If the lines frac{1-x}{2}=frac{7y+4}{2lambda} | vedclass

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(B) 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{-(x

Vedclass · Accepted Answer

$\frac{7}{4}$

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(B) 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: $\frac{-(x-1)}{2} = \frac{7(y+4/7)}{2\lambda} = \frac{2(z-5/2)}{2}$,which simplifies to $\frac{x-1}{-2} = \frac{y+4/7}{2\lambda/7} = \frac{z-5/2}{1}$.The direction ratios are $a_1 = -2, b_1 = \frac{2\lambda}{7}, c_1 = 1$.For the second line: $\frac{-7(x-1)}{3\lambda} = \frac{y-1}{7} = \frac{-(z-6)}{5}$,which simplifies to $\frac{x-1}{-3\lambda/7} = \frac{y-1}{7} = \frac{z-6}{-5}$.The direction ratios are $a_2 = -\frac{3\lambda}{7}, b_2 = 7, c_2 = -5$.Since the lines are at right angles,the dot product of their direction vectors must be zero: $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.Substituting the values: $(-2)(-\frac{3\lambda}{7}) + (\frac{2\lambda}{7})(7) + (1)(-5) = 0$.$\frac{6\lambda}{7} + 2\lambda - 5 = 0$.Multiply by $7$: $6\lambda + 14\lambda - 35 = 0$.$20\lambda = 35$.$\lambda = \frac{35}{20} = \frac{7}{4}$.

If the lines $\frac{1-x}{2}=\frac{7y+4}{2\lambda}=\frac{2z-5}{2}$ and $\frac{7-7x}{3\lambda}=\frac{y-1}{7}=\frac{6-z}{5}$ are at right angles,then the value of $\lambda$ is

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