If the foot of the perpendicular from the poi | vedclass

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(B) Given the plane $P: 2x + my + nz = 4$ and the foot of the perpendicular $B\left(-2, \frac{7}{2}, \frac{3}{2}\right)$ from point $A(-1, 4, 3)$.Sinc

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$\sqrt{26}$

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(B) Given the plane $P: 2x + my + nz = 4$ and the foot of the perpendicular $B\left(-2, \frac{7}{2}, \frac{3}{2}\right)$ from point $A(-1, 4, 3)$.Since $B$ lies on the plane,we have $2(-2) + m(\frac{7}{2}) + n(\frac{3}{2}) = 4$,which simplifies to $7m + 3n = 16$. $(1)$Also,the vector $\vec{AB} = B - A = (-2 - (-1), \frac{7}{2} - 4, \frac{3}{2} - 3) = (-1, -\frac{1}{2}, -\frac{3}{2})$.The normal vector to the plane is $\vec{n} = (2, m, n)$. Since $\vec{AB}$ is parallel to $\vec{n}$,we have $\frac{2}{-1} = \frac{m}{-1/2} = \frac{n}{-3/2} = k$.Thus,$k = -2$,so $m = (-1/2)(-2) = 1$ and $n = (-3/2)(-2) = 3$.We need the distance of point $A$ from the plane $P$ measured parallel to the line with direction ratios $(3, -1, -4)$. Let this line be $L$. The equation of line $L$ passing through $A(-1, 4, 3)$ is $\frac{x+1}{3} = \frac{y-4}{-1} = \frac{z-3}{-4} = \lambda$.Any point $C$ on this line is $(3\lambda - 1, -\lambda + 4, -4\lambda + 3)$.Since $C$ lies on the plane $2x + y + 3z = 4$,we substitute the coordinates of $C$ into the plane equation:$2(3\lambda - 1) + 1(-\lambda + 4) + 3(-4\lambda + 3) = 4$$6\lambda - 2 - \lambda + 4 - 12\lambda + 9 = 4$$-7\lambda + 11 = 4 \Rightarrow -7\lambda = -7 \Rightarrow \lambda = 1$.Thus,the point $C$ is $(3(1) - 1, -1 + 4, -4(1) + 3) = (2, 3, -1)$.The distance $AC = \sqrt{(2 - (-1))^2 + (3 - 4)^2 + (-1 - 3)^2} = \sqrt{3^2 + (-1)^2 + (-4)^2} = \sqrt{9 + 1 + 16} = \sqrt{26}$.

If the foot of the perpendicular from the point $A(-1, 4, 3)$ on the plane $P: 2x + my + nz = 4$ is $B\left(-2, \frac{7}{2}, \frac{3}{2}\right)$,then the distance of the point $A$ from the plane $P$,measured parallel to a line with direction ratios $3, -1, -4$,is equal to.

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