Find the length of the perpendicular drawn from point $(2, 3, 4)$ to line $\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3} .$

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Solution

$\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3} = t ⟹ x = 4 - 2 t, y = 6 t, z = 1 - 3 t$
Let the point on line be $(4 - 2 t, 6 t, 1 - 3 t)$
For this point to be perpendicular drawn from $(2, 3, 4)$
$⟹ - 2 (4 - 2 t - 2) + 6 (6 t - 3) - 3 (1 - 3 t - 4) = 0$
$⟹ t = \frac{13}{49}$
Length of perpendicular will be $\sqrt{(2 - \frac{170}{49})^{2} + (3 - \frac{78}{49})^{2} + (4 - \frac{10}{49})^{2}} = \frac{3 \sqrt{4949}}{49}$

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Find the length of the perpendicular drawn from point (2,3,4) to line 4−x2=y6=1−z3.

4−x2=y6=1−z3=t⟹x=4−2t,y=6t,z=1−3tLet the point on line be(4−2t,6t,1−3t)For this point to be perpendicular drawn from (2,3,4)⟹−2(4−2t−2)+6(6t−3)−3(1−3t−4)=0 ⟹t=1349Length of perpendicular will be √(2−17049)2+(3−7849)2+(4−1049)2=3√494949

Find the length of the perpendicular drawn from point $(2, 3, 4)$ to line $\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3} .$