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CBSE Sample Paper 2023-24 · Section D · 5 marks

Q35: Image of (1,6,3)(1, 6, 3) in a line, or where two aeroplanes are closest

Sample 2023-24Ch 11 · Three Dimensional GeometryCh 10 · Vector Algebra
The question

(a) Find the coordinates of the image of the point (1,6,3)(1, 6, 3) with respect to the line

r⃗=(j^+2k^)+λ(i^+2j^+3k^);\vec r = (\hat j + 2\hat k) + \lambda(\hat i + 2\hat j + 3\hat k);

where 'λ\lambda' is a scalar. Also, find the distance of the image from the yy-axis.

OR

(b) An aeroplane is flying along the line r⃗=λ(i^−j^+k^)\vec r = \lambda(\hat i - \hat j + \hat k)⁠; where 'λ\lambda' is a scalar and another aeroplane is flying along the line r⃗=i^−j^+μ(−2j^+k^)\vec r = \hat i - \hat j + \mu(-2\hat j + \hat k)⁠; where 'μ\mu' is a scalar. At what points on the lines should they reach, so that the distance between them is the shortest? Find the shortest possible distance between them.

Exam answer

(a)

Let P(1,6,3)P(1, 6, 3) be the given point, and let LL be the foot of the perpendicular from PP to the given line. The given line is

x−01=y−12=z−23=λ\frac{x - 0}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} = \lambda
Why?

A line r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b passes through the point with position vector a⃗\vec a and runs along b⃗\vec b⁠.

Here a⃗=j^+2k^\vec a = \hat j + 2\hat k⁠, the point (0,1,2)(0, 1, 2)⁠, and b⃗=i^+2j^+3k^\vec b = \hat i + 2\hat j + 3\hat k⁠, direction ratios 1, 2, 3.

Setting each fraction equal to λ\lambda gives x=λx = \lambda⁠, y=2λ+1y = 2\lambda + 1⁠, z=3λ+2z = 3\lambda + 2⁠.

so a general point on it is (λ,2λ+1,3λ+2)(\lambda, 2\lambda + 1, 3\lambda + 2)⁠.

Let the coordinates of LL be (λ,2λ+1,3λ+2)(\lambda, 2\lambda + 1, 3\lambda + 2)⁠.

So, the direction ratios of PLPL are λ−1\lambda - 1⁠, 2λ−52\lambda - 5 and 3λ−13\lambda - 1⁠.

Why?

The direction ratios of the segment from PP to LL are the differences of their coordinates, LL minus PP⁠:

λ−1, (2λ+1)−6,(3λ+2)−3\begin{aligned} &\lambda - 1,\ (2\lambda + 1) - 6, \\ &(3\lambda + 2) - 3 \end{aligned}

Simplify each one:

(2λ+1)−6=2λ−5(3λ+2)−3=3λ−1\begin{aligned} (2\lambda + 1) - 6 &= 2\lambda - 5 \\ (3\lambda + 2) - 3 &= 3\lambda - 1 \end{aligned}

The direction ratios of the given line are 1, 2 and 3, and the line is perpendicular to PLPL⁠. Therefore,

(λ−1)1+(2λ−5)2+(3λ−1)3=0⇒14λ−14=0⇒λ=1\begin{aligned} &(\lambda - 1)1 + (2\lambda - 5)2 \\ &\quad + (3\lambda - 1)3 = 0 \\ &\Rightarrow 14\lambda - 14 = 0 \Rightarrow \lambda = 1 \end{aligned}
Why?

Two lines with direction ratios a1,b1,c1a_1, b_1, c_1 and a2,b2,c2a_2, b_2, c_2 are perpendicular when

a1a2+b1b2+c1c2=0a_1a_2 + b_1b_2 + c_1c_2 = 0

Expanding: λ−1+4λ−10+9λ−3=14λ−14\lambda - 1 + 4\lambda - 10 + 9\lambda - 3 = 14\lambda - 14⁠.

