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Chapter 10 · Vector Algebra

Exercise 10.4 · Q12 (MCQ): Area of the rectangle ABCD from its position vectors

Exercise 10.4Area of a parallelogram by the vector product · Section 10.6.3
The question

Area of a rectangle having vertices A, B, C and D with position vectors −i^+12j^+4k^-\hat i + \frac12\hat j + 4\hat k⁠, i^+12j^+4k^\hat i + \frac12\hat j + 4\hat k⁠, i^−12j^+4k^\hat i - \frac12\hat j + 4\hat k and −i^−12j^+4k^-\hat i - \frac12\hat j + 4\hat k⁠, respectively is

(A) 12\frac12

(B) 1

(C) 2

(D) 4

Exam answer

AB→=2i^\overrightarrow{AB} = 2\hat i and BC→=−j^\overrightarrow{BC} = -\hat j are adjacent sides, so

Area=∣AB→×BC→∣=∣2i^×(−j^)∣=∣−2k^∣=2\begin{aligned} \text{Area} &= |\overrightarrow{AB}\times\overrightarrow{BC}| \\ &= |2\hat i\times(-\hat j)| = |-2\hat k| = 2 \end{aligned}

Answer: (C) 2

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

A rectangle is a parallelogram, so its area is the length of the vector product of two adjacent sides:

area=∣AB→×BC→∣.\text{area} = |\overrightarrow{AB}\times\overrightarrow{BC}|.

Here the sides come out as simple multiples of i^\hat i and j^\hat j⁠, so the product is quick.

Given, to find, and the plan
Given
The position vectors of A, B, C, D: −i^+12j^+4k^-\hat i + \frac12\hat j + 4\hat k⁠, i^+12j^+4k^\hat i + \frac12\hat j + 4\hat k⁠, i^−12j^+4k^\hat i - \frac12\hat j + 4\hat k⁠, −i^−12j^+4k^-\hat i - \frac12\hat j + 4\hat k⁠.
To find
The area of the rectangle ABCD, among the options.
Plan
(1) Find the adjacent sides AB→\overrightarrow{AB} and BC→\overrightarrow{BC}⁠. (2) Confirm ABCD is a rectangle. (3) Area =∣AB→×BC→∣= |\overrightarrow{AB}\times\overrightarrow{BC}|⁠. (4) Test every option.

Basics you need

1Vector joining two points

AB→=OB→−OA→\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}⁠: position vector of the end minus that of the start (Section 10.5.2). Basics: three numbers fix a point

2Area of a parallelogram

If a⃗\vec a and b⃗\vec b are adjacent sides, the area is ∣a⃗×b⃗∣|\vec a\times\vec b| (Section 10.6.3, Observation 9). Lesson: the cross product

3Products of i^,j^,k^\hat i, \hat j, \hat k

i^×j^=k^\hat i\times\hat j = \hat k and (λa⃗)×(μb⃗)=λμ(a⃗×b⃗)(\lambda\vec a)\times(\mu\vec b) = \lambda\mu(\vec a\times\vec b) (Section 10.6.3, Observation 4 and Property 3). Lesson: Examples 22–25

Solution

  1. Two adjacent sides

    AB→=(1+1)i^+(12−12)j^+(4−4)k^=2i^,BC→=(1−1)i^+(−12−12)j^+(4−4)k^=−j^.\begin{aligned} \overrightarrow{AB} &= (1 + 1)\hat i + \left(\tfrac12 - \tfrac12\right)\hat j \\ &\quad + (4 - 4)\hat k = 2\hat i, \\ \overrightarrow{BC} &= (1 - 1)\hat i + \left(-\tfrac12 - \tfrac12\right)\hat j \\ &\quad + (4 - 4)\hat k = -\hat j. \end{aligned}
    Each side is the position vector of its end minus that of its start. All four points have k^\hat k component 4, so they lie in the plane z=4z = 4⁠, parallel to the xy-plane, and every side has no k^\hat k part.
    AB→=OB→−OA→\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} (Section 10.5.2). Basics: three numbers fix a point

  2. ABCD is a rectangle

    DC→=(1+1)i^+(−12+12)j^=2i^=AB→\overrightarrow{DC} = (1 + 1)\hat i + \left(-\frac12 + \frac12\right)\hat j = 2\hat i = \overrightarrow{AB}⁠, and AB→⋅BC→=(2)(0)+(0)(−1)+(0)(0)=0\overrightarrow{AB}\cdot\overrightarrow{BC} = (2)(0) + (0)(-1) + (0)(0) = 0⁠.

