The foot of the perpendicular from the point | vedclass

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(A) The equation of the line passing through $B(4, 7, 1)$ and $C(3, 5, 3)$ is given by $\frac{x-4}{3-4} = \frac{y-7}{5-7} = \frac{z-1}{3-1} = \lambda$

Vedclass · Accepted Answer

$\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right)$

Vedclass · Answer

(A) The equation of the line passing through $B(4, 7, 1)$ and $C(3, 5, 3)$ is given by $\frac{x-4}{3-4} = \frac{y-7}{5-7} = \frac{z-1}{3-1} = \lambda$. This simplifies to $\frac{x-4}{-1} = \frac{y-7}{-2} = \frac{z-1}{2} = \lambda$. Any point $P$ on this line is given by $P(-\lambda+4, -2\lambda+7, 2\lambda+1)$. Since $AP$ is perpendicular to the line,the direction vector of $AP$ is $\vec{AP} = ((-\lambda+4-1), (-2\lambda+7-0), (2\lambda+1-3)) = (-\lambda+3, -2\lambda+7, 2\lambda-2)$. The direction vector of the line is $\vec{v} = (-1, -2, 2)$. Since $\vec{AP} \cdot \vec{v} = 0$,we have $(-1)(-\lambda+3) + (-2)(-2\lambda+7) + (2)(2\lambda-2) = 0$. $\lambda - 3 + 4\lambda - 14 + 4\lambda - 4 = 0$ $\Rightarrow 9\lambda - 21 = 0$ $\Rightarrow \lambda = \frac{21}{9} = \frac{7}{3}$. Substituting $\lambda = \frac{7}{3}$ into the coordinates of $P$: $x = -\frac{7}{3} + 4 = \frac{5}{3}$,$y = -2(\frac{7}{3}) + 7 = \frac{7}{3}$,$z = 2(\frac{7}{3}) + 1 = \frac{17}{3}$. Thus,the foot of the perpendicular is $\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right)$.

The foot of the perpendicular from the point $A(1, 0, 3)$ to the line joining the points $B(4, 7, 1)$ and $C(3, 5, 3)$ is

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