(λ−1)(1)=λ−1(2λ−5)(2)=4λ−10(3λ−1)(3)=9λ−3λ+4λ+9λ=14λ−1−10−3=−1414λ−14=014λ=14λ=1414=1\begin{aligned} (\lambda - 1)(1) &= \lambda - 1 \\ (2\lambda - 5)(2) &= 4\lambda - 10 \\ (3\lambda - 1)(3) &= 9\lambda - 3 \\[4pt] \lambda + 4\lambda + 9\lambda &= 14\lambda \\ -1 - 10 - 3 &= -14 \\[4pt] 14\lambda - 14 &= 0 \\ 14\lambda &= 14 \\ \lambda &= \frac{14}{14} = 1 \end{aligned}

Then λ=1\lambda = 1 in (λ,2λ+1,3λ+2)(\lambda, 2\lambda + 1, 3\lambda + 2) gives LL⁠.

x=1y=2(1)+1=3z=3(1)+2=5\begin{aligned} x &= 1 \\ y &= 2(1) + 1 = 3 \\ z &= 3(1) + 2 = 5 \end{aligned}

So, the coordinates of LL are (1,3,5)(1, 3, 5)⁠.

Let Q(x1,y1,z1)Q(x_1, y_1, z_1) be the image of P(1,6,3)P(1, 6, 3) in the given line. Then LL is the mid-point of PQPQ⁠.

Why?

The image is the mirror point: QQ lies on the line through PP and LL⁠, on the other side of the given line, at the same distance. So LL is halfway between PP and QQ⁠.

Mid-point formula, one coordinate at a time:

(x1+x22,y1+y22,z1+z22)\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right)
x1+12=1, y1+62=3\frac{x_1 + 1}{2} = 1,\ \frac{y_1 + 6}{2} = 3 and z1+32=5\text{and } \frac{z_1 + 3}{2} = 5 ⇒x1=1, y1=0, z1=7\Rightarrow x_1 = 1,\ y_1 = 0,\ z_1 = 7
Why?

Multiply each equation by 2: x1+1=2x_1 + 1 = 2⁠, y1+6=6y_1 + 6 = 6 and z1+3=10z_1 + 3 = 10⁠.

x1+1=2⇒x1=2−1=1y1+6=6⇒y1=6−6=0z1+3=10⇒z1=10−3=7\begin{aligned} x_1 + 1 = 2 &\Rightarrow x_1 = 2 - 1 = 1 \\ y_1 + 6 = 6 &\Rightarrow y_1 = 6 - 6 = 0 \\ z_1 + 3 = 10 &\Rightarrow z_1 = 10 - 3 = 7 \end{aligned}

A quick way to remember it: image =2L−P= 2L - P⁠, so (2−1,6−6,10−3)(2 - 1, 6 - 6, 10 - 3)⁠.

Hence, the image of P(1,6,3)P(1, 6, 3) in the given line is (1,0,7)(1, 0, 7)⁠.

The foot of the perpendicular from (1,0,7)(1, 0, 7) to the yy-axis is (0,0,0)(0, 0, 0)⁠, so the distance of the image from the yy-axis is

12+72=50=52\sqrt{1^2 + 7^2} = \sqrt{50} = 5\sqrt2
Why?

The foot of the perpendicular from (x,y,z)(x, y, z) to the yy-axis is (0,y,0)(0, y, 0)⁠, so the distance from the yy-axis is

x2+z2\sqrt{x^2 + z^2}

Here y=0y = 0⁠, so the foot is the origin. And 50=25×2=52\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt2⁠.

12+72=1+49=5050=25 2=52\begin{aligned} \sqrt{1^2 + 7^2} &= \sqrt{1 + 49} = \sqrt{50} \\ \sqrt{50} &= \sqrt{25}\,\sqrt2 = 5\sqrt2 \end{aligned}

Watch out: the yy-coordinate does not enter this distance.

The distance of the image from the yy-axis is 50=52\sqrt{50} = 5\sqrt2 units.

(b)

r⃗=λ(i^−j^+k^)…(i)\vec r = \lambda(\hat i - \hat j + \hat k)\quad\ldots(\text{i}) r⃗=i^−j^+μ(−2j^+k^)…(ii)\vec r = \hat i - \hat j + \mu(-2\hat j + \hat k)\quad\ldots(\text{ii})

The lines are not parallel, as i^−j^+k^\hat i - \hat j + \hat k and −2j^+k^-2\hat j + \hat k are not parallel. So there is a unique segment PQPQ⁠, PP on line (i) and QQ on line (ii), at right angles to both lines, and the shortest distance between the aeroplanes is PQPQ⁠.

