CBSE Sample Paper 2020-21 · Part A: Section I · 1 mark

Q11: Area of the triangle with sides $2\hat i$ and $-3\hat j$

Sample 2020-21Ch 10 · Vector Algebra

The question

Find the area of the triangle whose two sides are represented by the vectors $2\hat i$ and $-3\hat j$ ⁠.

Exam answer

Area of a triangle with sides $\vec a$ and $\vec b$ $= \frac{1}{2}|\vec a \times \vec b|$ ⁠.

Why?

$|\vec a \times \vec b|$ is the area of the parallelogram with sides $\vec a$ and $\vec b$ ⁠. A diagonal cuts that parallelogram into two equal triangles, so the triangle has half of it.

\begin{aligned} \text{Area} &= \frac{1}{2}|2\hat i \times (-3\hat j)| \\ &= \frac{1}{2}\lvert -6\hat k\rvert \quad [\hat i \times \hat j = \hat k] \\ &= \frac{1}{2} \times 6 = 3 \end{aligned}

Why?

Numbers come out of a cross product, and $\hat i \times \hat j = \hat k$ ⁠:

\begin{aligned} &2\hat i \times (-3\hat j) \\ &= (2)(-3)(\hat i \times \hat j) \\ &= -6\hat k \end{aligned}

The length of $-6\hat k$ is 6, because $|\hat k| = 1$ and a length is never negative.

Check: $2\hat i$ and $-3\hat j$ are perpendicular, so this is a right triangle with legs 2 and 3, and $\frac{1}{2} \times 2 \times 3 = 3$ ⁠.

$\therefore$ Area of the triangle $= 3$ sq. units.

Where this comes from

Chapter 10, Vector Algebra: Area of a triangle by 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: The two sides $2\hat i$ and $-3\hat j$ are given directly, so skip finding them. Their cross product is $-6\hat k$ ⁠, and half its length is 3.

Quick lesson · Worked NCERT solutions · Interactive lesson

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Q11: Area of the triangle with sides $2\hat i$ and $-3\hat j$