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If the line −x+13=−y−2−2k=z+32\frac{-x+1}{3} = \frac{-y-2}{-2k} = \frac{z+3}{2} and −1+x3k=−1+y1=−z+65\frac{-1+x}{3k} = \frac{-1 +y}{1} = \frac{-z+6}{5} are perpendicular, then the value of k is:

Solution

✅ Correct Option: 2

Standard symmetric form of a line:

x−x1a=y−y1b=z−z1c\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}

The given equations are:

−x+13=−y−2−2k=z+32\frac{-x + 1}{3} = \frac{-y - 2}{-2k} = \frac{z + 3}{2}

and

−1+x3k=−1+y1=−z+65\frac{-1 + x}{3k} = \frac{-1 + y}{1} = \frac{-z + 6}{5}


x−1−3=y+22k=z+32\frac{x - 1}{-3} = \frac{y + 2}{2k} = \frac{z + 3}{2}

The direction vector for the first line is:

d1⃗=⟨−3,2k,2⟩\vec{d_1} = \langle -3, 2k, 2 \rangle


x−13k=y−11=z−6−5\frac{x - 1}{3k} = \frac{y - 1}{1} = \frac{z - 6}{-5}

The direction vector for the second line is:

d2⃗=⟨3k,1,−5⟩\vec{d_2} = \langle 3k, 1, -5 \rangle


d1⃗⋅d2⃗=0\vec{d_1} \cdot \vec{d_2} = 0

(−3)(3k)+(2k)(1)+(2)(−5)=0(-3)(3k) + (2k)(1) + (2)(-5) = 0

−9k+2k−10=0-9k + 2k - 10 = 0

−7k−10=0-7k - 10 = 0

−7k=10-7k = 10

k=−107k = -\frac{10}{7}

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