Find the equation of the line passing through the point (-1,3,-2) and perpendicular to the lines x/1 =y/2=z/3 and x+2/5 - Maths - Three Dimensional Geometry

Question

Find the equation of the line passing through the point
(-1,3,-2) and perpendicular to the lines x/1 =y/2=z/3 and x+2/5 =
y-1/2= z+1/5

Pronay · Accepted Answer

We have to find the equation of line passing through P(-1,3,-2) and perpendicular to the lines

$\frac{x}{1} = \frac{y}{2} = \frac{z}{3} & \frac{x + 2}{5} = \frac{y - 1}{2} = \frac{z + 1}{5}$

To find the vector which is perpendicular to a given two lines can be calculated by cross product as

$| \hat{i} \hat{j} \hat{k} | | 1 2 3 | | 5 2 5 |$

=(10-6) $\hat{i}$ - (5-15) $\hat{j}$ +(2-10) $\hat{k}$

$= 4 \hat{i} + 10 \hat{j} - 8 \hat{k}$

Therefore the equation of the required line is-

$- 1 \hat{i} + 3 \hat{j} - 2 \hat{k} = λ (4 \hat{i} + 10 \hat{j} - 8 \hat{k})$

Find the equation of the line passing through the point (-1,3,-2) and perpendicular to the lines x/1 =y/2=z/3 and x+2/5 = y-1/2= z+1/5