Question

The plane $\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$ cuts the axes in $A,B$, $C$, then the area of the $\Delta ABC$ is:

• A. $\sqrt{29}$
• B. $\sqrt{41}$
• C. $\sqrt{61}$
• D. $2\sqrt{61}$

Answer 1

The correct option is C $\sqrt{61}$

The point at which the plane cut the x-axis can be found by putting $y=0$ and $z=0$.

We get the point as $A=(2,0,0)$.

Similarly, the the plane cut the y-axis and z-axis are given by $B=(0,3,0)$ and $C=(0,0,4)$.

The vector $AB$ is given by $-2i+3j$ and vector $AC$ is given by $-2i+4j$.

The area is given by the formula $\frac{|AB\times AC|}{2}$

$AB\times AC$ can be found by using the determinant with row entries $(i,j,k)$, $(-2,3,0)$ and $(-2,0,4)$.

We get, $AB\times AC=12i+8j+6k.$

Hence, $\frac{|AB\times AC|}{2}=\frac{\sqrt{{12}^2+8^2+6^2}}{2}=\sqrt{61}$.