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CBSE Board Examination 2023 · Section B · 2 marks

Q23: c⃗\vec c from AC→=54AB→\overrightarrow{AC} = \frac54\overrightarrow{AB}⁠, or are two lines perpendicular?

Board 2023Ch 10 · Vector AlgebraCh 11 · Three Dimensional Geometry
The question

(a) Position vectors of the points A, B and C as shown in the figure below are a⃗\vec a⁠, b⃗\vec b and c⃗\vec c respectively.

A B C
A, B and C on one line, with position vectors a⃗\vec a⁠, b⃗\vec b and c⃗\vec c⁠.

If AC→=54AB→\overrightarrow{AC} = \frac54\overrightarrow{AB}⁠, express c⃗\vec c in terms of a⃗\vec a and b⃗\vec b⁠.

OR

(b) Check whether the lines given by equations x=2λ+2x = 2\lambda + 2⁠, y=7λ+1y = 7\lambda + 1⁠, z=−3λ−3z = -3\lambda - 3 and x=−μ−2x = -\mu - 2⁠, y=2μ+8y = 2\mu + 8⁠, z=4μ+5z = 4\mu + 5 are perpendicular to each other or not.

Exam answer

(a) AC→=c⃗−a⃗\overrightarrow{AC} = \vec c - \vec a and AB→=b⃗−a⃗\overrightarrow{AB} = \vec b - \vec a⁠. According to the question,

Why?

The vector from one point to another is the position vector of the end point minus that of the starting point:

PQ→=q⃗−p⃗\overrightarrow{PQ} = \vec q - \vec p

So AC→\overrightarrow{AC} starts at A and ends at C.

c⃗−a⃗=54(b⃗−a⃗)\vec c - \vec a = \frac54\left(\vec b - \vec a\right) c⃗=54b⃗−54a⃗+a⃗\begin{aligned} \vec c &= \frac54\vec b - \frac54\vec a + \vec a \end{aligned}
Why?

Open the bracket, then add a⃗\vec a to both sides to get c⃗\vec c on its own.

c⃗−a⃗=54b⃗−54a⃗c⃗=54b⃗−54a⃗+a⃗\begin{aligned} \vec c - \vec a &= \frac54\vec b - \frac54\vec a \\ \vec c &= \frac54\vec b - \frac54\vec a + \vec a \end{aligned}

∴ c⃗=5b⃗4−a⃗4\therefore\ \vec c = \frac{5\vec b}{4} - \frac{\vec a}{4}

Why?

Collect the a⃗\vec a terms:

−54a⃗+a⃗=−54a⃗+44a⃗=−5+44a⃗=−14a⃗\begin{aligned} -\frac54\vec a + \vec a &= -\frac54\vec a + \frac44\vec a \\ &= \frac{-5 + 4}{4}\vec a \\ &= -\frac14\vec a \end{aligned}

Check with the figure: put A at 0 and B at 4 on a number line. Then C is at

5×44−04=5\frac{5 \times 4}{4} - \frac{0}{4} = 5

one unit beyond B, as drawn.

(b) The direction ratios of the lines are ⟨2,7,−3⟩\langle 2, 7, -3\rangle and ⟨−1,2,4⟩\langle -1, 2, 4\rangle⁠.

Why?

In x=2λ+2x = 2\lambda + 2⁠, y=7λ+1y = 7\lambda + 1⁠, z=−3λ−3z = -3\lambda - 3⁠, the numbers multiplying λ\lambda give the direction. Solving each for λ\lambda⁠:

x−2=2λ⇒λ=x−22y−1=7λ⇒λ=y−17z+3=−3λ⇒λ=z+3−3\begin{aligned} x - 2 = 2\lambda &\Rightarrow \lambda = \frac{x - 2}{2} \\ y - 1 = 7\lambda &\Rightarrow \lambda = \frac{y - 1}{7} \\ z + 3 = -3\lambda &\Rightarrow \lambda = \frac{z + 3}{-3} \end{aligned} x−22=y−17=z+3−3\frac{x - 2}{2} = \frac{y - 1}{7} = \frac{z + 3}{-3}

So the direction ratios are 2, 7, −3-3⁠. The coefficients of μ\mu in the second line give −1-1⁠, 2, 4 in the same way:

x+2−1=y−82=z−54\frac{x + 2}{-1} = \frac{y - 8}{2} = \frac{z - 5}{4}
a1a2+b1b2+c1c2=2(−1)+7(2)+(−3)(4)=−2+14−12=0\begin{aligned} &a_1a_2 + b_1b_2 + c_1c_2 \\ &= 2(-1) + 7(2) + (-3)(4) \\ &= -2 + 14 - 12 = 0 \end{aligned}
Why?

Two lines are perpendicular exactly when their directions are perpendicular, that is when

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

This sum is the dot product of the two direction vectors. Multiply matching ratios and add.

2(−1)=−27(2)=14(−3)(4)=−12−2+14=1212−12=0\begin{aligned} 2(-1) &= -2 \\ 7(2) &= 14 \\ (-3)(4) &= -12 \\[4pt] -2 + 14 &= 12 \\ 12 - 12 &= 0 \end{aligned}

∴\therefore The given lines are perpendicular to each other.

Where this comes from

Chapter 10, Vector Algebra: Position vectors on a line; perpendicular lines.

Question type: (a) The vector from A to B, and its direction cosines; (b) Perpendicular or parallel

How to spot it: click to see

(a) Ch 10, type 2. The vector from A to B, and its direction cosines

Spot it: two points are given, like A(1,2,−3)A(1, 2, -3) and B(−1,−2,1)B(-1, -2, 1)⁠.

Here: AC→\overrightarrow{AC} and AB→\overrightarrow{AB} are each end minus start, c⃗−a⃗\vec c - \vec a and b⃗−a⃗\vec b - \vec a⁠. Put them into AC→=54AB→\overrightarrow{AC} = \frac54\overrightarrow{AB} and solve for c⃗\vec c⁠.

(b) Ch 11, type 7. Perpendicular or parallel: test the direction ratios

Spot it: "Show that the lines are perpendicular (or parallel)", or "find kk so that the lines are at right angles".

Here: The numbers multiplying λ\lambda and μ\mu give the direction ratios 2,7,−32, 7, -3 and −1,2,4-1, 2, 4⁠. Their sum of products is −2+14−12=0-2 + 14 - 12 = 0⁠, so the lines are perpendicular.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 11, Three Dimensional Geometry.

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