Why?

The directions are not multiples of each other: the first has i^\hat i part 1, the second has i^\hat i part 0.

For two such lines, the shortest path from one to the other is the common perpendicular. So we look for the points PP and QQ that make PQ→\overrightarrow{PQ} perpendicular to both lines.

Let the position vectors of PP and QQ be λ(i^−j^+k^)\lambda(\hat i - \hat j + \hat k) and i^+(−1−2μ)j^+μk^\hat i + (-1 - 2\mu)\hat j + \mu\hat k⁠.

PQ→=OQ→−OP→=(1−λ)i^+(−1−2μ+λ)j^+(μ−λ)k^\begin{aligned} \overrightarrow{PQ} &= \overrightarrow{OQ} - \overrightarrow{OP} \\ &= (1 - \lambda)\hat i + (-1 - 2\mu + \lambda)\hat j \\ &\quad + (\mu - \lambda)\hat k \end{aligned}
Why?

Write a general point of each line in components:

OP→=(λ,−λ,λ)OQ→=(1,−1−2μ,μ)\begin{aligned} \overrightarrow{OP} &= (\lambda, -\lambda, \lambda) \\ \overrightarrow{OQ} &= (1, -1 - 2\mu, \mu) \end{aligned}

Then subtract component by component, QQ minus PP⁠.

i^:1−λj^:(−1−2μ)−(−λ)=−1−2μ+λk^:μ−λ\begin{aligned} \hat i&: 1 - \lambda \\ \hat j&: (-1 - 2\mu) - (-\lambda) \\ &= -1 - 2\mu + \lambda \\ \hat k&: \mu - \lambda \end{aligned}

PQ→\overrightarrow{PQ} is perpendicular to both i^−j^+k^\hat i - \hat j + \hat k and −2j^+k^-2\hat j + \hat k⁠:

(1−λ)(1)+(−1−2μ+λ)(−1)+(μ−λ)(1)=0⇒2+3μ−3λ=0\begin{aligned} &(1 - \lambda)(1) + (-1 - 2\mu + \lambda)(-1) \\ &\quad + (\mu - \lambda)(1) = 0 \\ &\Rightarrow 2 + 3\mu - 3\lambda = 0 \end{aligned}
Why?

Perpendicular vectors have dot product 0:

a⃗⋅b⃗=a1b1+a2b2+a3b3\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3

Expanding: 1−λ+1+2μ−λ+μ−λ1 - \lambda + 1 + 2\mu - \lambda + \mu - \lambda⁠, which collects to 2+3μ−3λ2 + 3\mu - 3\lambda⁠.

(1−λ)(1)=1−λ(−1−2μ+λ)(−1)=1+2μ−λ(μ−λ)(1)=μ−λnumbers:1+1=2μ:2μ+μ=3μλ:−λ−λ−λ=−3λ\begin{aligned} (1 - \lambda)(1) &= 1 - \lambda \\ (-1 - 2\mu + \lambda)(-1) &= 1 + 2\mu - \lambda \\ (\mu - \lambda)(1) &= \mu - \lambda \\[4pt] \text{numbers}&: 1 + 1 = 2 \\ \mu&: 2\mu + \mu = 3\mu \\ \lambda&: -\lambda - \lambda - \lambda = -3\lambda \end{aligned}
(1−λ)(0)+(−1−2μ+λ)(−2)+(μ−λ)(1)=0⇒2+5μ−3λ=0\begin{aligned} &(1 - \lambda)(0) + (-1 - 2\mu + \lambda)(-2) \\ &\quad + (\mu - \lambda)(1) = 0 \\ &\Rightarrow 2 + 5\mu - 3\lambda = 0 \end{aligned}
Why?

Same with the second direction (0,−2,1)(0, -2, 1)⁠: 0+2+4μ−2λ+μ−λ0 + 2 + 4\mu - 2\lambda + \mu - \lambda⁠, which collects to 2+5μ−3λ2 + 5\mu - 3\lambda⁠.

