First slide

Introduction to 3-D Geometry

Question

The plane $\frac{x}{2} + \frac{y}{3} + \frac{z}{1} = 1$ meet the co-ordinate axis at the points A, B, C respectively then the area bounded by the triangle ABC is

Easy

A

6 sq. units
B

$\frac{9}{2}$ sq. units
C

$\frac{5}{2}$ sq. units
D

$\frac{7}{2}$ sq. units

Solution

In the problem $A (2, 0, 0), B (0, 3, 0), C (0, 0, 1)$
Area of $Δ A B C = \frac{1}{2} \sqrt{Δ_{x}^{2} + Δ_{y}^{2} + Δ_{z}^{2}}$ where $Δ_{x} = |\begin{matrix} 0 & 3 & 0 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \end{matrix}| = 3$
$Δ_{y} = |\begin{matrix} 2 & 0 & 0 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \end{matrix}| = 2$
$Δ_{z} = |\begin{matrix} 0 & 3 & 3 \\ 2 & 0 & 0 \\ 1 & 1 & 1 \end{matrix}| = | (- 6) | = 6$
Required area = $\frac{1}{2} \sqrt{9 + 4 + 36} = \frac{7}{2}$

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