Prove that for all values of $\lambda$ and $\mu,$ the two planes intersect on the same line.

Question

Prove that for all values of $\lambda$ and $\mu,$ the planes $\frac{2x}{a}+\frac{y}{b}+\frac{2z}{c}-1+\lambda(\frac{x}{a}-\frac{2y}{b}-\frac{z}{c}-2)=0$ and $\frac{4x}{a}-\frac{3y}{b}-5+\mu(\frac{5y}{b}+\frac{4z}{c}+3)=0$ intersect on the same line.

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I have no idea how to prove this question.I can see that the first plane and the second plane both are a family of planes.Please help me.Thanks in advance.

Answer 1

The normal to the first plane is $$\left({2+\lambda\over a}, {1-2\lambda\over b}, {2-\lambda\over c}\right)$$

The normal to the second plane is $$\left({4\over a},{5u-3\over b},{4u\over c}\right)$$

The cross product of normals is $$\left({-3\lambda u-3\lambda-6u+6\over bc},{-4\lambda u-4\lambda - 8u+8\over ca},{5\lambda u+5\lambda+10u-10\over ab}\right)$$

which is $$(\lambda u+\lambda+2u-2)\left({-3\over bc},{-4\over ca},{5\over ab}\right)$$

Also, since both planes pass through point $(-a,-3b,3c)$ so the equation of the intersection is $$(-a,-3b,3c)+k\left({-3\over bc},{-4\over ca},{5\over ab}\right)$$

Answer 2

HINT: You have two family of planes

$$P_1+\lambda P_2=0$$

$$P_3+\mu P_4=0$$

The direction of line of intersection of $P_1$ and $P_2$ is along $\vec{n_1}\times \vec{n_2}$ where $\vec{n_1}$, $\vec{n_2}$ are normal vectors to $P_1$ and $P_2$ Respectively.

Similarly find line of intersection of $P_3$ and $P_4$. Observe whether both the lines have same direction.