The force 𝐅 equals three 𝐢 plus 𝑚𝐣 is acting at the point 𝐴 negative five, negative four in parallel to 𝐵𝐷, where the coordinates of 𝐵 and 𝐷 are five, six and nine, three, respectively. Determine the distance between the point 𝐵 and the line of action of 𝐅.
Let’s begin with a diagram to model the scenario. We have the points 𝐴 negative five, negative four; 𝐵 five, six; and 𝐷 nine, three. The force 𝐅 is acting in parallel to the line 𝐵𝐷. Since the 𝐢-component of 𝐅 is positive three, its direction must be to the right and downwards. Now, we need to determine its exact direction by calculating the direction of the line 𝐵𝐷. 𝐵𝐷 is given by the position vector of the point 𝐷 nine, three minus the position vector of the point 𝐵 five, six. This comes to four, negative three.
𝐅 is a vector which is parallel to this vector. So it is some scalar multiple, let’s call it 𝜇, of four, negative three. We already know that the 𝐢-component of 𝐅 is three. Therefore, four times 𝜇 is equal to three. And rearranging for 𝜇 gives us 𝜇 equals three over four. This gives us back the 𝐢-component of 𝐅 equal to three and the 𝐣-component three over four multiplied by negative three, which is negative nine over four. So the value of 𝑚 is negative nine over four.
Now, we need to determine the distance between the point 𝐵 and the line of action of 𝐅. This is equivalent to setting 𝐵 as a pivot point and finding the perpendicular distance 𝐿 between 𝐵 and the line of action of 𝐅. Recall that the magnitude of a moment 𝐌 of a force 𝐅 about a point is equal to the magnitude of 𝐅 multiplied by its perpendicular distance from the pivot point. And rearranging for 𝐿 gives us 𝐿 equals the magnitude of 𝐌 over the magnitude of 𝐅. We already have the vector 𝐅. So calculating its magnitude will be simple. But we need to find the moment of 𝐅 about the point 𝐵.
Recall that the moment 𝐌 of a force 𝐅 is equal to 𝐫 cross 𝐅, where 𝐫 is the vector from the pivot point to the point of action of 𝐅. In this case, 𝐫 is equal to the vector 𝐁 to 𝐀. 𝐁 to 𝐀 is equal to the position vector of 𝐴 negative five, negative four minus the position vector of 𝐵 five, six, which comes to negative 10, negative 10. 𝐫 cross 𝐅 is therefore equal to the determinant of the three-by-three matrix 𝐢, 𝐣, 𝐤, negative 10, negative 10, zero, three, negative nine over four, zero.
Both vectors lie in the 𝑥𝑦-plane and therefore have a 𝐤-component of zero. When taking their cross product, only the 𝐤-component will be nonzero. Taking this determinant by expanding along the top row gives us 90 over four minus negative 30 times 𝐤, which comes to 105 over two 𝐤. The magnitude of 𝐌 is therefore simply 105 over two. 𝐅 is equal to three, negative nine over four. So the magnitude of 𝐅 is equal to the square root of three squared plus negative nine over four all squared. The distance 𝐿 is then equal to 105 over two all over the square root of nine plus 81 over 16.
This simplifies and gives us our final answer. The distance between the point 𝐵 and the line of action of 𝐅, 𝐿, equals 14 length units.