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CBSE Board Examination 2023 · Section D · 5 marks

Q34: Line through (1,2,−4)(1, 2, -4) parallel to AB and the distance between them, or points on AB at 14 units from B

Board 2023Ch 11 · Three Dimensional GeometryCh 10 · Vector Algebra
The question

(a) Find the vector and the Cartesian equations of a line passing through the point (1,2,−4)(1, 2, -4) and parallel to the line joining the points A(3,3,−5)A(3, 3, -5) and B(1,0,−11)B(1, 0, -11)⁠. Hence, find the distance between the two lines.

OR

(b) Find the equations of the line passing through the points A(1,2,3)A(1, 2, 3) and B(3,5,9)B(3, 5, 9)⁠. Hence, find the coordinates of the points on this line which are at a distance of 14 units from point B.

Exam answer

(a) Direction ratios of AB: 1−3, 0−3, −11+51 - 3,\ 0 - 3,\ -11 + 5⁠, i.e. −2,−3,−6-2, -3, -6⁠. So AB is parallel to b⃗=2i^+3j^+6k^\vec b = 2\hat i + 3\hat j + 6\hat k⁠.

Why?

The direction ratios of the line through two points are the differences of their coordinates, here B minus A.

1−3=−20−3=−3−11−(−5)=−11+5=−6\begin{aligned} 1 - 3 &= -2 \\ 0 - 3 &= -3 \\ -11 - (-5) &= -11 + 5 = -6 \end{aligned}

−2,−3,−6-2, -3, -6 and 2,3,62, 3, 6 are proportional (multiply by −1-1⁠), so they give the same direction. The positive set is simpler.

The required line is parallel to AB, so it has the same direction b⃗\vec b⁠.

Vector equation of the required line through (1,2,−4)(1, 2, -4) is

r⃗=i^+2j^−4k^+λ(2i^+3j^+6k^)\begin{aligned} \vec r &= \hat i + 2\hat j - 4\hat k \\ &\quad + \lambda(2\hat i + 3\hat j + 6\hat k) \end{aligned}
Why?

A line through the point with position vector a⃗\vec a⁠, parallel to b⃗\vec b⁠, is

r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b

Here a⃗=i^+2j^−4k^\vec a = \hat i + 2\hat j - 4\hat k⁠.

and its Cartesian equation is

x−12=y−23=z+46\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6}
Why?

The Cartesian form through (x1,y1,z1)(x_1, y_1, z_1) with direction ratios a,b,ca, b, c is

x−x1a=y−y1b=z−z1c\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}

With z1=−4z_1 = -4⁠, z−z1z - z_1 becomes z+4z + 4⁠.

Equation of the line through A(3,3,−5)A(3, 3, -5) and B(1,0,−11)B(1, 0, -11) is

r⃗=3i^+3j^−5k^+μ(2i^+3j^+6k^)\begin{aligned} \vec r &= 3\hat i + 3\hat j - 5\hat k \\ &\quad + \mu(2\hat i + 3\hat j + 6\hat k) \end{aligned}
Why?

Line AB passes through A(3,3,−5)A(3, 3, -5) and has the same direction b⃗\vec b⁠. A different letter, μ\mu⁠, is used for its parameter.

Distance between the parallel lines is given by

d=∣(a⃗2−a⃗1)×b⃗∣∣b⃗∣d = \frac{\left|(\vec a_2 - \vec a_1)\times\vec b\right|}{|\vec b|}
Why?

This is the standard distance between the parallel lines r⃗=a⃗1+λb⃗\vec r = \vec a_1 + \lambda\vec b and r⃗=a⃗2+μb⃗\vec r = \vec a_2 + \mu\vec b⁠.

The size of the cross product is the area of a parallelogram with base ∣b⃗∣|\vec b|⁠. Dividing by the base leaves the height, which is the distance between the lines.

Here b⃗=2i^+3j^+6k^\vec b = 2\hat i + 3\hat j + 6\hat k⁠, a⃗1=i^+2j^−4k^\vec a_1 = \hat i + 2\hat j - 4\hat k⁠, a⃗2=3i^+3j^−5k^\vec a_2 = 3\hat i + 3\hat j - 5\hat k⁠.

a⃗2−a⃗1=2i^+j^−k^\vec a_2 - \vec a_1 = 2\hat i + \hat j - \hat k
Why?

Subtract component by component: 3−1=23 - 1 = 2⁠, 3−2=13 - 2 = 1 and −5−(−4)=−1-5 - (-4) = -1⁠.

(a⃗2−a⃗1)×b⃗=∣i^j^k^21−1236∣=(6+3)i^−(12+2)j^+(6−2)k^=9i^−14j^+4k^\begin{aligned} &(\vec a_2 - \vec a_1)\times\vec b \\ &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 1 & -1 \\ 2 & 3 & 6 \end{vmatrix} \\ &= (6 + 3)\hat i - (12 + 2)\hat j \\ &\quad + (6 - 2)\hat k \\ &= 9\hat i - 14\hat j + 4\hat k \end{aligned}
Why?

