If the lines frac{2x-4}{lambda} = frac{y-1}{2 | vedclass

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(C) First,rewrite the equations of the lines in the standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.For the first line: $\frac{2(x

Vedclass · Accepted Answer

$\frac{-6}{7}$

Vedclass · Answer

(C) First,rewrite the equations of the lines in the standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.For the first line: $\frac{2(x-2)}{\lambda} = \frac{y-1}{2} = \frac{z-3}{1} \Rightarrow \frac{x-2}{\lambda/2} = \frac{y-1}{2} = \frac{z-3}{1}$.The direction ratios are $\vec{v_1} = (\frac{\lambda}{2}, 2, 1)$.For the second line: $\frac{x-1}{1} = \frac{3(y-1/3)}{\lambda} = \frac{z-2}{1} \Rightarrow \frac{x-1}{1} = \frac{y-1/3}{\lambda/3} = \frac{z-2}{1}$.The direction ratios are $\vec{v_2} = (1, \frac{\lambda}{3}, 1)$.Since the lines are perpendicular,the dot product of their direction ratios must be zero: $\vec{v_1} \cdot \vec{v_2} = 0$.$(\frac{\lambda}{2})(1) + (2)(\frac{\lambda}{3}) + (1)(1) = 0$.$\frac{\lambda}{2} + \frac{2\lambda}{3} + 1 = 0$.Multiply by $6$: $3\lambda + 4\lambda + 6 = 0$.$7\lambda = -6$.$\lambda = \frac{-6}{7}$.

If the lines $\frac{2x-4}{\lambda} = \frac{y-1}{2} = \frac{z-3}{1}$ and $\frac{x-1}{1} = \frac{3y-1}{\lambda} = \frac{z-2}{1}$ are perpendicular to each other,then $\lambda=$

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