The question
(a) Find the shortest distance between the following lines:
r ⃗ = ( i ^ + j ^ − k ^ ) + s ( 2 i ^ + j ^ + k ^ ) \begin{aligned} \vec r &= (\hat i + \hat j - \hat k) \\ &\quad + s(2\hat i + \hat j + \hat k) \end{aligned} r = ( i ^ + j ^ − k ^ ) + s ( 2 i ^ + j ^ + k ^ ) r ⃗ = ( i ^ + j ^ + 2 k ^ ) + t ( 4 i ^ + 2 j ^ + 2 k ^ ) \begin{aligned} \vec r &= (\hat i + \hat j + 2\hat k) \\ &\quad + t(4\hat i + 2\hat j + 2\hat k) \end{aligned} r = ( i ^ + j ^ + 2 k ^ ) + t ( 4 i ^ + 2 j ^ + 2 k ^ ) OR
(b) Find the vector and the cartesian equations of the plane containing the point i ^ + 2 j ^ − k ^ \hat i + 2\hat j - \hat k i ^ + 2 j ^ − k ^
r ⃗ = ( i ^ + 2 j ^ + 2 k ^ ) + s ( 2 i ^ − 3 j ^ + 2 k ^ ) \begin{aligned} \vec r &= (\hat i + 2\hat j + 2\hat k) \\ &\quad + s(2\hat i - 3\hat j + 2\hat k) \end{aligned} r = ( i ^ + 2 j ^ + 2 k ^ ) + s ( 2 i ^ − 3 j ^ + 2 k ^ ) and
r ⃗ = ( 3 i ^ + j ^ − 2 k ^ ) + t ( i ^ − 3 j ^ + k ^ ) \begin{aligned} \vec r &= (3\hat i + \hat j - 2\hat k) \\ &\quad + t(\hat i - 3\hat j + \hat k) \end{aligned} r = ( 3 i ^ + j ^ − 2 k ^ ) + t ( i ^ − 3 j ^ + k ^ ) Part (b), the equation of a plane, is no longer in the CBSE syllabus (2025-26).
Exam answer Show all explanations
(a)
Here a ⃗ 1 = i ^ + j ^ − k ^ \vec a_1 = \hat i + \hat j - \hat k a 1 = i ^ + j ^ − k ^ a ⃗ 2 = i ^ + j ^ + 2 k ^ \vec a_2 = \hat i + \hat j + 2\hat k a 2 = i ^ + j ^ + 2 k ^ b ⃗ = 2 i ^ + j ^ + k ^ \vec b = 2\hat i + \hat j + \hat k b = 2 i ^ + j ^ + 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
As 4 i ^ + 2 j ^ + 2 k ^ = 2 ( 2 i ^ + j ^ + k ^ ) 4\hat i + 2\hat j + 2\hat k = 2(2\hat i + \hat j + \hat k) 4 i ^ + 2 j ^ + 2 k ^ = 2 ( 2 i ^ + j ^ + k ^ )
Why?
The direction vectors are multiples of each other, so they point the same way and the lines are parallel. The formula for skew lines cannot be used: its denominator ∣ b ⃗ 1 × b ⃗ 2 ∣ |\vec b_1 \times \vec b_2| ∣ b 1 × b 2 ∣
S.D. = ∣ ( a ⃗ 2 − a ⃗ 1 ) × b ⃗ ∣ ∣ b ⃗ ∣ \text{S.D.} = \frac{|(\vec a_2 - \vec a_1) \times \vec b|}{|\vec b|} S.D. = ∣ b ∣ ∣ ( a 2 − a 1 ) × b ∣ Why?
∣ ( a ⃗ 2 − a ⃗ 1 ) × b ⃗ ∣ |(\vec a_2 - \vec a_1) \times \vec b| ∣ ( a 2 − a 1 ) × b ∣ a ⃗ 2 − a ⃗ 1 \vec a_2 - \vec a_1 a 2 − a 1 b ⃗ \vec b b ∣ b ⃗ ∣ |\vec b| ∣ b ∣
a ⃗ 2 − a ⃗ 1 = 3 k ^ \vec a_2 - \vec a_1 = 3\hat k a 2 − a 1 = 3 k ^ Why?
