 + 16} = \sqrt{80} \\[4pt] \text{distance} &= \frac{|0|}{\sqrt{80}} = 0 \end{aligned} ∣ b 1 × b 2 ∣ distance = 8 2 + 0 2 + ( − 4 ) 2 = 64 + 0 + 16 = 80 = 80 ∣0∣ = 0 A shortest distance of 0 means the lines meet.
∴ \therefore ∴
For the point of intersection,
3 i ^ + 2 j ^ − 4 k ^ + λ ( i ^ + 2 j ^ + 2 k ^ ) = 5 i ^ − 2 j ^ + μ ( 3 i ^ + 2 j ^ + 6 k ^ ) \begin{aligned} &3\hat i + 2\hat j - 4\hat k + \lambda(\hat i + 2\hat j + 2\hat k) \\ &= 5\hat i - 2\hat j + \mu(3\hat i + 2\hat j + 6\hat k) \end{aligned} 3 i ^ + 2 j ^ − 4 k ^ + λ ( i ^ + 2 j ^ + 2 k ^ ) = 5 i ^ − 2 j ^ + μ ( 3 i ^ + 2 j ^ + 6 k ^ ) ⇒ 3 + λ = 5 + 3 μ … ( i ) 2 + 2 λ = − 2 + 2 μ … ( ii ) − 4 + 2 λ = 6 μ … ( iii ) \begin{aligned} \Rightarrow 3 + \lambda &= 5 + 3\mu \quad\ldots(\text{i}) \\ 2 + 2\lambda &= -2 + 2\mu \quad\ldots(\text{ii}) \\ -4 + 2\lambda &= 6\mu \quad\ldots(\text{iii}) \end{aligned} ⇒ 3 + λ 2 + 2 λ − 4 + 2 λ = 5 + 3 μ … ( i ) = − 2 + 2 μ … ( ii ) = 6 μ … ( iii ) Why?
At the common point the two position vectors are equal, so the i ^ \hat i i ^ j ^ \hat j j ^ k ^ \hat k k ^ λ \lambda λ μ \mu μ
( i ) ⇒ λ − 3 μ = 2 ( ii ) ⇒ λ − μ = − 2 ( ii ) − ( i ) ⇒ 2 μ = − 4 \begin{aligned} (\text{i}) &\Rightarrow \lambda - 3\mu = 2 \\ (\text{ii}) &\Rightarrow \lambda - \mu = -2 \\ (\text{ii}) - (\text{i}) &\Rightarrow 2\mu = -4 \end{aligned} ( i ) ( ii ) ( ii ) − ( i ) ⇒ λ − 3 μ = 2 ⇒ λ − μ = − 2 ⇒ 2 μ = − 4 Solving (i) and (ii), we get μ = − 2 \mu = -2 μ = − 2 λ = − 4 \lambda = -4 λ = − 4 − 4 − 8 = − 12 = 6 ( − 2 ) -4 - 8 = -12 = 6(-2) − 4 − 8 = − 12 = 6 ( − 2 )
Why?
