The equation of the plane pi through the line | vedclass

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Question

(A) The given equations of the planes are $\pi_1 = x+3y-6=0$ and $\pi_2 = 3x-y+4z=0$. The equation of the plane $\pi$ passing through the line of inte

Vedclass · Accepted Answer

$2x+y+2z-3=0$

Vedclass · Answer

(A) The given equations of the planes are $\pi_1 = x+3y-6=0$ and $\pi_2 = 3x-y+4z=0$. The equation of the plane $\pi$ passing through the line of intersection is $\pi_1+\lambda \pi_2=0$. $(x+3y-6)+\lambda(3x-y+4z) = 0$ $(1+3\lambda)x + (3-\lambda)y + 4\lambda z - 6 = 0$ ... $(i)$ The perpendicular distance from the origin $(0,0,0)$ to the plane $(i)$ is given as $1$. Using the distance formula $d = \frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}$,we have: $\frac{|-6|}{\sqrt{(1+3\lambda)^2 + (3-\lambda)^2 + (4\lambda)^2}} = 1$ $36 = (1+9\lambda^2+6\lambda) + (9+\lambda^2-6\lambda) + 16\lambda^2$ $36 = 26\lambda^2 + 10$ $26\lambda^2 = 26 \implies \lambda^2 = 1 \implies \lambda = \pm 1$. For $\lambda = 1$,the equation becomes $(1+3)x + (3-1)y + 4(1)z - 6 = 0$,which simplifies to $4x+2y+4z-6=0$ or $2x+y+2z-3=0$. Thus,the correct option is $(a)$.

The equation of the plane $\pi$ through the line of intersection of the planes $\pi_1 \equiv x+3y-6=0$ and $\pi_2 \equiv 3x-y+4z=0$ is $\pi_1+\lambda \pi_2=0$. If the plane $\pi$ is at unit distance from the origin,then an equation of the plane $\pi$ is

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