Subtract matching components of a ⃗ 2 = i ^ + j ^ + 2 k ^ \vec a_2 = \hat i + \hat j + 2\hat k a 2 = i ^ + j ^ + 2 k ^ a ⃗ 1 = i ^ + j ^ − k ^ \vec a_1 = \hat i + \hat j - \hat k a 1 = i ^ + j ^ − k ^
i ^ : 1 − 1 = 0 j ^ : 1 − 1 = 0 k ^ : 2 − ( − 1 ) = 2 + 1 = 3 \begin{aligned} \hat i&:\ 1 - 1 = 0 \\ \hat j&:\ 1 - 1 = 0 \\ \hat k&:\ 2 - (-1) = 2 + 1 = 3 \end{aligned} i ^ j ^ k ^ : 1 − 1 = 0 : 1 − 1 = 0 : 2 − ( − 1 ) = 2 + 1 = 3 ( 3 k ^ ) × ( 2 i ^ + j ^ + k ^ ) = ∣ i ^ j ^ k ^ 0 0 3 2 1 1 ∣ = − 3 i ^ + 6 j ^ \begin{aligned} &(3\hat k) \times (2\hat i + \hat j + \hat k) \\ &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 0 & 0 & 3 \\ 2 & 1 & 1 \end{vmatrix} = -3\hat i + 6\hat j \end{aligned} ( 3 k ^ ) × ( 2 i ^ + j ^ + k ^ ) = i ^ 0 2 j ^ 0 1 k ^ 3 1 = − 3 i ^ + 6 j ^ Why?
Expand the determinant along the first row:
i ^ : 0 ⋅ 1 − 3 ⋅ 1 = − 3 j ^ : − ( 0 ⋅ 1 − 3 ⋅ 2 ) = 6 k ^ : 0 ⋅ 1 − 0 ⋅ 2 = 0 \begin{aligned} \hat i&:\ 0 \cdot 1 - 3 \cdot 1 = -3 \\ \hat j&:\ -(0 \cdot 1 - 3 \cdot 2) = 6 \\ \hat k&:\ 0 \cdot 1 - 0 \cdot 2 = 0 \end{aligned} i ^ j ^ k ^ : 0 ⋅ 1 − 3 ⋅ 1 = − 3 : − ( 0 ⋅ 1 − 3 ⋅ 2 ) = 6 : 0 ⋅ 1 − 0 ⋅ 2 = 0 Each one in full:
i ^ : ( 0 ) ( 1 ) − ( 3 ) ( 1 ) = 0 − 3 = − 3 j ^ : − ( ( 0 ) ( 1 ) − ( 3 ) ( 2 ) ) = − ( 0 − 6 ) = − ( − 6 ) = 6 k ^ : ( 0 ) ( 1 ) − ( 0 ) ( 2 ) = 0 − 0 = 0 \begin{aligned} \hat i&:\ (0)(1) - (3)(1) = 0 - 3 = -3 \\ \hat j&:\ -\big((0)(1) - (3)(2)\big) \\ &\quad = -(0 - 6) = -(-6) = 6 \\ \hat k&:\ (0)(1) - (0)(2) = 0 - 0 = 0 \end{aligned} i ^ j ^ k ^ : ( 0 ) ( 1 ) − ( 3 ) ( 1 ) = 0 − 3 = − 3 : − ( ( 0 ) ( 1 ) − ( 3 ) ( 2 ) ) = − ( 0 − 6 ) = − ( − 6 ) = 6 : ( 0 ) ( 1 ) − ( 0 ) ( 2 ) = 0 − 0 = 0 Watch out for the minus sign in front of the j ^ \hat j j ^
S.D. = ∣ − 3 i ^ + 6 j ^ ∣ 4 + 1 + 1 = 9 + 36 6 = 3 5 6 \begin{aligned} \text{S.D.} &= \frac{\lvert -3\hat i + 6\hat j\rvert}{\sqrt{4 + 1 + 1}} \\ &= \frac{\sqrt{9 + 36}}{\sqrt6} = \frac{3\sqrt5}{\sqrt6} \end{aligned} S.D. = 4 + 1 + 1 ∣ − 3 i ^ + 6 j ^ ∣ = 6 9 + 36 = 6 3 5 Why?
