# Determine if the three planes intersect and a single

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Determine if the three planes intersect and a single point by finding the value of ´ n 1 ( ´ n 2 × ´ n 3 ) : [2]
2 x 3 y 5 = 0 [3] Eliminate x by using equations [2] and [3]: [2]x2 2 x + 4 y + 6 z = 0 [4] Subtract equation [3] from equation [4]: 2 x + 4 y + 6 z = 0 [4] −( 2 x 3 y 5 )= 0 [3] 7 y + 6 z + 5 = 0 [5] Eliminate x by using equations [1] and [2]: [2]x4 4 x + 8 y + 12 z = 0 [6] Subtract [1] from [6]: 4 x + 8 y + 12 z = 0 [ 6 ] −( 4 x + y 9 z )= 0 [1] 7 y + 21 z = 0 y =− 3 z Substitute y =− 3 z into equation [2]: 7 ( 3 z ) + 6 z + 5 = 0 15 z + 5 = 0 z = 1 3 Substitute z = 1 3 into y =− 3 z y =− 3 ( 1 3 ) y =− 1 Substitute y =− 1 and z = 1 3 into [2] x + 2 (− 1 )+ 3 ( 1 3 )= 0 x 2 + 1 = 0 x = 1 Therefore, the point of intersection for the three planes is ( 1, 1, 1 3 ) . 17. Find the shortest distance from P ( 4,2,6 ) to the plane 2 x 3 y + z 8 = 0
The coordinates of the normal to the plane are ´ n = ( 2, 3,1 ) .To find another direction vector from point P to the plane, find another point on the plane. Call this point R. To find R, choose any value for two of the variables in the equation and then solve for the third. Let x=0 and y=0 2 ( 0 ) 3 ( 0 ) + z 8 = 0 z = 8 Therefore, another point on the plane is R(0,0,8). Now, find the direction of PR ´ PR = R P = ( 0,0,8 ) ( 4,2,6 ) ¿ ( 4, 2,2 ) Now, use the projection of PR on ´ n to find the distance between P and the plane: | Pro j ´ n ´ PR | = | ´ PR . ´ n ´ n. ´ n | | ´ n | ¿ | ( 4, 2,2 ) . ( 2, 3,1 ) ( 2, 3,1 ) ( 2, 3,1 ) | | 2, 3,1 | ¿ | 4 ( 2 ) + ( 2 ) ( 3 ) + 2 ( 1 ) 2 ( 2 ) + ( 3 ) ( 3 ) + 1 ( 1 ) | 2 2 + ( 3 ) 2 + 1 2 ¿ | 8 + 6 + 2 24 + 9 + 1 | 4 + 9 + 1 ¿ 16 14 14 ¿ 4.28 units Therefore, the shortest distance between P and the plane is 4.28 units.