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Question

The area of the triangle with vertices

A(1,1,1)

B(2,3,1)

, and

C(3,1,2)

is:

(A)

\dfrac{\sqrt{21}}{2}

(B)

\dfrac{\sqrt{42}}{2}

(C)

\sqrt{21}

(D)

\dfrac{\sqrt{14}}{2}

Solution Path

Side vectors, cross product, area = half magnitude = 2*sqrt(6).

01Question Setup

1/4

Find the area of triangle with vertices

A(1,1,1)

B(2,3,1)

C(1,5,5)

Area

= ;?

02Cross Product Setup

2/4

\vec{AB} = \hat{i} + 2\hat{j}

\vec{AC} = 4\hat{j} + 4\hat{k}

. Compute

\vec{AB} \times \vec{AC}

\vec{AB} \times \vec{AC}

03Cross ProductKEY INSIGHT

3/4

\vec{AB} \times \vec{AC} = 8\hat{i} - 4\hat{j} + 4\hat{k}

|\vec{AB} \times \vec{AC}| = \sqrt{64+16+16} = \sqrt{96}

= 4\sqrt{6}

04Final Answer

4/4

Area

= \frac{1}{2}|\vec{AB} \times \vec{AC}| = 2\sqrt{6}

\boxed{2\sqrt{6}}

Concepts from this question1 concepts unlocked

★

Cross Product and Area of Parallelogram

STANDARD

The magnitude of the cross product |a x b| equals the area of the parallelogram formed by the two vectors. The direction of a x b is perpendicular to both a and b (right-hand rule).

\text{Area} = |\vec{a} \times \vec{b}|

JEE frequently asks for areas of triangles (half the parallelogram area) and parallelograms using cross products. Also essential for finding normals to planes.

Area of triangleArea of parallelogramNormal to a planeTorque in physics

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