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CBSE Sample Paper 2022-23 · Section A · 1 mark

Q3: Area of triangle ABC from AB→×AC→\overrightarrow{AB} \times \overrightarrow{AC}

Sample 2022-23Ch 10 · Vector Algebra
The question

The area of a triangle with vertices A, B, C is given by

(a) ∣AB→×AC→∣\left|\overrightarrow{AB} \times \overrightarrow{AC}\right|

(b) 12∣AB→×AC→∣\frac{1}{2}\left|\overrightarrow{AB} \times \overrightarrow{AC}\right|

(b) 14∣AC→×AB→∣\frac{1}{4}\left|\overrightarrow{AC} \times \overrightarrow{AB}\right|

(d) 18∣AC→×AB→∣\frac{1}{8}\left|\overrightarrow{AC} \times \overrightarrow{AB}\right|

Exam answer

∣AB→×AC→∣\left|\overrightarrow{AB} \times \overrightarrow{AC}\right| is the area of the parallelogram with adjacent sides AB and AC.

Why?

The magnitude of a cross product is

∣a⃗×b⃗∣=∣a⃗∣ ∣b⃗∣sin⁡θ|\vec a \times \vec b| = |\vec a|\,|\vec b|\sin\theta

where θ\theta is the angle between the vectors. In the parallelogram on a⃗\vec a and b⃗\vec b⁠, the height on the side a⃗\vec a is ∣b⃗∣sin⁡θ|\vec b|\sin\theta⁠. So this is base times height, the area of the parallelogram.

The triangle ABC is half of this parallelogram.

Why?

The diagonal BC cuts the parallelogram into two congruent triangles, and ABC is one of them. So the triangle has half the area.

For example, take A at the origin, AB→=4i^\overrightarrow{AB} = 4\hat i and AC→=3j^\overrightarrow{AC} = 3\hat j⁠. Then

AB→×AC→=4i^×3j^=(4)(3)(i^×j^)=12k^\begin{aligned} \overrightarrow{AB} \times \overrightarrow{AC} &= 4\hat i \times 3\hat j \\ &= (4)(3)(\hat i \times \hat j) \\ &= 12\hat k \end{aligned}

so ∣AB→×AC→∣=12\left|\overrightarrow{AB} \times \overrightarrow{AC}\right| = 12⁠, and the right triangle has area 12×4×3=6\frac{1}{2} \times 4 \times 3 = 6⁠, half of 12.

∴\therefore Area of △ABC=12∣AB→×AC→∣\triangle ABC = \frac{1}{2}\left|\overrightarrow{AB} \times \overrightarrow{AC}\right|

Why?

The order of the two vectors does not change the size: AC→×AB→\overrightarrow{AC} \times \overrightarrow{AB} is the negative of AB→×AC→\overrightarrow{AB} \times \overrightarrow{AC}⁠, with the same magnitude. So the other options differ only in the factor, and only 12\frac{1}{2} is right.

Answer: (b) 12∣AB→×AC→∣\dfrac{1}{2}\left|\overrightarrow{AB} \times \overrightarrow{AC}\right|

The paper labels the third option (b) again; it should be (c). The answer is the first option (b).

Where this comes from

Chapter 10, Vector Algebra: Area of a triangle using the cross product.

Question type: Area of a triangle

How to spot it: click to see

Type 9. Area of a triangle: half the cross product

Spot it: "the area of the triangle with vertices A, B and C".

Here: Every option is a multiple of ∣AB→×AC→∣|\overrightarrow{AB} \times \overrightarrow{AC}|⁠, the parallelogram on AB and AC. The triangle is half of it, so pick the 12\frac12⁠.

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