If a i j 4k and b i j 4k a times b Find the area of tr

If a = i + j + 4k and b = i + j + 4k a times b= Find the area of triangle with vertices (0, 0, 0), (5, 0, -3), (s, 1, -2) The plane that passes through the point (-5, -5, 1) and perpendicular to both y + 2z = -10 and 2x - y = -5 has? It\'s implicit equation Find the vector equation for the line of intersection of planes 5x + 3y - 4z = 3 and 5x + 3z = -1 r = + t planes passes through the 3 points (-5, 6, 4), (-9, 5, 1) and (-9, 4, 3). Find normal vector Find the angle of intersection of the plane. -3x - y - 2z = -4 -3x + 5y - 4z = 5 radians degree

Solution

1) a = i +j +4k ; b = i +j +4k

axb = 0 as they are same vectors and the angle between the two is 0 degrees.

2) A= (6,0,0); B = ( 5,0, -3) and C =( 5, 1, -2)

vector AB = ( -1 , 0 , -3) ; vector AC = ( -1 , 1 , -2)

||AB|| = sqrt(10) ; ||AC|| = sqrt(6)

AB.AC = 1 +6 = 7

Angle between AB and AC : cos(theta) = AB.AC /||AB||.||AC||

= 7/sqrt60

sin(theta) = sqrt( 1 - cos^2theta ) = sqrt( 1 - 49/60) = sqrt(11/60)

Area of traingle = ||AB||.||AC||sin(theta)/2

= sqrt(60)*sqrt(11/60)/2

= sqrt(11)/2