For v ⃗ = x i ^ + y j ^ + z k ^ \vec v = x\hat i + y\hat j + z\hat k v = x i ^ + y j ^ + z k ^ ∣ v ⃗ ∣ = x 2 + y 2 + z 2 |\vec v| = \sqrt{x^2 + y^2 + z^2} ∣ v ∣ = x 2 + y 2 + z 2 ∣ b ⃗ ∣ = 2 2 + 1 2 + 1 2 = 6 |\vec b| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt6 ∣ b ∣ = 2 2 + 1 2 + 1 2 = 6 45 = 9 × 5 = 3 5 \sqrt{45} = \sqrt{9 \times 5} = 3\sqrt5 45 = 9 × 5 = 3 5
∣ − 3 i ^ + 6 j ^ ∣ = ( − 3 ) 2 + ( 6 ) 2 + ( 0 ) 2 = 9 + 36 + 0 = 45 = 9 × 5 = 3 5 ∣ b ⃗ ∣ = ( 2 ) 2 + ( 1 ) 2 + ( 1 ) 2 = 4 + 1 + 1 = 6 \begin{aligned} |-3\hat i + 6\hat j| &= \sqrt{(-3)^2 + (6)^2 + (0)^2} \\ &= \sqrt{9 + 36 + 0} = \sqrt{45} \\ &= \sqrt{9 \times 5} = 3\sqrt5 \\[4pt] |\vec b| &= \sqrt{(2)^2 + (1)^2 + (1)^2} \\ &= \sqrt{4 + 1 + 1} = \sqrt6 \end{aligned} ∣ − 3 i ^ + 6 j ^ ∣ ∣ b ∣ = ( − 3 ) 2 + ( 6 ) 2 + ( 0 ) 2 = 9 + 36 + 0 = 45 = 9 × 5 = 3 5 = ( 2 ) 2 + ( 1 ) 2 + ( 1 ) 2 = 4 + 1 + 1 = 6 The same answer can be written as 30 2 \frac{\sqrt{30}}{2} 2 30 6 \sqrt6 6
3 5 6 = 3 5 × 6 6 × 6 = 3 30 6 = 30 2 ( divide by 3 ) \begin{aligned} \frac{3\sqrt5}{\sqrt6} &= \frac{3\sqrt5 \times \sqrt6}{\sqrt6 \times \sqrt6} \\ &= \frac{3\sqrt{30}}{6} \\ &= \frac{\sqrt{30}}{2} \quad (\text{divide by } 3) \end{aligned} 6 3 5 = 6 × 6 3 5 × 6 = 6 3 30 = 2 30 ( divide by 3 ) ∴ \therefore ∴ = 3 5 6 = \frac{3\sqrt5}{\sqrt6} = 6 3 5
(b)
The plane is parallel to the given lines, so the cross product of 2 i ^ − 3 j ^ + 2 k ^ 2\hat i - 3\hat j + 2\hat k 2 i ^ − 3 j ^ + 2 k ^ i ^ − 3 j ^ + k ^ \hat i - 3\hat j + \hat k i ^ − 3 j ^ + k ^
Why?
A plane parallel to two lines contains both their directions. The cross product of two vectors is perpendicular to each of them, so it is perpendicular to the plane: a normal.
( 2 i ^ − 3 j ^ + 2 k ^ ) × ( i ^ − 3 j ^ + k ^ ) = ∣ i ^ j ^ k ^ 2 − 3 2 1 − 3 1 ∣ = i ^ ( − 3 + 6 ) − j ^ ( 2 − 2 ) + k ^ ( − 6 + 3 ) = 3 i ^ − 3 k ^ \begin{aligned} &(2\hat i - 3\hat j + 2\hat k) \times (\hat i - 3\hat j + \hat k) \\ &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & -3 & 2 \\ 1 & -3 & 1 \end{vmatrix} \\ &= \hat i(-3 + 6) - \hat j(2 - 2) \\ &\quad + \hat k(-6 + 3) \\ &= 3\hat i - 3\hat k \end{aligned} ( 2 i ^ − 3 j ^ + 2 k ^ ) × ( i ^ − 3 j ^ + k ^ ) = i ^ 2 1 j ^ − 3 − 3 k ^ 2 1 = i ^ ( − 3 + 6 ) − j ^ ( 2 − 2 ) + k ^ ( − 6 + 3 ) = 3 i ^ − 3 k ^ Why?
Each bracket is a 2 by 2 determinant from the last two rows:
i ^ : ( − 3 ) ( 1 ) − ( 2 ) ( − 3 ) j ^ : ( 2 ) ( 1 ) − ( 2 ) ( 1 ) k ^ : ( 2 ) ( − 3 ) − ( − 3 ) ( 1 ) \begin{aligned} \hat i&:\ (-3)(1) - (2)(-3) \\ \hat j&:\ (2)(1) - (2)(1) \\ \hat k&:\ (2)(-3) - (-3)(1) \end{aligned} i ^ j ^ k ^ : ( − 3 ) ( 1 ) − ( 2 ) ( − 3 ) : ( 2 ) ( 1 ) − ( 2 ) ( 1 ) : ( 2 ) ( − 3 ) − ( − 3 ) ( 1 ) Work out each one:
i ^ : − 3 − ( − 6 ) = − 3 + 6 = 3 j ^ : 2 − 2 = 0 k ^ : − 6 − ( − 3 ) = − 6 + 3 = − 3 \begin{aligned} \hat i&:\ -3 - (-6) = -3 + 6 = 3 \\ \hat j&:\ 2 - 2 = 0 \\ \hat k&:\ -6 - (-3) = -6 + 3 = -3 \end{aligned} i ^ j ^ k ^ : − 3 − ( − 6 ) = − 3 + 6 = 3 : 2 − 2 = 0 : − 6 − ( − 3 ) = − 6 + 3 = − 3 The j ^ \hat j j ^ − j ^ ( 0 ) = 0 -\hat j(0) = 0 − j ^ ( 0 ) = 0
The plane passes through the point i ^ + 2 j ^ − k ^ \hat i + 2\hat j - \hat k i ^ + 2 j ^ − k ^
Why?
