1)Write all vectors perpendicular to both a = 3i – k and b = 2i + j +2k.
2). Given the planes 5x-2y+z=-2 and x+y-4z=1 , find an equation (of any type) for their line of intersection.
Solution
1) any vector parallel to a (cross product) b will be perpendicular to both a and b (3i-k) * (2i+j+2k) = 3k -6j -2j -i = -i-8j+3k so general form is t(-i-8j+3k) where t is any real number. 2) The planes are obviously not parallel to each other. any vector on a plane is perpendicular to it\'s normal. so the line of intersection of two planes is perpendicular to both the normals. so the line is parallel to (5i-2j+k) cross product ( i+j-4k). now, we have to find a common point of intersection. Then the line of intersection is passing through that point and parallel to the cross product of the both normals. point of intersection can be caluclated by eliminating one variable from 5x-2y+z=-2 and x+y-4z=1 (2x+2y-8z=2) add both the eqns 7x-7z=0=> x=z . so consider x=z=0. then y=1 so (0,1,0) is a point common to both the planes.. and direction of the line is parallel to the cross product of the both normals.
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Basic Civil Engineering first year Notes- Chapter 4 Building.pptx
1)Write all vectors perpendicular to both a - 3i k and b - 2i + j +2k.docx
1. 1)Write all vectors perpendicular to both a = 3i – k and b = 2i + j +2k.
2). Given the planes 5x-2y+z=-2 and x+y-4z=1 , find an equation (of any type) for their line of
intersection.
Solution
1) any vector parallel to a (cross product) b will be perpendicular to both a and b (3i-k) *
(2i+j+2k) = 3k -6j -2j -i = -i-8j+3k so general form is t(-i-8j+3k) where t is any real number. 2)
The planes are obviously not parallel to each other. any vector on a plane is perpendicular to it's
normal. so the line of intersection of two planes is perpendicular to both the normals. so the line
is parallel to (5i-2j+k) cross product ( i+j-4k). now, we have to find a common point of
intersection. Then the line of intersection is passing through that point and parallel to the cross
product of the both normals. point of intersection can be caluclated by eliminating one variable
from 5x-2y+z=-2 and x+y-4z=1 (2x+2y-8z=2) add both the eqns 7x-7z=0=> x=z . so consider
x=z=0. then y=1 so (0,1,0) is a point common to both the planes.. and direction of the line is
parallel to the cross product of the both normals.
