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CBSE Board Examination 2025 · Section C · 3 marks

Q29: Skew lines and their shortest distance, or the ratio in which the wicketkeeper divides bowler to fielder

Board 2025Ch 11 · Three Dimensional GeometryCh 10 · Vector Algebra
The question

(a) Verify that lines given by r⃗=(1−λ)i^+(λ−2)j^+(3−2λ)k^\vec r = (1 - \lambda)\hat i + (\lambda - 2)\hat j + (3 - 2\lambda)\hat k and r⃗=(μ+1)i^+(2μ−1)j^−(2μ+1)k^\vec r = (\mu + 1)\hat i + (2\mu - 1)\hat j - (2\mu + 1)\hat k are skew lines. Hence, find shortest distance between the lines.

OR

(b) During a cricket match, the position of the bowler, the wicket keeper and the leg slip fielder are in a line given by B⃗=2i^+8j^\vec B = 2\hat i + 8\hat j⁠, W⃗=6i^+12j^\vec W = 6\hat i + 12\hat j and F⃗=12i^+18j^\vec F = 12\hat i + 18\hat j respectively. Calculate the ratio in which the wicketkeeper divides the line segment joining the bowler and the leg slip fielder.

Exam answer

(a)

Rewriting the lines,

r⃗=(i^−2j^+3k^)+λ(−i^+j^−2k^)\begin{aligned} \vec r &= (\hat i - 2\hat j + 3\hat k) \\ &\quad + \lambda(-\hat i + \hat j - 2\hat k) \end{aligned} r⃗=(i^−j^−k^)+μ(i^+2j^−2k^)\begin{aligned} \vec r &= (\hat i - \hat j - \hat k) \\ &\quad + \mu(\hat i + 2\hat j - 2\hat k) \end{aligned}
Why?

Collect the terms without λ\lambda and the terms with λ\lambda⁠. For example (1−λ)i^=i^−λi^(1 - \lambda)\hat i = \hat i - \lambda\hat i and (3−2λ)k^=3k^−2λk^(3 - 2\lambda)\hat k = 3\hat k - 2\lambda\hat k⁠.

This gives the form r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b⁠: a⃗\vec a is a point on the line and b⃗\vec b is its direction. In the second line, −(2μ+1)k^=−k^−2μk^-(2\mu + 1)\hat k = -\hat k - 2\mu\hat k⁠.

So a⃗1=i^−2j^+3k^\vec a_1 = \hat i - 2\hat j + 3\hat k⁠, b⃗1=−i^+j^−2k^\vec b_1 = -\hat i + \hat j - 2\hat k⁠, a⃗2=i^−j^−k^\vec a_2 = \hat i - \hat j - \hat k and b⃗2=i^+2j^−2k^\vec b_2 = \hat i + 2\hat j - 2\hat k⁠.

The direction ratios −1,1,−2-1, 1, -2 and 1,2,−21, 2, -2 are not proportional, so the lines are not parallel.

Why?

Parallel lines have proportional direction ratios. Here −11=−1\frac{-1}{1} = -1 but 12≠−1\frac{1}{2} \ne -1⁠, so they are not proportional.

Skew lines are lines that are neither parallel nor intersecting. So we must also show that the lines do not meet.

a⃗2−a⃗1=j^−4k^\vec a_2 - \vec a_1 = \hat j - 4\hat k b⃗1×b⃗2=∣i^j^k^−11−212−2∣\vec b_1 \times \vec b_2 = \begin{vmatrix} \hat i & \hat j & \hat k \\ -1 & 1 & -2 \\ 1 & 2 & -2 \end{vmatrix} =i^(−2+4)−j^(2+2)+k^(−2−1)=2i^−4j^−3k^\begin{aligned} &= \hat i(-2 + 4) - \hat j(2 + 2) \\ &\quad + \hat k(-2 - 1) \\ &= 2\hat i - 4\hat j - 3\hat k \end{aligned}
Why?

