53.2k views
3 votes
Find an equation of the plane through the origin which is parallel to the plane x − 6y + 4z = 7.

1 Answer

3 votes
Linear equations in three-dimensional space is characterized by the standard formula: ax + by + cz + d = 0, where a, b and c are coefficients and d is a constant. Three-dimensional space is different than two-dimensional because you don't base it on the slope of the line. The basis of two lines being parallel in three-dimensional space is the determinant of the matrix being zero. The a coefficients are in the first column, b in the second, c in the third and the constant in the fourth column. From the given equation, a = 1, b = --6, c = 4 and d = 7. Then, let the equation parallel to this would be unknowns x, y, z and k.

1 -6 4 7
x y z k

To find the determinant on the matrix, cross multiply the diagonals. Find the sum of the diagonal facing down first, then subtract to this the sum of all the cross-multiplied terms of diagonals up.

(y + -6z + 4k) - (-6x + 4y + 7z) = 0
y - 6z + 4k + 6x + -4y + 7z = 0
6x -5y -13z +4k = 0

This is the general form of the parallel line. Since it passes the origin, let's substitute the coordinates (0,0,0) to the x,y and z variables, respectively. Then, we can determine the value of k.

6(0) - 5(0) - 13(0) + 4k = 0
k = 0

Therefore, the expression for the parallel line is: 6x - 5y - 13z = 0.
User Pradip Das
by
7.5k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories