Which of the following is the length of the perpendicular drawn from point five, three, negative four to the straight line 𝐫 equals two, negative three, negative four plus 𝑡 times two, three, six rounded to the nearest hundredth? (A) 5.76, (B) 5.77, (C) 0.82, (D) 0.17, (E) 6.77.
Okay, so here we have a point and a straight line. And we want to figure out the length of the perpendicular, we’ll call it 𝑑, from this point to the line. We’ll start by recalling this relationship, which tells us the distance between a point in space and a straight line. To solve for this distance, we’ll need to know a vector that goes from our point in space to a point, any point, on the line. And we’ll also need to know a vector that runs parallel to the line.
Drawn on our sketch, the vector 𝐒 could look like this. As far as the vector 𝐏𝐋, if we pick some point on our line, say here, and say that that is a known point, then the vector 𝐏𝐋 will look like this. Let’s say then that our known point in space is 𝑃 and our unknown point on the line is 𝐿. To be able to solve for the distance 𝑑, we’ll need to know the coordinates of 𝐿 and the components of 𝑆.
It turns out that we can get both of these bits of information from the equation of our line. Since this equation is given in what’s called vector form, that means this first vector goes from the origin of a coordinate frame to the point two, negative three, negative four on the line. This, then, is our point 𝐿. The next vector in this equation two, three, six is one that runs parallel to this line. And so we can call this Vector 𝐒.
Now that we know 𝑃, 𝐿, and 𝐒, our next step can be to calculate this vector 𝐏𝐋. As we’ve seen 𝐏𝐋 is a vector, and we find it by subtracting the coordinates of point 𝑃 from those of point 𝐿. Our vector 𝐏𝐋 then will have components two minus five, negative three minus three, negative four minus negative four. This is equal to a vector with components negative three, negative six, zero. Knowing this, we’re now ready to calculate the cross product of our 𝐏𝐋 and 𝐒 vectors.
𝐏𝐋 cross 𝐒 equals the determinant of this matrix. In the first row are our three unit vectors, and then below them are the 𝑥- and then the 𝑦- and 𝑧-components of our 𝐏𝐋 and 𝐒 vectors, respectively. The 𝐢-component of this cross product equals the determinant of this two-by-two matrix. Negative six times six minus zero times three is negative 36. Then for the 𝐣-component, negative the determinant of this matrix, we get negative three times six minus zero times two or negative 18. Finally, the 𝐤-component equal to the determinant of this two-by-two matrix, negative three times three minus negative six times two all comes out to three.
This, then, is our complete cross product. Note that we can write this as a vector with components negative 36, 18, three. Okay, we’re now ready to calculate 𝑑 by computing the magnitude of 𝐏𝐋 cross 𝐒 and dividing that by the magnitude of 𝐒. We can write the magnitude of 𝐏𝐋 cross 𝐒 as the square root of negative 36 squared plus 18 squared plus three squared. And then in the denominator, the magnitude of 𝐒 equals the square root of two squared plus three squared plus six squared. Entering this whole expression on our calculator, the answer we get is 5.76583 and so on and so forth.
Now one of our answer options is 5.76, while another is 5.77. At this point, looking back at our question statement, we see we want to round our answer to the nearest hundredth. In our numerical result then, we see that the six is in the hundreds place. The digit immediately to the right of this is greater than or equal to five. This means when we round this whole answer to the nearest hundredth, we will round the six up to a seven. Our rounded answer then is 5.77. This is given as option (B), and it’s the length of the perpendicular from a given point to the straight line.