Subtract the matching components:

a⃗2−a⃗1=(1−1)i^+(−1+2)j^+(−1−3)k^=0i^+1j^−4k^\begin{aligned} \vec a_2 - \vec a_1 &= (1 - 1)\hat i \\ &\quad + (-1 + 2)\hat j \\ &\quad + (-1 - 3)\hat k \\ &= 0\hat i + 1\hat j - 4\hat k \end{aligned}

The determinant is expanded along the first row:

i^: (1)(−2)−(−2)(2)j^: (−1)(−2)−(−2)(1)k^: (−1)(2)−(1)(1)\begin{aligned} \hat i&:\ (1)(-2) - (-2)(2) \\ \hat j&:\ (-1)(-2) - (-2)(1) \\ \hat k&:\ (-1)(2) - (1)(1) \end{aligned}

Work out each one:

i^: −2−(−4)=−2+4=2j^: 2−(−2)=2+2=4k^: −2−1=−3\begin{aligned} \hat i&:\ -2 - (-4) = -2 + 4 = 2 \\ \hat j&:\ 2 - (-2) = 2 + 2 = 4 \\ \hat k&:\ -2 - 1 = -3 \end{aligned}

With the signs ++⁠, −-⁠, ++⁠:

i^(2)−j^(4)+k^(−3)=2i^−4j^−3k^\begin{aligned} &\hat i(2) - \hat j(4) + \hat k(-3) \\ &= 2\hat i - 4\hat j - 3\hat k \end{aligned}

Watch out: the j^\hat j term has a minus sign in front.

(a⃗2−a⃗1)⋅(b⃗1×b⃗2)=0(2)+1(−4)+(−4)(−3)=−4+12=8≠0\begin{aligned} &(\vec a_2 - \vec a_1)\cdot(\vec b_1 \times \vec b_2) \\ &= 0(2) + 1(-4) + (-4)(-3) \\ &= -4 + 12 = 8 \ne 0 \end{aligned}

So the lines do not intersect. As they are neither parallel nor intersecting, the lines are skew.

Why?

b⃗1×b⃗2\vec b_1 \times \vec b_2 is perpendicular to both lines. Two non-parallel lines meet exactly when they lie in one plane, that is when

(a⃗2−a⃗1)⋅(b⃗1×b⃗2)=0(\vec a_2 - \vec a_1)\cdot(\vec b_1 \times \vec b_2) = 0

Here this product is 88⁠, not 00⁠, so the lines cannot meet. Multiply matching components of (0,1,−4)(0, 1, -4) and (2,−4,−3)(2, -4, -3)⁠:

(0)(2)=0(1)(−4)=−4(−4)(−3)=120−4+12=8\begin{aligned} (0)(2) &= 0 \\ (1)(-4) &= -4 \\ (-4)(-3) &= 12 \\[4pt] 0 - 4 + 12 &= 8 \end{aligned}
∣b⃗1×b⃗2∣=4+16+9=29|\vec b_1 \times \vec b_2| = \sqrt{4 + 16 + 9} = \sqrt{29}

Shortest distance

=∣(a⃗2−a⃗1)⋅(b⃗1×b⃗2)∣∣b⃗1×b⃗2∣=829\begin{aligned} &= \frac{|(\vec a_2 - \vec a_1)\cdot(\vec b_1 \times \vec b_2)|}{|\vec b_1 \times \vec b_2|} \\ &= \frac{8}{\sqrt{29}} \end{aligned}
Why?

The shortest distance between skew lines is measured along their common perpendicular, the direction of b⃗1×b⃗2\vec b_1 \times \vec b_2⁠. It is the length of the projection of a⃗2−a⃗1\vec a_2 - \vec a_1 on that direction, which is the formula used here.

The numerator is ∣8∣=8|8| = 8⁠, already found above.

The denominator is the length of 2i^−4j^−3k^2\hat i - 4\hat j - 3\hat k⁠:

∣b⃗1×b⃗2∣=(2)2+(−4)2+(−3)2=4+16+9=29\begin{aligned} |\vec b_1 \times \vec b_2| &= \sqrt{(2)^2 + (-4)^2 + (-3)^2} \\ &= \sqrt{4 + 16 + 9} \\ &= \sqrt{29} \end{aligned}

∴\therefore The lines are skew, and the shortest distance is 829\frac{8}{\sqrt{29}} units.

(b)

Let W divide BF in the ratio k:1k : 1⁠. By the section formula,

W⃗=kF⃗+1⋅B⃗k+1\vec W = \frac{k\vec F + 1\cdot\vec B}{k + 1}
Why?

