No idea. What is line 1? 3?
To be ⊥, v1 • v2 = 0
Determine the equation of line 3 through [1, 4 , -3] perpendicular to line 1.
To be ⊥, v1 • v2 = 0
that is, their dot product is zero
One such vector is [-6,1,4]
So the equation could be [x,y,z] = (1,4,-3) + k[-6,1,4]
note that [-6,1,4] dot [1,2,4] = 0
There would be an infinite number of those lines.
I don't know what your reference to "line 3" is, was there more to this question?
Thanks for the help and yes there was another part but I got that already
The direction vector of line 1 is [1, 2, 4], which means any vector parallel to line 1 is a scalar multiple of [1, 2, 4].
To find a vector perpendicular to [1, 2, 4], we can take the cross product of [1, 2, 4] with any other vector. Let's choose the vector [a, b, c] as an example.
The cross product of two vectors, [x1, y1, z1] and [x2, y2, z2], is given by the following formula:
[x1, y1, z1] x [x2, y2, z2] = [y1*z2 - y2*z1, z1*x2 - z2*x1, x1*y2 - x2*y1]
Substituting [1, 2, 4] and [a, b, c] into the formula, we get:
[1, 2, 4] x [a, b, c] = [2c - 4b, 4a - 4c, 2b - a]
Since the perpendicular line must pass through the point [1, 4, -3], we can determine the values of a, b, and c by substituting these coordinates into the equation:
2c - 4b = 1
4a - 4c = 4
2b - a = -3
Solving this system of equations will give us the values of a, b, and c.
Once we have the direction vector [a, b, c], we can write the equation of the new line as:
[x, y, z] = [1, 4, -3] + k[a, b, c]
Substituting the values of a, b, and c into the equation above will give us the final equation of the line perpendicular to line 1 and passing through the point [1, 4, -3].