Let P 1 : 3 y + z + 1 = 0 and P 2 : 2 x - y + 3 z - 7 = 0 and the equation of line A B is x - 1 2 = y - 3 - 1 = z - 4 3 in 3 D space. Shortest distance between the line of intersection of planes P 1 and P 2 and the line A B is equal to

Question

Let P 1 : 3 y + z + 1 = 0 and P 2 : 2 x - y + 3 z - 7 = 0 and the equation of line A B is x - 1 2 = y - 3 - 1 = z - 4 3 in 3 D space. Shortest distance between the line of intersection of planes P 1 and P 2 and the line A B is equal to, 7 10 units, 7 2 5 units, 6 10 units, 2 2 5 units

MARKS App · Accepted Answer

Plane P 1 is parallel to line A B ⇒ Shortest distance = Perpendicular distance of any point on line to plane P 1 = 0 = 0 + 9 + 4 + 1 9 + 1 = 14 10 = 7 2 5 units

Search any question & find its solution

Looking for more such questions to practice?