Question

Find the equation of the plane which passes through the line of intersection of the planes 3𝑥 + 4𝑦 − 5𝑧 = 9 and 2𝑥 + 6𝑦 + 6𝑧 = 7 and which is perpendicular to the plane 3𝑥 + 2𝑦 − 5𝑧 + 6 = 0

Answer 1

Any plane though the line of intersection of the planes:

$3𝑥 + 4𝑦 − 5𝑧 - 9=0 \text{ and } 2𝑥 + 6𝑦 + 6𝑧 - 7=0$

is given by;

$(3𝑥 + 4𝑦 − 5𝑧-9)+\lambda(2𝑥 + 6𝑦 + 6𝑧-7)=0\\ \Rightarrow x(3+2\lambda)+y(4+6\lambda)+z(-5+6\lambda)=(3+2\lambda,\ 4+6\lambda,\ -5+6\lambda)\cdot(x,\ y,\ z)=9+7\lambda\cdots\cdots(1)$and this is perpendicular to the plane $\displaystyle 3𝑥 + 2𝑦 − 5𝑧 =(3,\ 2,\ -5)\cdot(x,\ y,\ z)= -6$ given in the question.

Since they are perpendicular, then their dot product of the vectors is zero. That is;

$\displaystyle (3+2\lambda,\ 4+6\lambda,\ -5+6\lambda)\cdot(3,\ 2,\ -5)=3\times(3+2\lambda)+2\times(4+6\lambda)-5\times(-5+6\lambda)=0\\ \Rightarrow9+6\lambda+8+12\lambda+25-30\lambda=0\\ \Rightarrow\lambda=\frac{7}{2}$

Substituting $\displaystyle \lambda=\frac{7}{2}$ into $(1)$ yields;

$\displaystyle x\left[3+2\left(\frac{7}{2}\right)\right]+y\left[4+6\left(\frac{7}{2}\right)\right]+z\left[-5+6\left(\frac{7}{2}\right)\right]=9+7\left(\frac{7}{2}\right)$

$\displaystyle \Rightarrow20x+50y+32z=67$ which is the equation of the plane we want.