Competency-Based
Assessment Made Simple.

Step 1: Express the line in parametric form

Let $\frac{x+2}{5}=\frac{y+1}{2}=\frac{-z+4}{-3} = \lambda$. Then, we can express any point on the line as:

$x = 5\lambda - 2$

$y = 2\lambda - 1$

$z = -3\lambda + 4$

So, a general point on the line is $(5\lambda - 2, 2\lambda - 1, -3\lambda + 4)$.

Step 2: Define the foot of the perpendicular

Let the foot of the perpendicular from the point $P(1, 1, 4)$ to the line be $Q(5\lambda - 2, 2\lambda - 1, -3\lambda + 4)$.

Step 3: Find the direction ratios of PQ

The direction ratios of the line segment $PQ$ are given by:

$a = (5\lambda - 2) - 1 = 5\lambda - 3$

$b = (2\lambda - 1) - 1 = 2\lambda - 2$

$c = (-3\lambda + 4) - 4 = -3\lambda$

So, the direction ratios of $PQ$ are $(5\lambda - 3, 2\lambda - 2, -3\lambda)$.

Step 4: Use the perpendicularity condition

Since $PQ$ is perpendicular to the given line, the dot product of their direction ratios is zero. The direction ratios of the given line are $(5, 2, -3)$. Therefore:

$5(5\lambda - 3) + 2(2\lambda - 2) - 3(-3\lambda) = 0$

$25\lambda - 15 + 4\lambda - 4 + 9\lambda = 0$

$38\lambda - 19 = 0$

$\lambda = \frac{19}{38} = \frac{1}{2}$

Step 5: Find the coordinates of the foot of the perpendicular

Substitute $\lambda = \frac{1}{2}$ into the coordinates of point $Q$:

$x = 5(\frac{1}{2}) - 2 = \frac{5}{2} - 2 = \frac{1}{2}$

$y = 2(\frac{1}{2}) - 1 = 1 - 1 = 0$

$z = -3(\frac{1}{2}) + 4 = -\frac{3}{2} + 4 = \frac{5}{2}$

Therefore, the foot of the perpendicular is $(\frac{1}{2}, 0, \frac{5}{2})$.