Let the line x-23=y-1-5=z+22 lie in the plane x+3y- z+=0.

Question

Let the line $\dfrac{x-2}3=\dfrac{y-1}{-5}=\dfrac{z+2}2$ lie in the plane $x+3y-\alpha z+\beta=0$. Then $(\alpha,\beta)$ equals

Accepted Answer

As the line $${{x - 2} \over 3} = {{y - 1} \over { - 5}} = {{z + 2} \over 2}$$ lie in the plane

$$x + 3y - \alpha z + \beta  = 0$$

$$\therefore$$ $$Pt\left( {2,1, - 2} \right)$$ lies on the plane

i.e. $$2 + 3 + 2\alpha  + \beta  = 0$$

or $$\,\,\,\,2\alpha  + \beta  + 5 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,....\left( i \right)$$

Also normal to plane will be perpendicular to line,

$$\therefore$$ $$\,\,\,3 \times 1 - 5 \times 3 + 2 \times \left( { - \alpha } \right) = 0$$

$$ \Rightarrow \alpha  =  - 6$$

From equation $(i)$ then, $$\beta  = 7$$

$$\therefore$$ $$\left( {\alpha ,\beta } \right) = \left( { - 6,7} \right)$$

Solutions · 1