(ii) was divided by 2 first. From μ = − 2 \mu = -2 μ = − 2 λ = − 2 + μ = − 4 \lambda = -2 + \mu = -4 λ = − 2 + μ = − 4
Tidy each equation, one move per line:
( i ) : 3 + λ = 5 + 3 μ λ − 3 μ = 5 − 3 = 2 ( ii ) : 2 + 2 λ = − 2 + 2 μ 2 λ − 2 μ = − 2 − 2 = − 4 λ − μ = − 2 ( ÷ 2 ) \begin{aligned} (\text{i})&: 3 + \lambda = 5 + 3\mu \\ &\ \ \lambda - 3\mu = 5 - 3 = 2 \\[4pt] (\text{ii})&: 2 + 2\lambda = -2 + 2\mu \\ &\ \ 2\lambda - 2\mu = -2 - 2 = -4 \\ &\ \ \lambda - \mu = -2 \quad (\div 2) \end{aligned} ( i ) ( ii ) : 3 + λ = 5 + 3 μ λ − 3 μ = 5 − 3 = 2 : 2 + 2 λ = − 2 + 2 μ 2 λ − 2 μ = − 2 − 2 = − 4 λ − μ = − 2 ( ÷ 2 ) Subtract (i) from (ii) to remove λ \lambda λ
( λ − μ ) − ( λ − 3 μ ) = − 2 − 2 λ − μ − λ + 3 μ = − 4 2 μ = − 4 μ = − 2 λ = − 2 + μ = − 2 + ( − 2 ) = − 4 \begin{aligned} (\lambda - \mu) - (\lambda - 3\mu) &= -2 - 2 \\ \lambda - \mu - \lambda + 3\mu &= -4 \\ 2\mu &= -4 \\ \mu &= -2 \\[4pt] \lambda &= -2 + \mu \\ &= -2 + (-2) = -4 \end{aligned} ( λ − μ ) − ( λ − 3 μ ) λ − μ − λ + 3 μ 2 μ μ λ = − 2 − 2 = − 4 = − 4 = − 2 = − 2 + μ = − 2 + ( − 2 ) = − 4 Test in (iii):
Left: − 4 + 2 ( − 4 ) = − 4 − 8 = − 12 Right: 6 ( − 2 ) = − 12 ✓ \begin{aligned} \text{Left: } -4 + 2(-4) &= -4 - 8 = -12 \\ \text{Right: } 6(-2) &= -12\ \checkmark \end{aligned} Left: − 4 + 2 ( − 4 ) Right: 6 ( − 2 ) = − 4 − 8 = − 12 = − 12 ✓ Two equations fix λ \lambda λ μ \mu μ
Substituting μ = − 2 \mu = -2 μ = − 2
r ⃗ = 5 i ^ − 2 j ^ − 2 ( 3 i ^ + 2 j ^ + 6 k ^ ) = − i ^ − 6 j ^ − 12 k ^ \begin{aligned} \vec r &= 5\hat i - 2\hat j - 2(3\hat i + 2\hat j + 6\hat k) \\ &= -\hat i - 6\hat j - 12\hat k \end{aligned} r = 5 i ^ − 2 j ^ − 2 ( 3 i ^ + 2 j ^ + 6 k ^ ) = − i ^ − 6 j ^ − 12 k ^ Why?
− 2 ( 3 i ^ + 2 j ^ + 6 k ^ ) = − 6 i ^ − 4 j ^ − 12 k ^ \begin{aligned} &-2(3\hat i + 2\hat j + 6\hat k) \\ &= -6\hat i - 4\hat j - 12\hat k \end{aligned} − 2 ( 3 i ^ + 2 j ^ + 6 k ^ ) = − 6 i ^ − 4 j ^ − 12 k ^ Component by component: 5 − 6 = − 1 5 - 6 = -1 5 − 6 = − 1 − 2 − 4 = − 6 -2 - 4 = -6 − 2 − 4 = − 6 0 − 12 = − 12 0 - 12 = -12 0 − 12 = − 12
Check with the first line at λ = − 4 \lambda = -4 λ = − 4 3 − 4 = − 1 3 - 4 = -1 3 − 4 = − 1 2 − 8 = − 6 2 - 8 = -6 2 − 8 = − 6 − 4 − 8 = − 12 -4 - 8 = -12 − 4 − 8 = − 12
∴ \therefore ∴ ( − 1 , − 6 , − 12 ) (-1, -6, -12) ( − 1 , − 6 , − 12 )
(b)
Let P( − 1 , 3 , − 6 ) (-1, 3, -6) ( − 1 , 3 , − 6 ) 2 i ^ + j ^ − 2 k ^ 2\hat i + \hat j - 2\hat k 2 i ^ + j ^ − 2 k ^
x + 1 2 = y − 3 1 = z + 6 − 2 = λ \frac{x + 1}{2} = \frac{y - 3}{1} = \frac{z + 6}{-2} = \lambda 2 x + 1 = 1 y − 3 = − 2 z + 6 = λ Why?
