Question

Find the length of the perpendicular from the point (4, -5, 3) to the line $\frac{x-5}{3}=\frac{y+2}{-4}=\frac{z-6}{5}$

Answer 1

According to question,

A(4,-5,3) and Line(L)=$$\begin{array}{c}\frac{x-5}{3}=\frac{y+2}{-4}=\frac{z-6}{5}\Rightarrow l_1\\ Now,\\ \frac{x-5}{3}=\frac{y+2}{-4}=\frac{z-6}{5}=\lambda \\ then,\\ \frac{x-5}{3}=\lambda ,\therefore x=3\lambda +5\\ \frac{y+2}{-4}=\lambda \therefore y=-4\lambda -2\\ \frac{z-6}{5}=\lambda \therefore z=5\lambda +6\\ so,\\ d{r}^{\prime }sof{L}_{1}are(3,-4,5)\\ d{r}^{\prime }sofALare(3\lambda +1,-4\lambda +3,5\lambda -3)\\ asthed{r}^{\prime }sare{\perp }^{lar}toeachother,\\ a_1{a}_2+b_1{b}_2+c_1{c}_2=0\\ (3\lambda +1)\left(3\right)+(-4\lambda +3)(-4)+(5\lambda -3)\left(5\right)=0\\ 9\lambda +3+16\lambda -12+25\lambda +15=0\\ 50\lambda -24=0\\ \therefore \lambda =\frac{24}{50}=\frac{12}{25}\end{array}$$