(1−λ)(0)=0(−1−2μ+λ)(−2)=2+4μ−2λ(μ−λ)(1)=μ−λμ:4μ+μ=5μλ:−2λ−λ=−3λ\begin{aligned} (1 - \lambda)(0) &= 0 \\ (-1 - 2\mu + \lambda)(-2) &= 2 + 4\mu - 2\lambda \\ (\mu - \lambda)(1) &= \mu - \lambda \\[4pt] \mu&: 4\mu + \mu = 5\mu \\ \lambda&: -2\lambda - \lambda = -3\lambda \end{aligned}

Subtracting the first equation from the second,

2μ=0⇒μ=0, so 2−3λ=02\mu = 0 \Rightarrow \mu = 0,\ \text{so } 2 - 3\lambda = 0

On solving these equations, we get λ=23\lambda = \dfrac23 and μ=0\mu = 0⁠.

Why?

The terms 22 and −3λ-3\lambda are the same in both equations, so subtracting removes them and leaves 2μ=02\mu = 0⁠.

(2+5μ−3λ)−(2+3μ−3λ)=2−2+5μ−3μ−3λ+3λ=2μ\begin{aligned} &(2 + 5\mu - 3\lambda) - (2 + 3\mu - 3\lambda) \\ &= 2 - 2 + 5\mu - 3\mu - 3\lambda + 3\lambda \\ &= 2\mu \end{aligned}

So 2μ=02\mu = 0⁠, and μ=02=0\mu = \frac02 = 0⁠.

Putting μ=0\mu = 0 in 2+3μ−3λ=02 + 3\mu - 3\lambda = 0 gives 3λ=23\lambda = 2⁠.

2+3(0)−3λ=02−3λ=02=3λ(add 3λ)λ=23(divide by 3)\begin{aligned} 2 + 3(0) - 3\lambda &= 0 \\ 2 - 3\lambda &= 0 \\ 2 &= 3\lambda \quad (\text{add } 3\lambda) \\ \lambda &= \frac23 \quad (\text{divide by } 3) \end{aligned}

So, the points at which they should be, so that the distance between them is the shortest, are 23(i^−j^+k^)\frac23(\hat i - \hat j + \hat k) and i^−j^\hat i - \hat j⁠, i.e. (23,−23,23)\left(\frac23, -\frac23, \frac23\right) and (1,−1,0)(1, -1, 0)⁠.

Why?

Put λ=23\lambda = \frac23 in OP→\overrightarrow{OP} and μ=0\mu = 0 in OQ→\overrightarrow{OQ}⁠:

OP→=(23,−23,23)OQ→=(1,−1−2(0),0)=(1,−1,0)\begin{aligned} \overrightarrow{OP} &= \left(\tfrac23, -\tfrac23, \tfrac23\right) \\ \overrightarrow{OQ} &= (1, -1 - 2(0), 0) \\ &= (1, -1, 0) \end{aligned}
PQ→=13i^−13j^−23k^\overrightarrow{PQ} = \frac13\hat i - \frac13\hat j - \frac23\hat k
Why?

Put λ=23\lambda = \frac23 and μ=0\mu = 0 in the components of PQ→\overrightarrow{PQ}⁠:

1−23=13−1−0+23=−130−23=−23\begin{aligned} 1 - \tfrac23 &= \tfrac13 \\ -1 - 0 + \tfrac23 &= -\tfrac13 \\ 0 - \tfrac23 &= -\tfrac23 \end{aligned}
∣PQ→∣=19+19+49=69=23\begin{aligned} |\overrightarrow{PQ}| &= \sqrt{\frac19 + \frac19 + \frac49} \\ &= \sqrt{\frac69} = \sqrt{\frac23} \end{aligned}
Why?

The length of ai^+bj^+ck^a\hat i + b\hat j + c\hat k is a2+b2+c2\sqrt{a^2 + b^2 + c^2}⁠. Then 69=23\frac69 = \frac23⁠.