The normal to a x + b y + c z + d = 0 ax + by + cz + d = 0 a x + b y + cz + d = 0 a , b , c a, b, c a , b , c 2 , 1 , − 2 2, 1, -2 2 , 1 , − 2
A line 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
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 Let the coordinates of Q be ( 2 λ − 1 , λ + 3 , − 2 λ − 6 ) (2\lambda - 1, \lambda + 3, -2\lambda - 6) ( 2 λ − 1 , λ + 3 , − 2 λ − 6 ) 2 x + y − 2 z + 5 = 0 2x + y - 2z + 5 = 0 2 x + y − 2 z + 5 = 0
Why?
x + 1 = 2 λ ⇒ x = 2 λ − 1 y − 3 = λ ⇒ y = λ + 3 z + 6 = − 2 λ ⇒ z = − 2 λ − 6 \begin{aligned} x + 1 = 2\lambda &\Rightarrow x = 2\lambda - 1 \\ y - 3 = \lambda &\Rightarrow y = \lambda + 3 \\ z + 6 = -2\lambda &\Rightarrow z = -2\lambda - 6 \end{aligned} x + 1 = 2 λ y − 3 = λ z + 6 = − 2 λ ⇒ x = 2 λ − 1 ⇒ y = λ + 3 ⇒ z = − 2 λ − 6 Setting each fraction equal to λ \lambda λ x = 2 λ − 1 x = 2\lambda - 1 x = 2 λ − 1 y = λ + 3 y = \lambda + 3 y = λ + 3 z = − 2 λ − 6 z = -2\lambda - 6 z = − 2 λ − 6
2 ( 2 λ − 1 ) + ( λ + 3 ) − 2 ( − 2 λ − 6 ) + 5 = 0 ⇒ 4 λ − 2 + λ + 3 + 4 λ + 12 + 5 = 0 ⇒ 9 λ + 18 = 0 ⇒ λ = − 2 \begin{aligned} &2(2\lambda - 1) + (\lambda + 3) \\ &\quad - 2(-2\lambda - 6) + 5 = 0 \\ &\Rightarrow 4\lambda - 2 + \lambda + 3 \\ &\quad + 4\lambda + 12 + 5 = 0 \\ &\Rightarrow 9\lambda + 18 = 0 \Rightarrow \lambda = -2 \end{aligned} 2 ( 2 λ − 1 ) + ( λ + 3 ) − 2 ( − 2 λ − 6 ) + 5 = 0 ⇒ 4 λ − 2 + λ + 3 + 4 λ + 12 + 5 = 0 ⇒ 9 λ + 18 = 0 ⇒ λ = − 2 Why?
Collect the λ \lambda λ 4 λ + λ + 4 λ = 9 λ 4\lambda + \lambda + 4\lambda = 9\lambda 4 λ + λ + 4 λ = 9 λ − 2 + 3 + 12 + 5 = 18 -2 + 3 + 12 + 5 = 18 − 2 + 3 + 12 + 5 = 18
2 ( 2 λ − 1 ) = 4 λ − 2 − 2 ( − 2 λ − 6 ) = 4 λ + 12 − 2 + 3 = 1 1 + 12 = 13 13 + 5 = 18 9 λ + 18 = 0 9 λ = − 18 λ = − 18 9 = − 2 \begin{aligned} 2(2\lambda - 1) &= 4\lambda - 2 \\ -2(-2\lambda - 6) &= 4\lambda + 12 \\[4pt] -2 + 3 &= 1 \\ 1 + 12 &= 13 \\ 13 + 5 &= 18 \\[4pt] 9\lambda + 18 &= 0 \\ 9\lambda &= -18 \\ \lambda &= \frac{-18}{9} = -2 \end{aligned} 2 ( 2 λ − 1 ) − 2 ( − 2 λ − 6 ) − 2 + 3 1 + 12 13 + 5 9 λ + 18 9 λ λ = 4 λ − 2 = 4 λ + 12 = 1 = 13 = 18 = 0 = − 18 = 9 − 18 = − 2 Watch out: − 2 × ( − 6 ) = + 12 -2 \times (-6) = +12 − 2 × ( − 6 ) = + 12
Q = ( − 4 − 1 , − 2 + 3 , 4 − 6 ) Q = (-4 - 1,\ -2 + 3,\ 4 - 6) Q = ( − 4 − 1 , − 2 + 3 , 4 − 6 ) Why?
