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Exercise 6.5 · Q7

Q.Find the foot of the perpendicular drawn from the point (5,4,2)(5,4,2) to the line x+12=y−33=z−1−1\dfrac{x+1}{2}=\dfrac{y-3}{3}=\dfrac{z-1}{-1}. Also, find the equation of the perpendicular.

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Write the line's general point, form the vector from DD to that point, and require it perpendicular to the line's direction — solving that single linear equation in tt gives the foot.

Step 1. General point on the line x+12=y−33=z−1−1=t\frac{x+1}2=\frac{y-3}3=\frac{z-1}{-1}=t: F=(−1+2t, 3+3t, 1−t)F=(-1+2t,\ 3+3t,\ 1-t).

Step 2. Form DF⃗\vec{DF} with D=(5,4,2)D=(5,4,2):

DF⃗=(−1+2t−5, 3+3t−4, 1−t−2)=(2t−6, 3t−1, −t−1).\vec{DF}=(-1+2t-5,\ 3+3t-4,\ 1-t-2)=(2t-6,\ 3t-1,\ -t-1).

Step 3. Perpendicularity: DF⃗⋅b⃗=0\vec{DF}\cdot\vec b=0 with b⃗=(2,3,−1)\vec b=(2,3,-1):

2(2t−6)+3(3t−1)+(−1)(−t−1)=0 ⟹ 4t−12+9t−3+t+1=0 ⟹ 14t−14=0 ⟹ t=1.2(2t-6)+3(3t-1)+(-1)(-t-1)=0\ \Longrightarrow\ 4t-12+9t-3+t+1=0\ \Longrightarrow\ 14t-14=0\ \Longrightarrow\ t=1.

Step 4. Foot of the perpendicular. F=(−1+2, 3+3, 1−1)=(1,6,0)F=(-1+2,\ 3+3,\ 1-1)=(1,6,0). …

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