- Home
- Standard 12
- Mathematics
If for some $\alpha$ and $\beta$ in $R,$ the intersection of the following three planes $x+4 y-2 z=1$ ; $x+7 y-5 z=\beta$ ; $x+5 y+\alpha z=5$ is a line in $\mathrm{R}^{3},$ then $\alpha+\beta$ is equal to
$10$
$-10$
$2$
$0$
Solution
For planes to intersect on a line
$\Rightarrow$ there should be infinite solution of the given system of equations for infinite solutions
$\Delta=\left|\begin{array}{ccc}{1} & {4} & {-2} \\ {1} & {7} & {-5} \\ {1} & {5} & {\alpha}\end{array}\right|=0 \Rightarrow 3 \alpha+9=0 \Rightarrow \alpha=-3$
$\Delta_{z}=\left|\begin{array}{ccc}{1} & {4} & {1} \\ {1} & {7} & {\beta} \\ {1} & {5} & {5}\end{array}\right|=0 \Rightarrow 13-\beta=0 \Rightarrow \beta=13$
Also for $\alpha=-3$ and $b=13 \Delta_{x}=\Delta_{y}=0$
$\alpha+\beta=-3+13=10$
