(i) Find the shortest distance between the line x+1/7 = y+1/-6 = z+1/1

Question

(i) Find the shortest distance between the line \(\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}\) and

\(\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}\)

(ii) Find the equation of the Plane passing through one point(-1, 3, 2) and ± r to the planes x + 2y + 3z = 5 and 3x + 3y + z = 0

Answer 1

We have,

\(\bar{a_1}=-i-j-k,\bar{a_2}=3i+5j+7k,\\\bar{b_1}=7i-6j+k,\bar{b_2}=i-2j+k\\\bar{a_2}-\bar{a_1}=4i+6j+8k\)

(ii) Let the equation of the plane be ax + by + cz + d = 0….(1) 

Since (1) is perpendicular to 3x + 2y + 3z = 5,

 a + 2b + 3c = 0 

Since (1) is perpendicular to 3x + 3y + z = 0 

3a + 3b + c = 0 

Using (2) and (3) we have;

(1) = -7x + 8y – 3z + d = 0 

Since (1) passes through (-1, 3, 2) we have; 

⇒ -7 (-1) + 8(3) – 3(2) + d = 0 

⇒ 7 + 24 – 6 + d = 0 = d = -25 

Therefore the equation of the plane is 

(1) ⇒ -7x + 8y – 3z – 25 = 0