JEE Main 2021 · Mathematics · Three Dimensional Geometry · medium

The plane 2x+3y+6z=12 meets the coordinate axes at A, B and C. The area of triangle ABC (in square units) is

Numerical answer

The plane

2x+3y+6z=12

meets the coordinate axes at

A

B

and

C

. The area of triangle

ABC

(in square units) is

Show answer and solution

Answer: 14

Intercept form:

\dfrac x6+\dfrac y4+\dfrac z2=1

, so

A(6,0,0)

B(0,4,0)

C(0,0,2)

. Then

\vec{AB}=(-6,4,0)

and

\vec{AC}=(-6,0,2)

. Their cross product is

\vec{AB}\times\vec{AC}=(8,\,12,\,24).

Its magnitude is

\sqrt{64+144+576}=\sqrt{784}=28

. The area is

\tfrac12\cdot28=14

. As a check, the area also equals the (projected area of the triangle on a coordinate plane) divided by the direction cosine:

\tfrac12\cdot6\cdot4\div\tfrac{6}{7}=14

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