Put λ = − 2 \lambda = -2 λ = − 2 ( 2 λ − 1 , λ + 3 , − 2 λ − 6 ) (2\lambda - 1, \lambda + 3, -2\lambda - 6) ( 2 λ − 1 , λ + 3 , − 2 λ − 6 ) 2 ( − 5 ) + 1 − 2 ( − 2 ) + 5 = 0 2(-5) + 1 - 2(-2) + 5 = 0 2 ( − 5 ) + 1 − 2 ( − 2 ) + 5 = 0
x = 2 ( − 2 ) − 1 = − 4 − 1 = − 5 y = − 2 + 3 = 1 z = − 2 ( − 2 ) − 6 = 4 − 6 = − 2 2 ( − 5 ) + 1 − 2 ( − 2 ) + 5 = − 10 + 1 + 4 + 5 = − 9 + 4 + 5 = − 5 + 5 = 0 \begin{aligned} x &= 2(-2) - 1 = -4 - 1 = -5 \\ y &= -2 + 3 = 1 \\ z &= -2(-2) - 6 = 4 - 6 = -2 \\[4pt] &2(-5) + 1 - 2(-2) + 5 \\ &= -10 + 1 + 4 + 5 \\ &= -9 + 4 + 5 = -5 + 5 = 0 \end{aligned} x y z = 2 ( − 2 ) − 1 = − 4 − 1 = − 5 = − 2 + 3 = 1 = − 2 ( − 2 ) − 6 = 4 − 6 = − 2 2 ( − 5 ) + 1 − 2 ( − 2 ) + 5 = − 10 + 1 + 4 + 5 = − 9 + 4 + 5 = − 5 + 5 = 0 ∴ \therefore ∴ ( − 5 , 1 , − 2 ) (-5, 1, -2) ( − 5 , 1 , − 2 )
The equation of the perpendicular is x + 1 2 = y − 3 1 = z + 6 − 2 \frac{x + 1}{2} = \frac{y - 3}{1} = \frac{z + 6}{-2} 2 x + 1 = 1 y − 3 = − 2 z + 6
PQ 2 = ( − 5 + 1 ) 2 + ( 1 − 3 ) 2 + ( − 2 + 6 ) 2 = 16 + 4 + 16 = 36 \begin{aligned} \text{PQ}^2 &= (-5 + 1)^2 + (1 - 3)^2 \\ &\quad + (-2 + 6)^2 \\ &= 16 + 4 + 16 = 36 \end{aligned} PQ 2 = ( − 5 + 1 ) 2 + ( 1 − 3 ) 2 + ( − 2 + 6 ) 2 = 16 + 4 + 16 = 36 Why?
