Question

If the plane $\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=\frac{1}{2}$ intersect x,y and z-axis at A,B and C respectively then volume of the tetrahedron OABC where 'O' is the origin is

• A. 6
• B. 4
• C. 3
• D. $\frac{1}{2}$

Answer 1

Answer: (D)

$\frac{1}{2}$

$\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=\frac{1}{2}$

$\Rightarrow \frac{x}{1}+\frac{y}{3/2}+\frac{z}{2}=1$ (Intercept Form)

$A=(1,0,0)$

$B=(0,\frac{3}{2},0)$

$C=(0,0,2)$

$0=(0,0,0)$

$\overrightarrow{a}=\hat{i}$

$\overrightarrow{b}=\frac{3}{2}\hat{j}$

$\overrightarrow{c}=2\hat{k}$

Volume $=\frac{1}{2}\left[\right(\overrightarrow{a}\times \overrightarrow{b}).\overrightarrow{c}]=\frac{1}{6}[\frac{3}{2}\hat{k}.2\hat{k}=\frac{3}{6}=\frac{1}{2}]$