A plane through the point with position vector a ⃗ \vec a a n ⃗ \vec n n
( r ⃗ − a ⃗ ) ⋅ n ⃗ = 0 (\vec r - \vec a)\cdot\vec n = 0 ( r − a ) ⋅ n = 0 that is, r ⃗ ⋅ n ⃗ = a ⃗ ⋅ n ⃗ \vec r\cdot\vec n = \vec a\cdot\vec n r ⋅ n = a ⋅ n
r ⃗ ⋅ ( 3 i ^ − 3 k ^ ) = ( i ^ + 2 j ^ − k ^ ) ⋅ ( 3 i ^ − 3 k ^ ) = 3 + 0 + 3 = 6 \begin{aligned} &\vec r\cdot(3\hat i - 3\hat k) \\ &= (\hat i + 2\hat j - \hat k)\cdot(3\hat i - 3\hat k) \\ &= 3 + 0 + 3 = 6 \end{aligned} r ⋅ ( 3 i ^ − 3 k ^ ) = ( i ^ + 2 j ^ − k ^ ) ⋅ ( 3 i ^ − 3 k ^ ) = 3 + 0 + 3 = 6 i.e. r ⃗ ⋅ ( i ^ − k ^ ) = 2 \vec r\cdot(\hat i - \hat k) = 2 r ⋅ ( i ^ − k ^ ) = 2
Why?
Dot product: 1 × 3 + 2 × 0 + ( − 1 ) ( − 3 ) = 6 1 \times 3 + 2 \times 0 + (-1)(-3) = 6 1 × 3 + 2 × 0 + ( − 1 ) ( − 3 ) = 6
( 1 ) ( 3 ) + ( 2 ) ( 0 ) + ( − 1 ) ( − 3 ) = 3 + 0 + 3 = 6 3 i ^ − 3 k ^ 3 = i ^ − k ^ , 6 3 = 2 \begin{aligned} &(1)(3) + (2)(0) + (-1)(-3) \\ &= 3 + 0 + 3 = 6 \\[4pt] &\frac{3\hat i - 3\hat k}{3} = \hat i - \hat k,\quad \frac63 = 2 \end{aligned} ( 1 ) ( 3 ) + ( 2 ) ( 0 ) + ( − 1 ) ( − 3 ) = 3 + 0 + 3 = 6 3 3 i ^ − 3 k ^ = i ^ − k ^ , 3 6 = 2 Putting r ⃗ = x i ^ + y j ^ + z k ^ \vec r = x\hat i + y\hat j + z\hat k r = x i ^ + y j ^ + z k ^
Why?
( x i ^ + y j ^ + z k ^ ) ⋅ ( i ^ − k ^ ) = x − z (x\hat i + y\hat j + z\hat k)\cdot(\hat i - \hat k) = x - z ( x i ^ + y j ^ + z k ^ ) ⋅ ( i ^ − k ^ ) = x − z x − z = 2 x - z = 2 x − z = 2
( x ) ( 1 ) + ( y ) ( 0 ) + ( z ) ( − 1 ) = x + 0 − z = x − z x − z = 2 x − z − 2 = 0 \begin{aligned} &(x)(1) + (y)(0) + (z)(-1) \\ &= x + 0 - z = x - z \\[4pt] &x - z = 2 \\ &x - z - 2 = 0 \end{aligned} ( x ) ( 1 ) + ( y ) ( 0 ) + ( z ) ( − 1 ) = x + 0 − z = x − z x − z = 2 x − z − 2 = 0 ∴ \therefore ∴ x − z − 2 = 0 x - z - 2 = 0 x − z − 2 = 0
Where this comes from Chapter 11, Three Dimensional Geometry : Shortest distance between parallel lines.
Question type: (a) The shortest distance between two lines
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: 4 i ^ + 2 j ^ + 2 k ^ = 2 ( 2 i ^ + j ^ + k ^ ) 4\hat i + 2\hat j + 2\hat k = 2(2\hat i + \hat j + \hat k) 4 i ^ + 2 j ^ + 2 k ^ = 2 ( 2 i ^ + j ^ + k ^ ) a ⃗ 2 − a ⃗ 1 = 3 k ^ \vec a_2 - \vec a_1 = 3\hat k a 2 − a 1 = 3 k ^
Quick lesson · Worked NCERT solutions · Interactive lesson
Also uses Chapter 10, Vector Algebra .