The foot of the perpendicular from the origin | vedclass

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(C) Let the plane $P$ be $2x + ay + 4z = k$. Since the foot of the perpendicular from the origin $(0, 0, 0)$ to the plane is $(2, a, 4)$,the normal ve

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$(3, 0, 4)$

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(C) Let the plane $P$ be $2x + ay + 4z = k$. Since the foot of the perpendicular from the origin $(0, 0, 0)$ to the plane is $(2, a, 4)$,the normal vector to the plane is $\vec{n} = 2\hat{i} + a\hat{j} + 4\hat{k}$.The equation of the plane is $2(x - 2) + a(y - a) + 4(z - 4) = 0$,which simplifies to $2x + ay + 4z = 4 + a^2 + 16 = 20 + a^2$.The intercepts are $A = (\frac{20 + a^2}{2}, 0, 0)$,$B = (0, \frac{20 + a^2}{a}, 0)$,and $C = (0, 0, \frac{20 + a^2}{4})$.The volume of the tetrahedron $OABC$ is $V = \frac{1}{6} |x_A y_B z_C| = \frac{1}{6} \cdot \frac{(20 + a^2)^3}{8a} = 144$.$(20 + a^2)^3 = 144 	imes 48 	imes a = 6912a$.Testing $a = 2$: $(20 + 4)^3 = 24^3 = 13824$ and $6912 	imes 2 = 13824$. Thus,$a = 2$.The equation of the plane is $2x + 2y + 4z = 20 + 4 = 24$,or $x + y + 2z = 12$.Checking the points:$A(2, 2, 4) \Rightarrow 2 + 2 + 2(4) = 12$ (Lies on plane).$B(0, 4, 4) \Rightarrow 0 + 4 + 2(4) = 12$ (Lies on plane).$C(3, 0, 4) \Rightarrow 3 + 0 + 2(4) = 11 
eq 12$ (Does $NOT$ lie on plane).$D(0, 6, 3) \Rightarrow 0 + 6 + 2(3) = 12$ (Lies on plane).Therefore,the point $(3, 0, 4)$ does not lie on the plane.

The foot of the perpendicular from the origin $O$ to a plane $P$,which meets the coordinate axes at points $A, B, C$,is $(2, a, 4)$,where $a \in N$. If the volume of the tetrahedron $OABC$ is $144 \text{ unit}^3$,then which of the following points does $NOT$ lie on the plane $P$?

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