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CBSE Board Examination 2024 · Section B · 2 marks

Q25: Find AD⃗\vec{AD} and the area of parallelogram ABCD from AB⃗\vec{AB} and DB⃗\vec{DB}

Board 2024Ch 10 · Vector Algebra
The question

In the given figure, ABCD is a parallelogram. If AB⃗=2i^−4j^+5k^\vec{AB} = 2\hat i - 4\hat j + 5\hat k and DB⃗=3i^−6j^+2k^\vec{DB} = 3\hat i - 6\hat j + 2\hat k⁠, then find AD⃗\vec{AD} and hence find the area of parallelogram ABCD.

A B C D
Exam answer

By the triangle law, AD⃗+DB⃗=AB⃗\vec{AD} + \vec{DB} = \vec{AB}⁠.

Why?

Going from A to D and then from D to B ends at the same place as going straight from A to B. This lets us find AD⃗\vec{AD} from the two vectors we are given.

AD⃗=AB⃗−DB⃗=(2i^−4j^+5k^)−(3i^−6j^+2k^)=−i^+2j^+3k^\begin{aligned} \vec{AD} &= \vec{AB} - \vec{DB} \\ &= (2\hat i - 4\hat j + 5\hat k) \\ &\quad - (3\hat i - 6\hat j + 2\hat k) \\ &= -\hat i + 2\hat j + 3\hat k \end{aligned}
Why?

Subtract component by component: 2−3=−12 - 3 = -1⁠, −4−(−6)=2-4 - (-6) = 2 and 5−2=35 - 2 = 3⁠.

i^:2−3=−1j^:−4−(−6)=−4+6=2k^:5−2=3\begin{aligned} \hat i&: 2 - 3 = -1 \\ \hat j&: -4 - (-6) = -4 + 6 = 2 \\ \hat k&: 5 - 2 = 3 \end{aligned}
AD⃗×AB⃗=∣i^j^k^−1232−45∣=i^(10+12)−j^(−5−6)+k^(4−4)=22i^+11j^\begin{aligned} &\vec{AD}\times\vec{AB} \\ &= \begin{vmatrix} \hat i & \hat j & \hat k \\ -1 & 2 & 3 \\ 2 & -4 & 5 \end{vmatrix} \\ &= \hat i(10 + 12) - \hat j(-5 - 6) \\ &\quad + \hat k(4 - 4) \\ &= 22\hat i + 11\hat j \end{aligned}
Why?

Row 2 holds the components of AD⃗\vec{AD} and row 3 those of AB⃗\vec{AB}⁠. Expand along the first row:

i^:2(5)−3(−4)=10+12j^:(−1)(5)−3(2)=−5−6k^:(−1)(−4)−2(2)=4−4\begin{aligned} \hat i&: 2(5) - 3(-4) = 10 + 12 \\ \hat j&: (-1)(5) - 3(2) = -5 - 6 \\ \hat k&: (-1)(-4) - 2(2) = 4 - 4 \end{aligned}

Watch out: the j^\hat j term carries a minus sign in front of its bracket.

Work out each bracket, then apply the signs ++⁠, −-⁠, ++⁠:

i^:+(10+12)=22j^:−(−5−6)=−(−11)=11k^:+(4−4)=0\begin{aligned} \hat i&: +(10 + 12) = 22 \\ \hat j&: -(-5 - 6) = -(-11) = 11 \\ \hat k&: +(4 - 4) = 0 \end{aligned}

AD and AB are adjacent sides of the parallelogram, so

Area=∣AD⃗×AB⃗∣=222+112=605=121×5=115\begin{aligned} \text{Area} &= |\vec{AD}\times\vec{AB}| \\ &= \sqrt{22^2 + 11^2} = \sqrt{605} \\ &= \sqrt{121\times 5} = 11\sqrt5 \end{aligned}
Why?

For a parallelogram with adjacent sides a⃗\vec a and b⃗\vec b⁠, the area is the length of the cross product:

Area=∣a⃗×b⃗∣\text{Area} = |\vec a\times\vec b|

AD⃗\vec{AD} and AB⃗\vec{AB} both start at A, so they are adjacent sides.

The length of 22i^+11j^+0k^22\hat i + 11\hat j + 0\hat k is 222+112+02\sqrt{22^2 + 11^2 + 0^2}⁠, and 484+121=605484 + 121 = 605⁠.

222=22×22=484112=11×11=121484+121+0=605\begin{aligned} 22^2 &= 22 \times 22 = 484 \\ 11^2 &= 11 \times 11 = 121 \\ 484 + 121 + 0 &= 605 \end{aligned}

Since 605=121×5605 = 121\times 5 and 121=11\sqrt{121} = 11⁠, the root simplifies to 11511\sqrt5⁠.

605=121×5=121×5=115\begin{aligned} \sqrt{605} &= \sqrt{121 \times 5} \\ &= \sqrt{121} \times \sqrt5 \\ &= 11\sqrt5 \end{aligned}

∴ AD⃗=−i^+2j^+3k^\therefore\ \vec{AD} = -\hat i + 2\hat j + 3\hat k⁠, and the area of ABCD =115= 11\sqrt5 sq. units.

Where this comes from

Chapter 10, Vector Algebra: Cross product: area of a parallelogram.

Question type: Cross product, and the area of a parallelogram

How to spot it: click to see

Type 8. Cross product, and the area of a parallelogram

Spot it: "find a⃗×b⃗\vec a\times\vec b", "find ∣a⃗×b⃗∣|\vec a\times\vec b|", or "the area of the parallelogram whose adjacent sides are".

Here: First get the side AD→=AB→−DB→\overrightarrow{AD} = \overrightarrow{AB} - \overrightarrow{DB}⁠. AD and AB are adjacent sides, so the area is ∣AD→×AB→∣|\overrightarrow{AD} \times \overrightarrow{AB}|⁠, from the determinant.

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