Expand the determinant along the first row:

i^:1(6)−(−1)(3)=9j^:−[2(6)−(−1)(2)]=−14k^:2(3)−1(2)=4\begin{aligned} \hat i&: 1(6) - (-1)(3) = 9 \\ \hat j&: -[2(6) - (-1)(2)] = -14 \\ \hat k&: 2(3) - 1(2) = 4 \end{aligned}

One at a time, covering the row of i^,j^,k^\hat i, \hat j, \hat k and one column:

i^:∣1−136∣=(1)(6)−(−1)(3)=6+3=9j^:−∣2−126∣=−((2)(6)−(−1)(2))=−(12+2)=−14k^:∣2123∣=(2)(3)−(1)(2)=6−2=4\begin{aligned} \hat i&: \begin{vmatrix} 1 & -1 \\ 3 & 6 \end{vmatrix} \\ &= (1)(6) - (-1)(3) \\ &= 6 + 3 = 9 \\[4pt] \hat j&: -\begin{vmatrix} 2 & -1 \\ 2 & 6 \end{vmatrix} \\ &= -\big((2)(6) - (-1)(2)\big) \\ &= -(12 + 2) = -14 \\[4pt] \hat k&: \begin{vmatrix} 2 & 1 \\ 2 & 3 \end{vmatrix} \\ &= (2)(3) - (1)(2) \\ &= 6 - 2 = 4 \end{aligned}

Watch out: the j^\hat j term takes a minus sign.

d=81+196+164+9+36\begin{aligned} d &= \frac{\sqrt{81 + 196 + 16}}{\sqrt{4 + 9 + 36}} \end{aligned}
Why?

The magnitudes are the roots of the sums of squares:

92+(−14)2+42=81+196+16=293\begin{aligned} &9^2 + (-14)^2 + 4^2 \\ &= 81 + 196 + 16 = 293 \end{aligned} 81+196=277277+16=293\begin{aligned} 81 + 196 &= 277 \\ 277 + 16 &= 293 \end{aligned} 22+32+62=49=722^2 + 3^2 + 6^2 = 49 = 7^2 4+9+36=4949=7\begin{aligned} 4 + 9 + 36 &= 49 \\ \sqrt{49} &= 7 \end{aligned}

So d=29349=2937d = \frac{\sqrt{293}}{\sqrt{49}} = \frac{\sqrt{293}}{7}⁠.

293 is not a perfect square, so 293\sqrt{293} stays as it is.

∴ d=2937\therefore\ d = \frac{\sqrt{293}}{7} units.

(b) Direction ratios of AB: 3−1, 5−2, 9−33 - 1,\ 5 - 2,\ 9 - 3⁠, i.e. 2,3,62, 3, 6⁠. Equation of line AB is

x−12=y−23=z−36\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{6}
Why?

The direction ratios are B minus A. The line passes through A(1,2,3)A(1, 2, 3)⁠, so use the Cartesian form with (x1,y1,z1)=(1,2,3)(x_1, y_1, z_1) = (1, 2, 3)⁠.

3−1=25−2=39−3=6\begin{aligned} 3 - 1 &= 2 \\ 5 - 2 &= 3 \\ 9 - 3 &= 6 \end{aligned}

or r⃗=i^+2j^+3k^\vec r = \hat i + 2\hat j + 3\hat k +λ(2i^+3j^+6k^)+ \lambda(2\hat i + 3\hat j + 6\hat k)⁠.

Let the coordinates of a required point on AB be (2λ+1,3λ+2,6λ+3)(2\lambda + 1, 3\lambda + 2, 6\lambda + 3) for some λ\lambda⁠. According to the question,

Why?

Set each fraction equal to λ\lambda⁠. For example, x−12=λ\frac{x - 1}{2} = \lambda gives x=2λ+1x = 2\lambda + 1⁠. Every point of the line has this form for some value of λ\lambda⁠.

The point must be 14 units from B(3,5,9)B(3, 5, 9)⁠. Use the distance formula, with both sides squared to avoid the root.

(2λ−2)2+(3λ−3)2+(6λ−6)2=142⇒ 49(λ−1)2=196⇒ λ2−2λ−3=0⇒ (λ−3)(λ+1)=0\begin{aligned} &(2\lambda - 2)^2 + (3\lambda - 3)^2 \\ &\quad + (6\lambda - 6)^2 = 14^2 \\ \Rightarrow\ &49(\lambda - 1)^2 = 196 \\ \Rightarrow\ &\lambda^2 - 2\lambda - 3 = 0 \\ \Rightarrow\ &(\lambda - 3)(\lambda + 1) = 0 \end{aligned}
Why?

