Intersection of 3D lines of form $ax+by+cz+d=0=a_1x+b_1y+c_1z+d_1$

Question

If the line $\frac{x}{1}=\frac{y}{2}=\frac{y}{3}$ intersects line $3β^2x + 3(1 – 2α)y + z = 3 = -\frac{1}{2}(6α^2x + 3(1 – 2β)y + 2z)$

Then value of $\alpha+\beta=$

Answer 1

The given line is (in parametric form)

$r = t (1, 2, 3) $

Substitute that in the first plane which is

$3 \beta^2 x + 3 (1 - 2\alpha) y + z = 3 $

You get

$t_1 = \dfrac{3}{ 3 \beta^2 + 6 (1 - 2\alpha) + 3 } = \dfrac{1}{\beta^2 + 2(1 - 2 \alpha) + 1 } $

Substitute the parametric equation of the line in the second plane, which is

$ 6 \alpha^2 x + 3 (1 - 2 \beta) y + 2 z = - 6 $

You get

$t_2 = \dfrac{-6}{ 6 \alpha^2 + 6 (1 - 2\beta) + 6 } = \dfrac{-1}{\alpha^2 + 1 - 2 \beta + 1 } $

Since the intersection is one point, we must have $t_1 = t_2$, therefore,

$\beta^2 + 2 (1 - 2\alpha) + 1 = - ( \alpha^2 - 2\beta + 2 ) $

Thus

$ \alpha^2 + \beta^2 - 4 \alpha - 2 \beta + 5 = 0 $

Completing the squares in $\alpha$ and $\beta$

$ (\alpha - 2 )^2 - 4 + (\beta - 1)^2 - 1 + 5 = 0 $

Thus

$ (\alpha - 2)^2 + (\beta - 1)^2 = 0 $

From which $\alpha = 2 $ and $\beta = 1$, and therefore, $\alpha + \beta = 3 $