Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The equations of the given lines can be rewritten in standard symmetric form $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$.

For the first line:
$\frac{-(x-1)}{2}=\frac{7\left(y+\frac{4}{7}\right)}{2\lambda }=\frac{2\left(z-\frac{5}{2}\right)}{2}$

$\Rightarrow \frac{x-1}{-2}=\frac{y+\frac{4}{7}}{\frac{2\lambda }{7}}=\frac{z-\frac{5}{2}}{1}$

The direction ratios of the first line are $a_1=-2$, $b_1=\frac{2\lambda }{7}$, $c_1=1$.

For the second line:
$\frac{-7(x-1)}{3\lambda }=\frac{y-1}{7}=\frac{-(z-6)}{5}$

$\Rightarrow \frac{x-1}{-\frac{3\lambda }{7}}=\frac{y-1}{7}=\frac{z-6}{-5}$

The direction ratios of the second line are $a_2=-\frac{3\lambda }{7}$, $b_2=7$, $c_2=-5$.

Since the two lines are at right angles, their direction vectors are perpendicular, which gives the condition $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$.

Substituting the direction ratios:
$(-2)\left(-\frac{3\lambda }{7}\right)+\left(\frac{2\lambda }{7}\right)(7)+(1)(-5)=0$

$\frac{6\lambda }{7}+2\lambda -5=0$

$\frac{6\lambda +14\lambda }{7}=5$

$\frac{20\lambda }{7}=5$

$\lambda =\frac{35}{20}=\frac{7}{4}$

Answer: $\frac{7}{4}$