Distance formula, with P( − 1 , 3 , − 6 ) (-1, 3, -6) ( − 1 , 3 , − 6 ) ( − 5 , 1 , − 2 ) (-5, 1, -2) ( − 5 , 1 , − 2 )
PQ 2 = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 + ( z 2 − z 1 ) 2 \begin{aligned} \text{PQ}^2 &= (x_2 - x_1)^2 \\ &\quad + (y_2 - y_1)^2 \\ &\quad + (z_2 - z_1)^2 \end{aligned} PQ 2 = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 + ( z 2 − z 1 ) 2 ( − 5 + 1 ) 2 = ( − 4 ) 2 = 16 ( 1 − 3 ) 2 = ( − 2 ) 2 = 4 ( − 2 + 6 ) 2 = 4 2 = 16 16 + 4 = 20 , 20 + 16 = 36 PQ = 36 = 6 \begin{aligned} (-5 + 1)^2 &= (-4)^2 = 16 \\ (1 - 3)^2 &= (-2)^2 = 4 \\ (-2 + 6)^2 &= 4^2 = 16 \\[4pt] 16 + 4 &= 20,\ 20 + 16 = 36 \\ \text{PQ} &= \sqrt{36} = 6 \end{aligned} ( − 5 + 1 ) 2 ( 1 − 3 ) 2 ( − 2 + 6 ) 2 16 + 4 PQ = ( − 4 ) 2 = 16 = ( − 2 ) 2 = 4 = 4 2 = 16 = 20 , 20 + 16 = 36 = 36 = 6 Check with the distance of a point from a plane:
∣ 2 ( − 1 ) + 3 − 2 ( − 6 ) + 5 ∣ 4 + 1 + 4 = ∣ − 2 + 3 + 12 + 5 ∣ 9 = 18 3 = 6 \begin{aligned} &\frac{|2(-1) + 3 - 2(-6) + 5|}{\sqrt{4 + 1 + 4}} \\ &= \frac{|-2 + 3 + 12 + 5|}{\sqrt9} \\ &= \frac{18}{3} = 6 \end{aligned} 4 + 1 + 4 ∣2 ( − 1 ) + 3 − 2 ( − 6 ) + 5∣ = 9 ∣ − 2 + 3 + 12 + 5∣ = 3 18 = 6 ∴ \therefore ∴ PQ = 36 = 6 \text{PQ} = \sqrt{36} = 6 PQ = 36 = 6
Where this comes from Chapter 11, Three Dimensional Geometry : Shortest distance; point of intersection of lines.
Question type: (a) The shortest distance between two lines + the foot of a perpendicular (combined)
How to spot it: click to see (a) Type 8. The shortest distance between two lines
Spot it: "Find the shortest distance between the lines". First look at the two directions.
Here: The directions i ^ + 2 j ^ + 2 k ^ \hat i + 2\hat j + 2\hat k i ^ + 2 j ^ + 2 k ^ 3 i ^ + 2 j ^ + 6 k ^ 3\hat i + 2\hat j + 6\hat k 3 i ^ + 2 j ^ + 6 k ^ ( 8 i ^ − 4 k ^ ) ⋅ ( 2 i ^ − 4 j ^ + 4 k ^ ) (8\hat i - 4\hat k)\cdot(2\hat i - 4\hat j + 4\hat k) ( 8 i ^ − 4 k ^ ) ⋅ ( 2 i ^ − 4 j ^ + 4 k ^ )
(a) Type 10. Beyond the book, but in CBSE papers: the foot of a perpendicular
Spot it: "Find the foot of the perpendicular from a point to a line", "the image of a point in a line", or "do the lines meet?". This book has none of these, but recent CBSE sample papers ask all three.
Here: A distance of 0 means the lines meet: set 3 + λ = 5 + 3 μ 3 + \lambda = 5 + 3\mu 3 + λ = 5 + 3 μ 2 + 2 λ = − 2 + 2 μ 2 + 2\lambda = -2 + 2\mu 2 + 2 λ = − 2 + 2 μ μ = − 2 \mu = -2 μ = − 2 k ^ \hat k k ^ ( − 1 , − 6 , − 12 ) (-1, -6, -12) ( − 1 , − 6 , − 12 )
Quick lesson · Worked NCERT solutions · Interactive lesson
Also uses Chapter 10, Vector Algebra .