(13)2=19(−13)2=19(−23)2=4919+19+49=1+1+49=6969=6÷39÷3=23\begin{aligned} \left(\tfrac13\right)^2 &= \tfrac19 \\ \left(-\tfrac13\right)^2 &= \tfrac19 \\ \left(-\tfrac23\right)^2 &= \tfrac49 \\[4pt] \tfrac19 + \tfrac19 + \tfrac49 &= \tfrac{1 + 1 + 4}{9} = \tfrac69 \\ \tfrac69 &= \tfrac{6 \div 3}{9 \div 3} = \tfrac23 \end{aligned}

Check with the shortest distance formula, where n⃗=b⃗1×b⃗2\vec n = \vec b_1 \times \vec b_2⁠:

SD=∣(a⃗2−a⃗1)⋅n⃗∣∣n⃗∣\text{SD} = \frac{|(\vec a_2 - \vec a_1)\cdot\vec n|}{|\vec n|}

Here a⃗2−a⃗1=i^−j^\vec a_2 - \vec a_1 = \hat i - \hat j and

n⃗=i^−j^−2k^SD=∣1+1+0∣6=26\begin{aligned} \vec n &= \hat i - \hat j - 2\hat k \\ \text{SD} &= \frac{|1 + 1 + 0|}{\sqrt6} = \frac{2}{\sqrt6} \end{aligned}

The cross product, with b⃗1=(1,−1,1)\vec b_1 = (1, -1, 1) and b⃗2=(0,−2,1)\vec b_2 = (0, -2, 1)⁠:

n⃗=∣i^j^k^1−110−21∣i^:(−1)(1)−(1)(−2)=−1+2=1j^:−((1)(1)−(1)(0))=−(1−0)=−1k^:(1)(−2)−(−1)(0)=−2−0=−2\begin{aligned} \vec n &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & -1 & 1 \\ 0 & -2 & 1 \end{vmatrix} \\[4pt] \hat i&: (-1)(1) - (1)(-2) \\ &= -1 + 2 = 1 \\ \hat j&: -\big((1)(1) - (1)(0)\big) \\ &= -(1 - 0) = -1 \\ \hat k&: (1)(-2) - (-1)(0) \\ &= -2 - 0 = -2 \end{aligned} (i^−j^)⋅n⃗=(1)(1)+(−1)(−1)+(0)(−2)=1+1+0=2∣n⃗∣=12+(−1)2+(−2)2=1+1+4=6\begin{aligned} &(\hat i - \hat j)\cdot\vec n \\ &= (1)(1) + (-1)(-1) + (0)(-2) \\ &= 1 + 1 + 0 = 2 \\[4pt] |\vec n| &= \sqrt{1^2 + (-1)^2 + (-2)^2} \\ &= \sqrt{1 + 1 + 4} = \sqrt6 \end{aligned}

and 26=46=23\frac{2}{\sqrt6} = \sqrt{\frac46} = \sqrt{\frac23}⁠, the same.

(2=42 = \sqrt4⁠, so 26=46\frac{2}{\sqrt6} = \sqrt{\frac46}⁠, and 46=4÷26÷2=23\frac46 = \frac{4 \div 2}{6 \div 2} = \frac23⁠.)

The shortest distance =23= \sqrt{\dfrac23} units.

Where this comes from

Chapter 11, Three Dimensional Geometry: Image of a point in a line; shortest distance.

Question type: (a) the foot of a perpendicular; (b) The shortest distance between two lines

How to spot it: click to see

(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: Write the line as x1=y−12=z−23\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}⁠, so the foot is L(λ,2λ+1,3λ+2)L(\lambda, 2\lambda + 1, 3\lambda + 2)⁠. PL perpendicular to 1,2,31, 2, 3 gives L(1,3,5)L(1, 3, 5)⁠, the midpoint of P and its image (1,0,7)(1, 0, 7)⁠; the image's foot on the y-axis is (0,0,0)(0, 0, 0)⁠.

(b) Type 8. The shortest distance between two lines

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

Here: i^−j^+k^\hat i - \hat j + \hat k and −2j^+k^-2\hat j + \hat k are not parallel, and the points are asked too. So write P with λ\lambda and Q with μ\mu⁠, make PQ→\overrightarrow{PQ} perpendicular to both directions, and get λ=23\lambda = \frac23⁠, μ=0\mu = 0⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 10, Vector Algebra.

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