The planes _1, _2 and _3 have equations x+y+z=4, x-y+z=2 and 2x+y+kz=m, respectively, where k,m in R

Question

The planes _1, _2 and _3 have equations x+y+z=4, x-y+z=2 and 2x+y+kz=m, respectively, where k,m in R. - Find a vector equation of the line of intersection of _1 and _2. - Find the common point of all three planes when k=5 and m=13. - Determine the conditions on k and m for the three planes to have exactly one common point, a common line, or no common point.

Accepted Answer

$\mathbf r=(3,1,0)+t(-1,0,1)$. For $k=5,m=13$: $(1,1,2)$. Unique point if $k\ne2$; common line if $k=2,m=7$; no common point if $k=2,m\ne7$.

A changing third plane

A changing third plane

Solution:

Solution locked