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Question: The equation of the line through the point (0, 1, 2) and perpendicular to the line $\frac{x-1}{2}=\f...

The equation of the line through the point (0, 1, 2) and perpendicular to the line x−12=y+13=z−1−2\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2} is:

A

x−3=y−14=z−23\frac{x}{-3}=\frac{y-1}{4}=\frac{z-2}{3}

B

x3=y−14=z−23\frac{x}{3}=\frac{y-1}{4}=\frac{z-2}{3}

C

x3=y−1−4=z−23\frac{x}{3}=\frac{y-1}{-4}=\frac{z-2}{3}

D

x3=y−14=z−2−3\frac{x}{3}=\frac{y-1}{4}=\frac{z-2}{-3}

Answer

x−3=y−14=z−23\frac{x}{-3}=\frac{y-1}{4}=\frac{z-2}{3}

Explanation

Solution

The given line has direction ratios v1⃗=⟨2,3,−2⟩\vec{v_1} = \langle 2, 3, -2 \rangle. The required line passes through P(0,1,2)P(0, 1, 2) and has direction ratios v2⃗=⟨a,b,c⟩\vec{v_2} = \langle a, b, c \rangle. For perpendicularity, their dot product must be zero: v1⃗⋅v2⃗=0⇒2a+3b−2c=0\vec{v_1} \cdot \vec{v_2} = 0 \Rightarrow 2a + 3b - 2c = 0. For option A, ⟨a,b,c⟩=⟨−3,4,3⟩\langle a, b, c \rangle = \langle -3, 4, 3 \rangle. 2(−3)+3(4)−2(3)=−6+12−6=02(-3) + 3(4) - 2(3) = -6 + 12 - 6 = 0. This option satisfies the condition.