Question

Algebraic sum of intercepts made by the plane x+3y-4z+6=0 on the axes is

• A. 7
• B. 0
• C. $\frac{13}{2}$
• D. $-\frac{13}{2}$

Answer 1

Answer: (D)

$-\frac{13}{2}$

Intercepts made by plane $x+3y-4z+6=0$ on

$x,y$ and $z-$axis is found by putting $(1,0,0),(0,1,0)$ and $(0,0,1)$ in equation respectively

Then $x-$ intercept $=-6$

$y-$ intercept $=-2$

$z-$ intercept $=\frac{3}{2}$

Algebraic sum $=-6-2+\frac{3}{2}$
$=-\frac{13}{2}$