    A(−1, ½, 4) B(1, ½, 4) C(1, −½, 4) D(−1, −½, 4) AB = 2î BC = −ĵ area = |AB × BC| = |−2k̂| = 2 2 × 1 = 2
    The four points drawn in the plane z=4z = 4 with one scale (x to the right, y up): AB→=2i^\overrightarrow{AB} = 2\hat i of length 2 and BC→=−j^\overrightarrow{BC} = -\hat j of length 1 meet at a right angle, so ABCD is a 2 by 1 rectangle.
    DC→=AB→\overrightarrow{DC} = \overrightarrow{AB} means one pair of opposite sides is equal and parallel, so ABCD is a parallelogram; the zero scalar product means the angle at B is a right angle. A parallelogram with a right angle is a rectangle, as the question says.
    Equal vectors; perpendicular vectors have scalar product 0 (Sections 10.3 and 10.6.1). Lesson: the dot product

  3. The area

    area=∣AB→×BC→∣=∣2i^×(−j^)∣=∣−2(i^×j^)∣=∣−2k^∣=2.\begin{aligned} \text{area} &= |\overrightarrow{AB}\times\overrightarrow{BC}| = |2\hat i\times(-\hat j)| \\ &= |-2(\hat i\times\hat j)| = |-2\hat k| = 2. \end{aligned}
    Scalars come out of the vector product: 2i^×(−j^)=(2)(−1)(i^×j^)=−2k^2\hat i\times(-\hat j) = (2)(-1)(\hat i\times\hat j) = -2\hat k⁠, whose length is 2. As a check, for a rectangle the area is length times breadth, ∣AB→∣×∣BC→∣=2×1=2|\overrightarrow{AB}| \times |\overrightarrow{BC}| = 2 \times 1 = 2⁠.
    Area of a parallelogram =∣a⃗×b⃗∣= |\vec a\times\vec b|⁠; i^×j^=k^\hat i\times\hat j = \hat k (Section 10.6.3, Observations 4 and 9). Lesson: the cross product

  4. Test every option

    The area is 2, option (C). Two of the other values come from common slips:

    (B) 1 is 12∣AB→×BC→∣\frac12|\overrightarrow{AB}\times\overrightarrow{BC}|⁠, the area of the triangle ABC, half the rectangle. (D) 4 is the length of the product of the diagonals: AC→=2i^−j^\overrightarrow{AC} = 2\hat i - \hat j⁠, BD→=−2i^−j^\overrightarrow{BD} = -2\hat i - \hat j⁠, and AC→×BD→=−4k^\overrightarrow{AC}\times\overrightarrow{BD} = -4\hat k⁠, twice the area. (A) 12\frac12 is only a quarter of the area; it is neither the rectangle nor the triangle ABC.

    For any parallelogram, the vector product of the diagonals is twice the vector product of the sides (in size), so using the diagonals without halving gives 4. Using the triangle formula with its 12\frac12 gives 1.
    Area of a triangle =12∣a⃗×b⃗∣= \frac12|\vec a\times\vec b| (Section 10.6.3, Observation 8). Lesson: the self-check quiz
Result

The area of the rectangle is 2: the answer is (C). (Textbook answer: (C).)

Check it with numbers

Sides: ∣AB→∣=2|\overrightarrow{AB}| = 2⁠, ∣BC→∣=1|\overrightarrow{BC}| = 1⁠, ∣CD→∣=2|\overrightarrow{CD}| = 2⁠, ∣DA→∣=1|\overrightarrow{DA}| = 1⁠. Diagonal: ∣AC→∣=4+1=5|\overrightarrow{AC}| = \sqrt{4 + 1} = \sqrt5⁠, and 22+12=52^2 + 1^2 = 5⁠, so Pythagoras confirms the right angle at B.

Length times breadth: 2×1=22 \times 1 = 2⁠, the same as ∣AB→×BC→∣|\overrightarrow{AB}\times\overrightarrow{BC}|⁠.

Common mistakes
  1. Crossing the position vectors. OA→×OB→\overrightarrow{OA}\times\overrightarrow{OB} is not a side product; the sides are differences of position vectors.
  2. Adding a 12\frac12⁠. The 12\frac12 is for a triangle; a rectangle is a parallelogram, area ∣AB→×BC→∣|\overrightarrow{AB}\times\overrightarrow{BC}|⁠, giving 2, not 1.
  3. Using a diagonal as a side. AC→\overrightarrow{AC} and BD→\overrightarrow{BD} are diagonals; their vector product has length 4, twice the area.
What to take away
  1. Area of a rectangle or parallelogram =∣side×adjacent side∣= |\text{side}\times\text{adjacent side}|⁠.
  2. Sides are differences of position vectors, taken around the figure in order.
  3. When the sides are multiples of i^\hat i and j^\hat j⁠, use i^×j^=k^\hat i\times\hat j = \hat k directly.

Every fact used, and where it comes from

FactUsed inSource
Vector joining two pointsSteps 1, 2, 4Section 10.5.2 (Basics: three numbers fix a point)
Equal vectors; scalar product zero for perpendicular vectorsStep 2Section 10.3; Section 10.6.1, Observation 2
Area of a parallelogram =∣a⃗×b⃗∣= |\vec a\times\vec b|⁠; of a triangle =12∣a⃗×b⃗∣= \frac12|\vec a\times\vec b|Steps 3, 4Section 10.6.3, Observations 8 and 9 (Lesson: the cross product)
i^×j^=k^\hat i\times\hat j = \hat k⁠; scalars come out of the productStep 3Section 10.6.3, Observation 4 and Property 3
The question and the answerResultExercise 10.4 Q12; NCERT answer key (Lesson: the self-check quiz)
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