The differences from B are 2λ+1−3=2λ−22\lambda + 1 - 3 = 2\lambda - 2⁠, then 3λ−33\lambda - 3 and 6λ−66\lambda - 6⁠. Each is a multiple of λ−1\lambda - 1⁠:

2λ+1−3=2λ−2=2(λ−1)3λ+2−5=3λ−3=3(λ−1)6λ+3−9=6λ−6=6(λ−1)\begin{aligned} 2\lambda + 1 - 3 &= 2\lambda - 2 = 2(\lambda - 1) \\ 3\lambda + 2 - 5 &= 3\lambda - 3 = 3(\lambda - 1) \\ 6\lambda + 3 - 9 &= 6\lambda - 6 = 6(\lambda - 1) \end{aligned}

Square each one: (2(λ−1))2=4(λ−1)2\big(2(\lambda - 1)\big)^2 = 4(\lambda - 1)^2⁠, and so on.

4(λ−1)2+9(λ−1)2+36(λ−1)2=49(λ−1)2\begin{aligned} &4(\lambda - 1)^2 + 9(\lambda - 1)^2 \\ &\quad + 36(\lambda - 1)^2 = 49(\lambda - 1)^2 \end{aligned}

(4+9+36=494 + 9 + 36 = 49⁠, and 142=19614^2 = 196⁠.)

Divide by 49: (λ−1)2=4(\lambda - 1)^2 = 4⁠, so λ2−2λ+1=4\lambda^2 - 2\lambda + 1 = 4⁠.

49(λ−1)2=196(λ−1)2=19649=4λ2−2λ+1=4λ2−2λ+1−4=0λ2−2λ−3=0\begin{aligned} 49(\lambda - 1)^2 &= 196 \\ (\lambda - 1)^2 &= \frac{196}{49} = 4 \\ \lambda^2 - 2\lambda + 1 &= 4 \\ \lambda^2 - 2\lambda + 1 - 4 &= 0 \\ \lambda^2 - 2\lambda - 3 &= 0 \end{aligned}

To factorise, find two numbers with product −3-3 and sum −2-2⁠. They are −3-3 and 11⁠.

Solving, λ=3\lambda = 3 and λ=−1\lambda = -1⁠.

Why?

From (λ−3)(λ+1)=0(\lambda - 3)(\lambda + 1) = 0⁠: λ−3=0\lambda - 3 = 0 gives λ=3\lambda = 3⁠, and λ+1=0\lambda + 1 = 0 gives λ=−1\lambda = -1⁠.

Put each value in (2λ+1,3λ+2,6λ+3)(2\lambda + 1, 3\lambda + 2, 6\lambda + 3)⁠. For λ=3\lambda = 3 this is (7,11,21)(7, 11, 21)⁠, and for λ=−1\lambda = -1 it is (−1,−1,−3)(-1, -1, -3)⁠.

λ=3:2(3)+1=73(3)+2=116(3)+3=21λ=−1:2(−1)+1=−13(−1)+2=−16(−1)+3=−3\begin{aligned} &\lambda = 3\text{:} \\ &2(3) + 1 = 7 \\ &3(3) + 2 = 11 \\ &6(3) + 3 = 21 \\[4pt] &\lambda = -1\text{:} \\ &2(-1) + 1 = -1 \\ &3(-1) + 2 = -1 \\ &6(-1) + 3 = -3 \end{aligned}

Check: from B to (7,11,21)(7, 11, 21) the differences are 4, 6 and 12, and

16+36+144=196=14\sqrt{16 + 36 + 144} = \sqrt{196} = 14

From B to (−1,−1,−3)(-1, -1, -3) the differences are −4-4⁠, −6-6 and −12-12⁠, whose squares are the same 16, 36 and 144, so this distance is 14 too.

There are two points because there is one on each side of B.

∴\therefore Required points are (7,11,21)(7, 11, 21) and (−1,−1,−3)(-1, -1, -3)⁠.

Where this comes from

Chapter 11, Three Dimensional Geometry: Line equations; distance between parallel lines.

Question type: (a) Write the line + The shortest distance between two lines (combined); (b) The line through two points

How to spot it: click to see

(a) Type 4. Write the line: a point and a direction

Spot it: "Find the equation of the line through the point … parallel to (or in the direction of) …".

Here: The new line goes through (1,2,−4)(1, 2, -4) parallel to AB. B minus A gives −2,−3,−6-2, -3, -6⁠, so take b⃗=2i^+3j^+6k^\vec b = 2\hat i + 3\hat j + 6\hat k and write r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b⁠.

(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: For the distance, both lines have the same b⃗\vec b⁠, so they are parallel: use the parallel formula with a⃗2−a⃗1=2i^+j^−k^\vec a_2 - \vec a_1 = 2\hat i + \hat j - \hat k⁠, which gives 2937\frac{\sqrt{293}}{7}⁠.

(b) Type 2. The line through two points: subtract, then divide

Spot it: two points are given, and you need the direction of the line through them.

Here: B minus A gives 2,3,62, 3, 6⁠, so AB is x−12=y−23=z−36\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{6}⁠. Then write its points as (2λ+1,3λ+2,6λ+3)(2\lambda + 1, 3\lambda + 2, 6\lambda + 3) and set the distance from B equal to 14.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 10, Vector Algebra.

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