Taking the ratio as k:1k : 1 leaves only one unknown. The point dividing the join of PP and QQ internally in the ratio m:nm : n has position vector

mQ⃗+nP⃗m+n\frac{m\vec Q + n\vec P}{m + n}

Here P=BP = B⁠, Q=FQ = F⁠, m=km = k and n=1n = 1⁠. Watch out: the far end FF gets the weight kk⁠.

6i^+12j^=(12k+2)i^+(18k+8)j^k+1\begin{aligned} &6\hat i + 12\hat j \\ &= \frac{(12k + 2)\hat i + (18k + 8)\hat j}{k + 1} \end{aligned}
Why?

Substitute the position vectors:

kF⃗+B⃗=k(12i^+18j^)+(2i^+8j^)\begin{aligned} k\vec F + \vec B &= k(12\hat i + 18\hat j) \\ &\quad + (2\hat i + 8\hat j) \end{aligned}

and collect the i^\hat i terms and the j^\hat j terms.

=12ki^+18kj^+2i^+8j^=(12k+2)i^+(18k+8)j^\begin{aligned} &= 12k\hat i + 18k\hat j + 2\hat i + 8\hat j \\ &= (12k + 2)\hat i + (18k + 8)\hat j \end{aligned}

Comparing the coefficients of i^\hat i⁠,

6(k+1)=12k+2⇒6k=4⇒k=23\begin{aligned} 6(k + 1) &= 12k + 2 \\ \Rightarrow 6k &= 4 \Rightarrow k = \frac23 \end{aligned}
Why?

Multiply both sides by k+1k + 1⁠. Two vectors are equal only when their i^\hat i parts and their j^\hat j parts are equal. Then 6k+6=12k+26k + 6 = 12k + 2 gives 4=6k4 = 6k⁠.

(12k+2)k+1=612k+2=6(k+1)12k+2=6k+612k−6k=6−26k=4k=46=4÷26÷2=23\begin{aligned} \frac{(12k + 2)}{k + 1} &= 6 \\ 12k + 2 &= 6(k + 1) \\ 12k + 2 &= 6k + 6 \\ 12k - 6k &= 6 - 2 \\ 6k &= 4 \\ k &= \frac46 = \frac{4 \div 2}{6 \div 2} = \frac23 \end{aligned}

The coefficients of j^\hat j give 12(k+1)=18k+812(k + 1) = 18k + 8⁠, that is k=23k = \frac23 again.

Why?
12(k+1)=18k+812k+12=18k+812−8=18k−12k4=6kk=46=23\begin{aligned} 12(k + 1) &= 18k + 8 \\ 12k + 12 &= 18k + 8 \\ 12 - 8 &= 18k - 12k \\ 4 &= 6k \\ k &= \frac46 = \frac23 \end{aligned}

12k+12=18k+812k + 12 = 18k + 8 gives 6k=46k = 4⁠. Both parts giving the same kk confirms that W lies on BF.

k:1=23:1=2:3k : 1 = \frac23 : 1 = 2 : 3 (multiply both by 33⁠). As kk is positive, W lies between B and F.

∴\therefore The wicketkeeper divides the segment in the ratio 2:32 : 3⁠.

Where this comes from

Chapter 11, Three Dimensional Geometry: Skew lines and shortest distance; section formula.

Question type: (a) The shortest distance between two lines; (b) Section formula

How to spot it: click to see

(a) Ch 11, type 8. The shortest distance between two lines

Spot it: "Find the shortest distance between the lines". First look at the two directions.

Here: Collect the λ\lambda and μ\mu terms to read b⃗1=−i^+j^−2k^\vec b_1 = -\hat i + \hat j - 2\hat k and b⃗2=i^+2j^−2k^\vec b_2 = \hat i + 2\hat j - 2\hat k⁠: not in proportion, so use the skew formula. The top is 8≠08 \ne 0⁠, so the lines do not meet, and d=829d = \frac{8}{\sqrt{29}}⁠.

(b) Ch 10, type 3. Section formula: the point that divides PQ

Spot it: "divides the line joining P and Q in the ratio 2 : 1", internally or externally.

Here: W lies on BF: let it divide BF in the ratio k:1k : 1⁠, put W⃗=kF⃗+B⃗k+1\vec W = \frac{k\vec F + \vec B}{k + 1}⁠, and compare the i^\hat i parts to get k=23k = \frac23⁠, that is 2:32 : 3⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 10, Vector Algebra.

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