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If the lines $\frac{x-1}{5}=\frac{y+1}{3}=\frac{3-z}{\lambda}$ and $\frac{x+1}{4}=\frac{1-3 y}{15}=z+1$ are perpendicular
to each other, then $\lambda=$
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to each other, then $\lambda=$
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Verified Answer
The correct answer is:
5
Given lines are $\frac{x-1}{5}=\frac{y+1}{3}=\frac{z-3}{-\lambda}$ and $\frac{x+1}{4}=\frac{y-\frac{1}{3}}{-5}=\frac{z+1}{1}$
d.r. of the given lines are $5,3,-\lambda$ and $4,-5,1$
These lines are $\perp$ er
$\therefore 5(4)+3(-5)+(-\lambda)(1)=0 \Rightarrow 20-15-\lambda=0 \Rightarrow \lambda=5$
d.r. of the given lines are $5,3,-\lambda$ and $4,-5,1$
These lines are $\perp$ er
$\therefore 5(4)+3(-5)+(-\lambda)(1)=0 \Rightarrow 20-15-\lambda=0 \Rightarrow \lambda=5$
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