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(B) The given line is $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$.Since the plane is perpendicular to this line,the direction ratios of the normal to the pl

Vedclass · Accepted Answer

$x+2y+3z=20$

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(B) The given line is $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$.Since the plane is perpendicular to this line,the direction ratios of the normal to the plane are the same as the direction ratios of the line,which are $(1, 2, 3)$.The general equation of a plane with normal $(a, b, c)$ is $ax+by+cz+d=0$.Substituting the direction ratios,we get $1x+2y+3z+d=0$.Since the plane passes through the point $(2, 3, 4)$,we substitute these coordinates into the equation:$1(2)+2(3)+3(4)+d=0$$2+6+12+d=0$$20+d=0$$d=-20$.Therefore,the equation of the plane is $x+2y+3z-20=0$,or $x+2y+3z=20$.

The equation of the plane perpendicular to the line $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ and passing through the point $(2, 3, 4)$ is:

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